Number series: definitions, properties, signs of convergence, examples, solutions. Number series: definitions, properties, signs of convergence, examples, solutions Series for the d'Alembert sign

Jean Leron d'Alembert was a famous French mathematician of the 18th century. In general, d'Alembert specialized in differential equations and, based on his research, worked on ballistics so that His Majesty’s cannonballs would fly better. At the same time, I didn’t forget about the number series; it was not for nothing that the ranks of Napoleon’s troops later converged and diverged so clearly.

Before formulating the sign itself, let's consider an important question:
When should D'Alembert's convergence test be used?

Let's start with a review first. Let's remember the cases when you need to use the most popular limit of comparison. The limiting criterion for comparison is applied when in the general term of the series:
1) The denominator contains a polynomial.
2) Polynomials are in both the numerator and denominator.
3) One or both polynomials can be under the root.

The main prerequisites for the application of d'Alembert's test are as follows:

1) The common term of the series (“stuffing” of the series) includes some number to a degree, for example, , and so on. Moreover, it doesn’t matter at all where this thing is located, in the numerator or in the denominator - what matters is that it is present there.

2) The common term of the series includes the factorial. What is factorial? Nothing complicated, factorial is just a condensed representation of the product:

…

…

! When using d'Alembert's test, we will have to describe the factorial in detail. As in the previous paragraph, the factorial can be located at the top or bottom of the fraction.

3) If in the general term of the series there is a “chain of factors”, for example, . This case is rare, but! When studying such a series, a mistake is often made - see Example 6.

Along with powers and/or factorials, polynomials are often found in the filling of a series; this does not change the situation - you need to use D'Alembert's sign.

In addition, in a common term of a series both a degree and a factorial can occur simultaneously; there may be two factorials, two degrees, it is important that there be at least something from the points considered - and this is precisely the prerequisite for using the D'Alembert sign.

D'Alembert's sign: Let's consider positive number series. If there is a limit on the ratio of the subsequent term to the previous one: , then:
a) When row converges
b) When row diverges
c) When the sign does not give an answer. You need to use another sign. Most often, one is obtained in the case when they try to apply the d'Alembert test where it is necessary to use the limiting comparison test.

Who still has problems with limits or misunderstandings of limits, refer to the topic Limits. Examples of solutions. Without an understanding of the limit and the ability to reveal uncertainty, unfortunately, one cannot advance further. And now the long-awaited examples.

Example 1
We see that in the general term of the series we have , and this is a sure prerequisite for using d'Alembert's test. First, the full solution and sample design, comments below.

We use d'Alembert's sign:

converges.

(1) We compose the ratio of the next member of the series to the previous one: . From the condition we see that the general term of the series is . In order to get the next member of the series it is necessary instead of substituting: .
(2) We get rid of the four-story fraction. If you have some experience with the solution, you can skip this step.
(3) Open the parentheses in the numerator. In the denominator we take the four out of the power.
(4) Reduce by . We take the constant beyond the limit sign. In the numerator we present similar terms in parentheses.
(5) Uncertainty is eliminated in the standard way - by dividing the numerator and denominator by “en” to the highest power.
(6) We divide the numerators term by term by the denominators, and indicate the terms that tend to zero.
(7) We simplify the answer and make a note that with the conclusion that, according to D’Alembert’s criterion, the series under study converges.

In the example considered, in the general term of the series we encountered a polynomial of the 2nd degree. What to do if there is a polynomial of the 3rd, 4th or higher degree? The fact is that if a polynomial of a higher degree is given, then difficulties will arise with opening the brackets. In this case, you can use the “turbo” solution method.

Example 2 Let's take a similar series and examine it for convergence
First the complete solution, then comments:

We use d'Alembert's sign:

Thus, the series under study converges.

(1) We create the relation .
(2) We get rid of the four-story fraction.
(3) Consider the expression in the numerator and the expression in the denominator. We see that in the numerator we need to open the brackets and raise them to the fourth power: , which we absolutely don’t want to do. In addition, for those who are not familiar with Newton's binomial, this task may not be feasible at all. Let's analyze the higher degrees: if we open the brackets at the top, we get the highest degree. Below we have the same senior degree: . By analogy with the previous example, it is obvious that when dividing the numerator and denominator term by term, we end up with one in the limit. Or, as mathematicians say, polynomials and - same order of growth. Thus, it is quite possible to outline the ratio with a simple pencil and immediately indicate that this thing tends to one. We deal with the second pair of polynomials in the same way: and , they too same order of growth, and their ratio tends to unity.

In fact, such a “hack” could have been pulled off in Example No. 1, but for a polynomial of the 2nd degree such a solution still looks somehow undignified. Personally, I do this: if there is a polynomial (or polynomials) of the first or second degree, I use the “long” way to solve Example 1. If I come across a polynomial of the 3rd or more high degrees, I use the “turbo” method similar to Example 2.

Example 3 .

Let's look at typical examples with factorials:

Example 4 Examine the series for convergence

The common term of the series includes both the degree and the factorial. It is clear as day that d'Alembert's sign must be used here. Let's decide.

Thus, the series under study diverges.

(1) We create the relation . We repeat again. By condition, the common member of the series is: . In order to get the next term in the series, instead you need to substitute, Thus: .
(2) We get rid of the four-story fraction.
(3) Pinch off the seven from the degree. We describe factorials in detail. How to do this - see the beginning of the lesson.
(4) We cut everything that can be cut.
(5) We move the constant beyond the limit sign. Open the parentheses in the numerator.
(6) We eliminate uncertainty in the standard way - by dividing the numerator and denominator by “en” to the highest power.

Example 5 Examine the series for convergence. The complete solution is below.

Example 6 Examine the series for convergence

Sometimes there are series that contain a “chain” of factors in their filling; we have not yet considered this type of series. How to study a series with a “chain” of factors? Use d'Alembert's sign. But first, to understand what is happening, let’s describe the series in detail:

From the expansion we see that each next member of the series has an additional factor added to the denominator, therefore, if the common member of the series is , then the next member of the series is:
. This is where they often automatically make a mistake, formally writing according to the algorithm that

An approximate sample solution might look like this: Let's use D'Alembert's sign:
Thus, the series under study converges.
RADICAL CAUCHY SIGN

Augustin Louis Cauchy is an even more famous French mathematician. Any student can tell you Cauchy's biography. technical specialty. In the most picturesque colors. It is no coincidence that this name is carved on the first floor of the Eiffel Tower.

Cauchy's convergence test for positive number series is somewhat similar to D'Alembert's test just discussed.

Radical Cauchy's sign: Let's consider positive number series. If there is a limit: , then:
a) When row converges. In particular, the series converges at .
b) When row diverges. In particular, the series diverges at .
c) When the sign does not give an answer. You need to use another sign. It is interesting to note that if Cauchy's test does not give us an answer to the question of the convergence of a series, then D'Alembert's test will not give us an answer either. But if d’Alembert’s test does not give an answer, then Cauchy’s test may well “work.” That is, the Cauchy sign is in this sense a stronger sign.

When should you use the radical Cauchy sign? The radical Cauchy test is usually used in cases where the common term of the series FULLY is in the degree depending on "en". Or when the root “good” is extracted from a common member of the series. There are also exotic cases, but we won’t worry about them.

Example 7 Examine the series for convergence

We see that the general term of the series is completely under a power depending on , which means we need to use the radical Cauchy test:

Thus, the series under study diverges.

(1) We formulate the common term of the series under the root.
(2) We rewrite the same thing, only without the root, using the property of degrees.
(3) In the indicator, we divide the numerator by the denominator term by term, indicating that
(4) As a result, we have uncertainty. This is where you could go the long way: cube, cube, then divide the numerator and denominator by “en” to the highest power. But in in this case There is a more effective solution: you can divide the numerator and denominator term by term directly under the constant power. To eliminate uncertainty, divide the numerator and denominator by (the highest power).
(5) We actually perform term-by-term division and indicate the terms that tend to zero.
(6) We bring the answer to mind, mark what we have and conclude that the series diverges.

Here's a simpler example for independent decision:

Example 8 Examine the series for convergence

And a couple more typical examples.

The full solution and sample design is below.

Example 9 Examine the series for convergence
We use the radical Cauchy test:

Thus, the series under study converges.

(1) Place the common term of the series under the root.
(2) We rewrite the same thing, but without the root, while opening the brackets using the abbreviated multiplication formula: .
(3) In the indicator, we divide the numerator by the denominator term by term and indicate that .
(4) An uncertainty of the form . Here you can directly divide the numerator by the denominator in parentheses by “en” to the highest degree. We encountered something similar when studying second wonderful limit. But here the situation is different. If the coefficients at higher powers were identical, for example: , then the trick with term-by-term division would no longer work, and it would be necessary to use the second remarkable limit. But we have these coefficients different(5 and 6), therefore it is possible (and necessary) to divide term by term (by the way, on the contrary - the second remarkable limit for different coefficients at higher powers no longer work).
(5) We actually perform term-by-term division and indicate which terms tend to zero.
(6) The uncertainty has been eliminated, the simplest limit remains: Why in infinitely large tends to zero? Because the base of the degree satisfies the inequality. If anyone has doubts about the fairness of the limit, then I won’t be lazy, I’ll pick up a calculator:
If , then
If , then
If , then
If , then
If , then
… etc. to infinity - that is, in the limit:
(7) We indicate that we conclude that the series converges.

Example 10 Examine the series for convergence

This is an example for you to solve on your own.

Sometimes a provocative example is offered for a solution, for example:. Here in exponent no "en", only a constant. Here you need to square the numerator and denominator (you get polynomials), and then follow the algorithm from the article Rows for dummies. In such an example, either the necessary test for convergence of the series or the limiting test for comparison should work.
INTEGRAL CAUCHY SIGN

I will disappoint those who did not understand the first course material well. In order to apply the Cauchy integral test, you must be more or less confident in finding derivatives, integrals, and also have the skill of calculation improper integral first kind. In textbooks on mathematical analysis The integral Cauchy test is given mathematically strictly; let’s formulate the test in a very primitive, but understandable way. And immediately examples for clarification.

Integral Cauchy test: Let's consider positive number series. Does this series converge or diverge?

Example 11 Examine the series for convergence

Almost a classic. Natural logarithm and some bullshit.

The main prerequisite for using the Cauchy integral test is is the fact that in the general term of the series there is a certain function and its derivative. From topic Derivative you probably remember the simplest table thing: , and we have just such a canonical case.

How to use the integral attribute? First, we take the integral icon and rewrite the upper and lower limits from the “counter” of the series: . Then, under the integral, we rewrite the “filling” of the series with the letter “he”: . Something is missing..., oh, yes, you also need to stick a differential icon in the numerator: .

Now we need to calculate improper integral. In this case, two cases are possible:

1) If it turns out that the integral converges, then our series will also converge.

2) If it turns out that the integral diverges, then our series will also diverge.

I repeat, if the material is neglected, then reading the paragraph will be difficult and unclear, since the use of a feature essentially comes down to calculating improper integral first kind.

The complete solution and example format should look something like this:

We use the integral sign:

Thus, the series under study diverges together with the corresponding improper integral.

Example 12 Examine the series for convergence

Solution and sample design at the end of the lesson

In the examples considered, the logarithm could also be under the root; this would not change the solution method.

And two more examples for starters

Example 13 Examine the series for convergence

According to the general “parameters,” the general term of the series seems to be suitable for using the limiting criterion for comparison. You just need to open the brackets and immediately submit to the candidate to completely compare this series with a convergent series. However, I was cheating a little, the parentheses may not be opened, but still the solution through the limiting comparison criterion will look rather pretentious.

Therefore, we use the integral Cauchy test:

The integrand function is continuous on

converges together with the corresponding improper integral.

! Note:the resulting number isis not sum of the series!!!

Example 14 Examine the series for convergence

The solution and sample design are at the end of the section that comes to an end.

In order to completely and irrevocably master the topic of number series, visit the topics.

Solutions and answers:

Example 3:We use d'Alembert's sign:

Thus, the series under study diverges.
Note: It was also possible to use the “turbo” solution method: immediately circle the ratio with a pencil, indicate that it tends to unity and make a note: “of the same order of growth.”

Example 5: We use d'Alembert's sign: Thus, the series under study converges.

Example 8:

Thus, the series under study converges.

Example 10:
We use the radical Cauchy test.

Thus, the series under study diverges.
Note: Here the base is degree, so

Example 12: We use an integral sign.

A finite number is obtained, which means the series under study converges

Example 14: We use the integral sign
The integrand is continuous on .

Thus, the series under study diverges together with the corresponding improper integral.
Note: A series can also be examined usinglimiting criterion for comparison . To do this, you need to open the brackets under the root and compare the series under study with the divergent series.

Alternating rows. Leibniz's sign. Examples of solutions

In order to understand the examples of this lesson, you need to have a good understanding of positive number series: understand what a series is, know the necessary sign for the convergence of a series, be able to apply comparison tests, d'Alembert's test, Cauchy's test. The topic can be raised almost from scratch by consistently studying the articles Rows for dummies And D'Alembert's sign. Cauchy's signs. Logically, this lesson is the third in a row, and it will allow you not only to understand the alternating rows, but also to consolidate the material already covered! There will be little novelty, and mastering the alternating rows will not be difficult. Everything is simple and accessible.

What is an alternating series? This is clear or almost clear from the name itself. Just a simple example. Let’s look at the series and describe it in more detail:

And now there will be a killer comment. The members of an alternating series have alternating signs: plus, minus, plus, minus, plus, minus, etc. to infinity.
Alignment provides a multiplier: if even, there will be a plus sign, if odd, there will be a minus sign. In mathematical jargon, this thing is called a “flasher.” Thus, an alternating series is “identified” by minus one to the degree “en”.

In practical examples, the alternation of the terms of the series can be provided not only by the multiplier, but also by its siblings: , , , …. For example:

The pitfall is “deceptions”: , , etc. - such multipliers do not provide sign change. It is absolutely clear that for any natural: , , . Rows with deceptions are slipped not only to especially gifted students, they arise from time to time “by themselves” during the solution functional series.

How to examine an alternating series for convergence? Use Leibniz's test. I don’t want to say anything about the German giant of thought Gottfried Wilhelm Leibniz, since in addition to his mathematical works, he wrote several volumes on philosophy. Dangerous for the brain.

Leibniz's test: If the members of an alternating series monotonously decrease in modulus, then the series converges. Or in two points:

2) The terms of the series decrease in absolute value: . Moreover, they decrease monotonically.

If completed both conditions, then the series converges.

Brief information about the module is given in the manualHot formulas school course mathematicians , but for convenience once again:

What does “modulo” mean? The module, as we remember from school, “eats” the minus sign. Let's return to the row. Mentally erase all the signs with an eraser and let's look at the numbers. We will see that every next series member less than the previous one. Thus, the following phrases mean the same thing:

– Members of the series regardless of sign are decreasing.
– Members of the series decrease modulo.
– Members of the series decrease in absolute value.
– Module the common term of the series tends to zero: End of help

Now let's talk a little about monotony. Monotony is boring consistency.

Members of the series strictly monotonous decrease in modulus if EVERY NEXT member of the series modulo LESS than previous: . The series has a strict monotonicity of decrease; it can be described in detail:

Or we can say briefly: each next member of the series modulo less than the previous one: .

Members of the series not strictly monotonous decrease in modulo if EACH FOLLOWING member of the series modulo is NOT GREATER than the previous one: . Let's consider a series with a factorial: There is a loose monotonicity here, since the first two terms of the series are identical in modulus. That is, each next member of the series modulo no more than the previous one: .

Under the conditions of Leibniz's theorem, decreasing monotonicity must be satisfied (it does not matter whether it is strict or non-strict). In this case, members of the series can even increase in modulus for some time, but the “tail” of the series must necessarily be monotonically decreasing. There is no need to be afraid of what I have piled up; practical examples will put everything in its place:

Example 1 Examine the series for convergence

The common term of the series includes a factor, which means that you need to use the Leibniz criterion

1) Checking the row for alternation. Usually at this point in the decision the series is described in detail and the verdict “The series is alternating” is pronounced.

2) Do the terms of the series decrease in absolute value? It is necessary to solve the limit, which is most often very simple.

– the terms of the series do not decrease in modulus. By the way, there is no longer any need to discuss the monotony of the decrease. Conclusion: the series diverges.

How to figure out what is equal? Very simple. As you know, the module destroys cons, so in order to create one, you just need to remove the flashing light from the roof. In this case, the common term of the series is . We stupidly remove the “flashing light”: .

Example 2 Examine the series for convergence

We use Leibniz's criterion:

1) The series is alternating.

2) – the terms of the series decrease in absolute value. Each next member of the series is less in absolute value than the previous one: thus, the decrease is monotonous.

Conclusion: the series converges.

Everything would be very simple - but this is not the end of the solution!

If a series converges according to Leibniz's test, then it is also said that the series converges conditionally.

If a series composed of modules also converges, then they say that the series converges absolutely.

Therefore, the second stage of solving a typical problem is on the agenda - studying the sign of the alternating series for absolute convergence.

It's not my fault - that's just the theory of number series =)

Let us examine our series for absolute convergence.
Let's compose a series of modules - again we simply remove the factor that ensures the alternation of signs: - diverges (harmonic series).

Thus our series is not absolutely convergent.
Series under study converges only conditionally.

Note that in Example No. 1 there is no need to study non-absolute convergence, since at the first step it was concluded that the series diverges.

We collect buckets, shovels, cars and leave the sandbox to look at the world with wide open eyes from the cabin of my excavator:

Example 3 Examine the series for convergence. We use the Leibniz criterion:

1)
This series is alternating.

2) – the terms of the series decrease in absolute value. Each next member of the series is less in absolute value than the previous one: this means that the decrease is monotonous. Conclusion: The series converges.

Analyzing the filling of the series, we come to the conclusion that here it is necessary to use the limiting criterion for comparison. It is more convenient to open the parentheses in the denominator:

Let's compare this series with a convergent series. We use the limiting criterion for comparison.

A finite number different from zero is obtained, which means that the series converges with the series . Series under study converges absolutely.

Example 4 Examine the series for convergence

Example 5 Examine the series for convergence

These are examples for you to decide on your own. Full solution and sample design at the end of the section.

As you can see, alternating rows are simple and boring! But don’t rush to close the page, in just a couple of screens we will look at a case that baffles many. In the meantime, a couple more examples for practice and repetition.

Example 6 Examine the series for convergence

We use Leibniz's criterion.
1) The series is alternating.
2)
The terms of the series decrease in modulus. Each next member of the series is less in absolute value than the previous one, which means the decrease is monotonous. Conclusion: the series converges.

Please note that I have not described the members of the series in detail. It is always advisable to describe them, but due to irresistible laziness in “difficult” cases, you can limit yourself to the phrase “The series is alternating in sign.” By the way, there is no need to treat this point formally, we always check(at least mentally) that the series actually alternates. A quick glance fails, and a mistake is made automatically. Remember about the “deceptions”, , , if they exist, then you need to get rid of them, getting a “regular” series with positive terms.

The second subtlety concerns the phrase about monotony, which I also shortened as much as possible. You can do this, and almost always your task will be accepted. I’ll say something completely bad - personally, I often keep silent about monotony, and such a number passes. But be prepared to describe everything in detail, down to detailed chains of inequalities (see example at the beginning of the lesson). In addition, sometimes the monotony is not strict, and this also needs to be monitored to replace the word “less” with the word “no more”.

We examine the series for absolute convergence:

Obviously, you need to use the radical Cauchy test:

Thus, the series converges. Series under study converges absolutely.

Example 7 Examine the series for convergence

This is an example for an independent solution. Often there are alternating rows that cause difficulties.

Example 8 Examine the series for convergence

We use Leibniz's criterion:
1) The series is alternating.

The point is that there are no standard, everyday techniques for solving such limits. Where does this limit go? To zero, to infinity? What is important here is WHAT grows faster at infinity– numerator or denominator.

NOTE: the concept of order of growth of a function is covered in detail in the articleMethods for solving limits . We have sequence limits, but this does not change the essence.

If the numerator at grows faster than the factorial, then . If, at infinity, the factorial grows faster than the numerator, then, on the contrary, it will “pull” the limit to zero: . Or maybe this limit is equal to some non-zero number?

Let's try to write down the first few terms of the series:
you can substitute some polynomial of the thousandth degree, this again will not change the situation - sooner or later the factorial will still “overtake” such a terrible polynomial. Factorial more high order growth than any power sequence.

– The factorial is growing faster than product of any quantity exponential and power sequences (our case).

– Any an exponential sequence grows faster than any power sequence, for example: , . Exponential sequence higher order of growth than any power sequence. Similar to the factorial, the exponential sequence “drags” the product of any number of any power sequences or polynomials: .

– Is there anything “cooler” than factorial? Eat! A power exponential sequence (“en” to the power “en”) grows faster than the factorial. In practice it is rare, but the information will not be superfluous. End of help

Thus, the second point of the study (do you still remember this? =)) can be written as follows:
2) , since the growth order is higher than .
The terms of the series decrease in modulus, starting from some number, in this case, each next member of the series is less in absolute value than the previous one, thus the decrease is monotonous.

Conclusion: the series converges.

Here is exactly the curious case when the terms of the series first increase in absolute value, which is why we had an erroneous initial opinion about the limit. But, starting from some number "en", the factorial is overtaken by the numerator, and the “tail” of the series becomes monotonically decreasing, which is fundamentally important for fulfilling the conditions of Leibniz’s theorem. What exactly this “en” equals is quite difficult to find out.

According to the corresponding theorem, from the absolute convergence of the series, the conditional convergence of the series follows. Conclusion: Study series converges absolutely.

And finally, a couple of examples for you to decide for yourself. One from the same opera (re-read the help), but simpler. Another one for gourmets is to consolidate the integral sign of convergence.

Example 9 Examine the series for convergence

Example 10 Examine the series for convergence

After a high-quality study of numerical positive and alternating series, with a clear conscience you can move on to functional series, which are no less monotonous and monotonous are interesting.

Solutions and answers:

Example 4: We use the Leibniz criterion:

1) This series is alternating.
2)
The terms of the series do not decrease in modulus. Conclusion: The series diverges.. , in this case, each next member of the series is less in absolute value than the previous one, thus the decrease is monotonous.

Thus, the series diverges along with the corresponding improper integral. Series under study converges only conditionally.

This article collects and structures the information necessary to solve almost any example on the topic of number series, from finding the sum of a series to examining it for convergence.

Review of the article.

Let's start with the definitions of positive and alternating series and the concept of convergence. Next, we will consider standard series, such as the harmonic series, the generalized harmonic series, and recall the formula for finding the sum of an infinitely decreasing geometric progression. After this, we will move on to the properties of convergent series, dwell on the necessary condition for the convergence of the series and state sufficient criteria for the convergence of the series. We will dilute the theory with solutions to typical examples with detailed explanations.

Page navigation.

Basic definitions and concepts.

Let us have a number sequence where .

Here is an example of a number sequence: .

Number series is the sum of the terms of a numerical sequence of the form .

As an example number series you can give the sum of an infinitely decreasing geometric progression with denominator q = -0.5: .

Called common member of the number series or the kth member of the series.

For the previous example, the general term of the number series has the form .

Partial sum of a number series is a sum of the form , where n is some natural number. also called the nth partial sum of a number series.

For example, the fourth partial sum of the series There is .

Partial amounts form an infinite sequence partial amounts number series.

For our series, the nth partial sum is found using the formula for the sum of the first n terms of a geometric progression , that is, we will have the following sequence of partial sums: .

The number series is called convergent, if there is a finite limit to the sequence of partial sums. If the limit of the sequence of partial sums of a number series does not exist or is infinite, then the series is called divergent.

The sum of a convergent number series is called the limit of the sequence of its partial sums, that is, .

In our example, therefore, the series converges, and its sum is equal to sixteen thirds: .

An example of a divergent series is the sum of a geometric progression with a denominator greater than one: . The nth partial sum is determined by the expression , and the limit of partial sums is infinite: .

Another example of a divergent number series is a sum of the form . In this case, the nth partial sum can be calculated as . The limit of partial sums is infinite .

Sum of the form called harmonic number series.

Sum of the form , where s is some real number, is called generalized by harmonic number series.

The above definitions are sufficient to justify the following very frequently used statements; we recommend that you remember them.

THE HARMONIC SERIES IS DIVERGENT.

Let us prove the divergence of the harmonic series.

Let's assume that the series converges. Then there is a finite limit of its partial sums. In this case, we can write and , which leads us to the equality .

On the other side,

The following inequalities are beyond doubt. Thus, . The resulting inequality indicates to us that the equality cannot be achieved, which contradicts our assumption about the convergence of the harmonic series.

Conclusion: the harmonic series diverges.

THE SUM OF GEOMETRIC PROGRESSION OF THE KIND WITH DENOMINATOR q IS A CONVERGING NUMERIC SERIES IF , AND A DIVERGING SERIES FOR .

Let's prove it.

We know that the sum of the first n terms of a geometric progression is found by the formula .

When fair

which indicates the convergence of the number series.

For q = 1 we have the number series . Its partial sums are found as , and the limit of partial sums is infinite , which indicates the divergence of the series in this case.

If q = -1, then the number series will take the form . Partial sums take value for odd n, and for even n. From this we can conclude that there is no limit on partial sums and the series diverges.

When fair

which indicates the divergence of the number series.

GENERALLY, THE HARMONIC SERIES CONVERGES AT s > 1 AND DIVERGES AT .

Proof.

For s = 1 we obtain a harmonic series, and above we established its divergence.

At s the inequality holds for all natural k. Due to the divergence of the harmonic series, it can be argued that the sequence of its partial sums is unlimited (since there is no finite limit). Then the sequence of partial sums of a number series is all the more unlimited (each member of this series is greater than the corresponding member of the harmonic series); therefore, the generalized harmonic series diverges as s.

It remains to prove the convergence of the series for s > 1.

Let's write down the difference:

Obviously, then

Let us write down the resulting inequality for n = 2, 4, 8, 16, …

Using these results, you can do the following with the original number series:

Expression is the sum of a geometric progression whose denominator is . Since we are considering the case for s > 1, then. That's why
. Thus, the sequence of partial sums of a generalized harmonic series for s > 1 is increasing and at the same time limited from above by the value , therefore, it has a limit, which indicates the convergence of the series. The proof is complete.

The number series is called positive sign, if all its terms are positive, that is, .

The number series is called signalternating, if the signs of its neighboring members are different. An alternating number series can be written as or , Where .

The number series is called alternating sign, if it contains an infinite number of both positive and negative terms.

An alternating number series is a special case of an alternating number series.

Rows

are positive, alternating and alternating, respectively.

For an alternating series, there is the concept of absolute and conditional convergence.

absolutely convergent, if a series of absolute values of its members converges, that is, a positive number series converges.

For example, number series And absolutely converge, since the series converges , which is the sum of an infinitely decreasing geometric progression.

An alternating series is called conditionally convergent, if the series diverges and the series converges.

An example of a conditionally convergent number series is the series . Number series , composed of the absolute values of the terms of the original series, divergent, since it is harmonic. At the same time, the original series is convergent, which is easily established using . Thus, the numerical sign is an alternating series conditionally convergent.

Properties of convergent number series.

Example.

Prove the convergence of the number series.

Solution.

Let's write the series in a different form . The number series converges, since the generalized harmonic series is convergent for s > 1, and due to the second property of convergent number series, the series with the numerical coefficient will also converge.

Example.

Does the number series converge?

Solution.

Let's transform the original series: . Thus, we have obtained the sum of two number series and , and each of them converges (see the previous example). Consequently, by virtue of the third property of convergent number series, the original series also converges.

Example.

Prove the convergence of a number series and calculate its amount.

Solution.

This number series can be represented as the difference of two series:

Each of these series represents the sum of an infinitely decreasing geometric progression and is therefore convergent. The third property of convergent series allows us to assert that the original number series converges. Let's calculate its sum.

The first term of the series is one, and the denominator of the corresponding geometric progression is equal to 0.5, therefore, .

The first term of the series is 3, and the denominator of the corresponding infinitely decreasing geometric progression is 1/3, so .

Let's use the results obtained to find the sum of the original number series:

A necessary condition for the convergence of a series.

If a number series converges, then the limit of its kth term is equal to zero: .

When examining any number series for convergence, the first thing to check is the fulfillment of the necessary convergence condition. Failure to fulfill this condition indicates the divergence of the number series, that is, if , then the series diverges.

On the other hand, you need to understand that this condition is not sufficient. That is, the fulfillment of equality does not indicate the convergence of the number series. For example, for a harmonic series the necessary condition for convergence is satisfied, and the series diverges.

Example.

Examine a number series for convergence.

Solution.

Let's check the necessary condition for the convergence of a number series:

Limit The nth term of the number series is not equal to zero, therefore, the series diverges.

Sufficient signs of convergence of a positive series.

When using sufficient features to study number series for convergence, you constantly encounter problems, so we recommend turning to this section if you have any difficulties.

Necessary and sufficient condition for the convergence of a positive number series.

For the convergence of a positive number series it is necessary and sufficient that the sequence of its partial sums be bounded.

Let's start with the signs of comparing series. Their essence lies in comparing the numerical series under study with a series whose convergence or divergence is known.

The first, second and third signs of comparison.

The first sign of comparison of series.

Let and be two positive number series and the inequality holds for all k = 1, 2, 3, ... Then the convergence of the series implies the convergence, and the divergence of the series implies the divergence of .

The first comparison criterion is used very often and is a very powerful tool for studying number series for convergence. The main problem is selecting a suitable series for comparison. A series for comparison is usually (but not always) chosen so that the exponent of its kth term is equal to the difference between the exponents of the numerator and denominator of the kth term of the numerical series under study. For example, let the difference between the exponents of the numerator and the denominator be equal to 2 – 3 = -1, therefore, for comparison, we select a series with the kth term, that is, a harmonic series. Let's look at a few examples.

Example.

Establish convergence or divergence of a series.

Solution.

Since the limit of the general term of the series is equal to zero, then the necessary condition for the convergence of the series is satisfied.

It is easy to see that the inequality is true for all natural k. We know that the harmonic series is divergent; therefore, by the first criterion of comparison, the original series is also divergent.

Example.

Examine the number series for convergence.

Solution.

Prerequisite convergence of the number series is satisfied, since . The inequality is obvious for any natural value of k. The series converges, since the generalized harmonic series is convergent for s > 1. Thus, the first sign of comparison of series allows us to state the convergence of the original number series.

Example.

Determine the convergence or divergence of a number series.

Solution.

, therefore, the necessary condition for the convergence of the number series is satisfied. Which row should I choose for comparison? A number series suggests itself, and in order to decide on s, we carefully examine the number sequence. The terms of a number sequence increase towards infinity. Thus, starting from some number N (namely, from N = 1619), the terms of this sequence will be greater than 2. Starting from this number N, the inequality is true. A number series converges due to the first property of convergent series, since it is obtained from a convergent series by discarding the first N – 1 terms. Thus, by the first property of comparison, the series is convergent, and by virtue of the first property of convergent number series, the series will also converge.

The second sign of comparison.

Let and be positive number series. If , then the convergence of the series implies the convergence of . If , then the divergence of the number series implies the divergence of .

Consequence.

If and , then the convergence of one series implies the convergence of the other, and the divergence implies divergence.

We examine the series for convergence using the second comparison criterion. As a series we take a convergent series. Let's find the limit of the ratio of the kth terms of the number series:

Thus, according to the second criterion of comparison, from the convergence of a number series, the convergence of the original series follows.

Example.

Examine the convergence of a number series.

Solution.

Let us check the necessary condition for the convergence of the series . The condition is met. To apply the second comparison criterion, let's take the harmonic series. Let's find the limit of the ratio of the kth terms:

Consequently, from the divergence of the harmonic series, the divergence of the original series according to the second criterion of comparison follows.

For information, we present the third criterion for comparing series.

The third sign of comparison.

Let and be positive number series. If the condition is satisfied from some number N, then convergence of the series implies convergence, and divergence of the series implies divergence.

D'Alembert's sign.

Comment.

D'Alembert's test is valid if the limit is infinite, that is, if , then the series converges if , then the series diverges.

If , then d'Alembert's test does not provide information about the convergence or divergence of the series and additional research is required.

Example.

Examine a number series for convergence using d'Alembert's test.

Solution.

Let's check the fulfillment of the necessary condition for the convergence of a number series; calculate the limit using:

The condition is met.

Let's use d'Alembert's sign:

Thus, the series converges.

Radical Cauchy sign.

Let be a positive number series. If , then the number series converges, if , then the series diverges.

Comment.

Cauchy's radical test is valid if the limit is infinite, that is, if , then the series converges if , then the series diverges.

If , then the radical Cauchy test does not provide information about the convergence or divergence of the series and additional research is required.

It is usually fairly easy to discern cases where it is best to use the radical Cauchy test. A typical case is when the general term of a number series is an exponential power expression. Let's look at a few examples.

Example.

Examine a positive number series for convergence using the radical Cauchy test.

Solution.

. Using the radical Cauchy test we obtain .

Therefore, the series converges.

Example.

Does the number series converge? .

Solution.

Let us use the radical Cauchy test , therefore, the number series converges.

Integral Cauchy test.

Let be a positive number series. Let's create a function of continuous argument y = f(x) similar to the function. Let the function y = f(x) be positive, continuous and decreasing on the interval , where ). Then in case of convergence improper integral the number series under study converges. If the improper integral diverges, then the original series also diverges.

When checking the decrease of the function y = f(x) on an interval, the theory from section may be useful to you.

Example.

Examine a number series with positive terms for convergence.

Solution.

The necessary condition for the convergence of the series is satisfied, since . Let's consider the function. It is positive, continuous and decreasing on the interval. The continuity and positivity of this function is beyond doubt, but let us dwell on the decrease in a little more detail. Let's find the derivative:
. It is negative on the interval, therefore, the function decreases on this interval.

Signs of series convergence.
D'Alembert's sign. Cauchy's signs

Work, work - and understanding will come later
J.L. d'Alembert

Congratulations to everyone on the start school year! Today is September 1, and in honor of the holiday, I decided to introduce readers to what you have been looking forward to and eager to know for a long time - signs of convergence of numerical positive series. The First of September holiday and my congratulations are always relevant, it’s okay if it’s actually summer outside, you’re now retaking the exam for the third time, study if you’ve visited this page!

For those who are just starting to study series, I recommend that you first read the article Number series for dummies. Actually, this cart is a continuation of the banquet. So, today in the lesson we will look at examples and solutions on the topics:

One of the common comparison signs that is found in practical examples is the D'Alembert sign. Cauchy's signs are less common, but also very popular. As always, I will try to present the material simply, accessible and understandable. The topic is not the most difficult, and all tasks are to a certain extent standard.

D'Alembert's convergence test

Jean Leron d'Alembert was a famous French mathematician of the 18th century. In general, d’Alembert specialized in differential equations and, based on his research, studied ballistics so that His Majesty’s cannonballs would fly better. At the same time, I didn’t forget about the number series; it was not for nothing that the ranks of Napoleon’s troops later converged and diverged so clearly.

Before formulating the sign itself, let's consider an important question:
When should D'Alembert's convergence test be used?

1) The denominator contains a polynomial.
2) Polynomials are in both the numerator and denominator.
3) One or both polynomials can be under the root.
4) Of course, there can be more polynomials and roots.

The main prerequisites for the application of d'Alembert's test are as follows:

1) The common term of the series (“filling” of the series) includes some number to a degree, for example, , , and so on. Moreover, it doesn’t matter at all where this thing is located, in the numerator or in the denominator - what matters is that it is present there.

2) The common term of the series includes the factorial. We crossed swords with factorials back in the lesson Number sequence and its limit. However, it won’t hurt to spread out the self-assembled tablecloth again:

…

…

! When using d'Alembert's test, we will have to describe the factorial in detail. As in the previous paragraph, the factorial can be located at the top or bottom of the fraction.

3) If in the general term of the series there is a “chain of factors”, for example, . This case is rare, but! When studying such a series, a mistake is often made - see Example 6.

Along with powers and/or factorials, polynomials are often found in the filling of a series; this does not change the situation - you need to use D'Alembert's sign.

D'Alembert's sign: Let's consider positive number series. If there is a limit on the ratio of the subsequent term to the previous one: , then:
a) When row converges
b) When row diverges
c) When the sign does not give an answer. You need to use another sign. Most often, one is obtained in the case when they try to apply the D'Alembert test where it is necessary to use the limiting comparison test.

For those who still have problems with limits or misunderstandings of limits, refer to the lesson Limits. Examples of solutions. Without an understanding of the limit and the ability to reveal uncertainty, unfortunately, one cannot advance further.

And now the long-awaited examples.

Example 1

We see that in the general term of the series we have , and this is a sure prerequisite for using d'Alembert's test. First, the full solution and sample design, comments below.

We use d'Alembert's sign:

converges.
(1) We compose the ratio of the next member of the series to the previous one: . From the condition we see that the general term of the series is . In order to get the next member of the series you need INSTEAD to substitute: .
(2) We get rid of the four-story fraction. If you have some experience with the solution, you can skip this step.
(3) Open the parentheses in the numerator. In the denominator we take the four out of the power.
(4) Reduce by . We take the constant beyond the limit sign. In the numerator we present similar terms in parentheses.
(5) Uncertainty is eliminated in the standard way - by dividing the numerator and denominator by “en” to the highest power.
(6) We divide the numerators term by term by the denominators, and indicate the terms that tend to zero.
(7) We simplify the answer and make a note that with the conclusion that, according to D’Alembert’s criterion, the series under study converges.

Example 2

Let's take a similar series and examine it for convergence

First the complete solution, then comments:

We use d'Alembert's sign:

Thus, the series under study converges.

(1) We create the relation .

(3) Consider the expression in the numerator and the expression in the denominator. We see that in the numerator we need to open the brackets and raise them to the fourth power: , which we absolutely don’t want to do. And for those who are not familiar with Newton's binomial, this task will be even more difficult. Let's analyze the higher degrees: if we open the brackets at the top , then we will get a senior degree. Below we have the same senior degree: . By analogy with the previous example, it is obvious that when dividing the numerator and denominator term by term, we end up with one in the limit. Or, as mathematicians say, polynomials And - same order of growth. Thus, it is quite possible to outline the relation with a simple pencil and immediately indicate that this thing is tending to one. We deal with the second pair of polynomials in the same way: and , they too same order of growth, and their ratio tends to unity.

In fact, such a “hack” could have been pulled off in Example No. 1, but for a polynomial of the 2nd degree such a solution still looks somehow undignified. Personally, I do this: if there is a polynomial (or polynomials) of the first or second degree, I use the “long” method for solving Example 1. If I come across a polynomial of the 3rd or higher degrees, I use the “turbo” method similar to Example 2.

Example 3

Examine the series for convergence

Let's look at typical examples with factorials:

Example 4

Examine the series for convergence

The common term of the series includes both the degree and the factorial. It is clear as day that d'Alembert's sign must be used here. Let's decide.

Thus, the series under study diverges.
(1) We create the relation . We repeat again. By condition, the common term of the series is: . In order to get the next term in the series, instead you need to substitute, Thus: .
(2) We get rid of the four-story fraction.
(3) Pinch off the seven from the degree. We describe factorials in detail. How to do this - see the beginning of the lesson or the article on number sequences.
(4) We cut everything that can be cut.
(5) We move the constant beyond the limit sign. Open the parentheses in the numerator.
(6) We eliminate uncertainty in the standard way - by dividing the numerator and denominator by “en” to the highest power.

Example 5

Examine the series for convergence

Full solution and sample design at the end of the lesson

Example 6

Examine the series for convergence

From the expansion we see that each next member of the series has an additional factor added to the denominator, therefore, if the common member of the series , then the next member of the series:
. This is where they often automatically make a mistake, formally writing according to the algorithm that

A sample solution might look like this:

We use d'Alembert's sign:

Thus, the series under study converges.

Radical Cauchy's sign

Augustin Louis Cauchy is an even more famous French mathematician. Any engineering student can tell you Cauchy's biography. In the most picturesque colors. It is no coincidence that this name is carved on the first floor of the Eiffel Tower.

Cauchy's convergence test for positive number series is somewhat similar to D'Alembert's test just discussed.

Radical Cauchy's sign: Let's consider positive number series. If there is a limit: , then:
a) When row converges. In particular, the series converges at .
b) When row diverges. In particular, the series diverges at .
c) When the sign does not give an answer. You need to use another sign. It is interesting to note that if Cauchy's test does not give us an answer to the question of the convergence of a series, then D'Alembert's test will not give an answer either. But if d’Alembert’s test does not give an answer, then Cauchy’s test may well “work.” That is, the Cauchy sign is in this sense a stronger sign.

When should you use the radical Cauchy sign? The radical Cauchy test is usually used in cases where the root “good” is extracted from a common member of the series. As a rule, this pepper is in a degree which depends on. There are also exotic cases, but we won’t worry about them.

Example 7

Examine the series for convergence

We see that the fraction is completely under a power depending on “en”, which means we need to use the radical Cauchy test:

Thus, the series under study diverges.

(1) We formulate the common term of the series under the root.

(2) We rewrite the same thing, only without the root, using the property of degrees.
(3) In the indicator, we divide the numerator by the denominator term by term, indicating that
(4) As a result, we have uncertainty. Here you could go the long way: cube, cube, then divide the numerator and denominator by “en” cubed. But in this case there is a more effective solution: this technique can be used directly under the constant degree. To eliminate uncertainty, divide the numerator and denominator by (the highest power of the polynomials).

(5) We perform term-by-term division and indicate the terms that tend to zero.
(6) We bring the answer to mind, mark what we have and conclude that the series diverges.

Here is a simpler example for you to solve on your own:

Example 8

Examine the series for convergence

And a couple more typical examples.

Full solution and sample design at the end of the lesson

Example 9

Examine the series for convergence
We use the radical Cauchy test:

Thus, the series under study converges.

(1) Place the common term of the series under the root.

(2) We rewrite the same thing, but without the root, while opening the brackets using the abbreviated multiplication formula: .
(3) In the indicator, we divide the numerator by the denominator term by term and indicate that .
(4) An uncertainty of the form is obtained, and here, too, division can be performed directly under the degree. But with one condition: the coefficients of the higher powers of the polynomials must be different. Ours are different (5 and 6), and therefore it is possible (and necessary) to divide both floors into . If these coefficients are the same, for example (1 and 1): , then such a trick does not work and you need to use second wonderful limit. If you remember, these subtleties were discussed in the last paragraph of the article Methods for solving limits.

(5) We actually perform term-by-term division and indicate which terms tend to zero.
(6) The uncertainty has been eliminated, we are left with the simplest limit: . Why in infinitely large tends to zero? Because the base of the degree satisfies the inequality. If anyone has doubts about the fairness of the limit , then I won’t be lazy, I’ll pick up a calculator:
If , then
If , then
If , then
If , then
If , then
… etc. to infinity - that is, in the limit:

Just like that infinitely decreasing geometric progression on your fingers =)
! Never use this technique as evidence! Because just because something is obvious, that doesn’t mean it’s right.

(7) We indicate that we conclude that the series converges.

Example 10

Examine the series for convergence

This is an example for you to solve on your own.

Integral Cauchy test

Or just an integral sign. I will disappoint those who did not understand the first course material well. In order to apply the Cauchy integral test, you must be more or less confident in finding derivatives, integrals, and also have the skill of calculation improper integral first kind.

In textbooks on mathematical analysis integral Cauchy test given mathematically strictly, but too confusingly, so I will formulate the sign not too strictly, but clearly:

Let's consider positive number series. If there is an improper integral, then the series converges or diverges along with this integral.

And just some examples for clarification:

Example 11

Examine the series for convergence

Almost a classic. Natural logarithm and some bullshit.

The main prerequisite for using the Cauchy integral test is is the fact that the general term of the series contains factors similar to a certain function and its derivative. From topic

D'Alembert's convergence test Radical Cauchy convergence test Integral Cauchy convergence test

Before formulating the sign itself, let's consider an important question:
When should D'Alembert's convergence test be used?

The main prerequisites for the application of d'Alembert's test are as follows:

2) The common term of the series includes the factorial. We crossed swords with factorials back in class Number sequence and its limit. However, it won’t hurt to spread out the self-assembled tablecloth again:

…

…

! When using d'Alembert's test, we will have to describe the factorial in detail. As in the previous paragraph, the factorial can be located at the top or bottom of the fraction.

3) If in the general term of the series there is a “chain of factors”, for example, . This case is rare, but! When studying such a series, a mistake is often made - see Example 6.

Along with powers and/or factorials, polynomials are often found in the filling of a series; this does not change the situation - you need to use D'Alembert's sign.

In addition, in a common term of a series both a degree and a factorial can occur simultaneously; there may be two factorials, two degrees, it is important that there be at least something the considered points - and this is precisely the prerequisite for using the D'Alembert sign.

D'Alembert's sign: Let's consider positive number series. If there is a limit on the ratio of the subsequent term to the previous one: , then:
a) When row converges. In particular, the series converges at .
b) When row diverges. In particular, the series diverges at .
c) When the sign does not give an answer. You need to use another sign. Most often, one is obtained in the case when they try to apply the d'Alembert test where it is necessary to use the limiting comparison test.

And now the long-awaited examples.

Example 1

We see that in the general term of the series we have , and this is a sure prerequisite for using d'Alembert's test. First, the full solution and sample design, comments below.

We use d'Alembert's sign:

converges.

Example 2

Let's take a similar series and examine it for convergence

First the complete solution, then comments:

We use d'Alembert's sign:

Thus, the series under study converges.

In fact, such a “hack” could have been pulled off in Example No. 1, but for a polynomial of the 2nd degree such a solution still looks somehow undignified. Personally, I do this: if there is a polynomial (or polynomials) of the first or second degree, I use the “long” method for solving Example 1. If I come across a polynomial of the 3rd or higher degrees, I use the “turbo” method similar to Example 2.

Example 3

Examine the series for convergence

Complete solution and sample design at the end of the lesson about number sequences.
(4) We cut everything that can be cut.
(5) We move the constant beyond the limit sign. Open the parentheses in the numerator.
(6) We eliminate uncertainty in the standard way - by dividing the numerator and denominator by “en” to the highest power.

Example 5

Examine the series for convergence

Full solution and sample design at the end of the lesson

Example 6

Examine the series for convergence

A sample solution might look like this:

We use d'Alembert's sign:

Thus, the series under study converges.

Before formulating the sign itself, let's consider an important question:
When should D'Alembert's convergence test be used?

The main prerequisites for the application of d'Alembert's test are as follows:

1) The common term of the series (“stuffing” of the series) includes some number to a degree, for example, , and so on. Moreover, it does not matter at all where these functions are located, in the numerator or in the denominator - what matters is that they are present there.

2) The common term of the series includes the factorial. What is factorial?

…

…

! When using d'Alembert's test, we will have to describe the factorial in detail. As in the previous paragraph, the factorial can be located at the top or bottom of the fraction.

3) If in the general term of the series there is a “chain of factors”, for example, . This case is rare.

Along with powers and/or factorials, polynomials are often found in the filling of a series; this does not change the situation - you need to use D'Alembert's sign.

Without an understanding of the limit and the ability to reveal uncertainty, unfortunately, one cannot advance further.

Example:
Solution: We see that in the general term of the series we have , and this is a sure prerequisite for using d'Alembert's test.

We use d'Alembert's sign:

converges.

Radical Cauchy sign.

Cauchy's convergence test for positive number series is somewhat similar to D'Alembert's test just discussed.

! It is interesting to note that if Cauchy's test does not give us an answer to the question of the convergence of a series, then D'Alembert's test will not give us an answer either. But if d’Alembert’s test does not give an answer, then Cauchy’s test may well “work.” That is, the Cauchy sign is in this sense a stronger sign.

!!! When should you use the radical Cauchy sign? The radical Cauchy test is usually used in cases where the common term of the series FULLY is in the degree depending on "en". Or when the root “good” is extracted from a common member of the series. There are also exotic cases, but we won’t worry about them.

Example: Examine the series for convergence

Solution: We see that the general term of the series is completely under a power depending on , which means we need to use the radical Cauchy test:

Thus, the series under study diverges.

Integral Cauchy test.

In order to apply the Cauchy integral test, you must be more or less confident in finding derivatives, integrals, and also have the skill of calculation improper integral first kind.

I will formulate it in my own words (for ease of understanding).

Integral Cauchy test: Let's consider positive number series. This series converges or diverges together with the corresponding improper integral.

! !! The main prerequisite for using the Cauchy integral test is is the fact that in the general term of the series there is a certain function and its derivative.

Example: Examine the series for convergence

Solution: From topic Derivative you probably remember the simplest table thing: , and we have just such a canonical case.

Now we need to calculate the improper integral. In this case, two cases are possible:

1) If it turns out that the integral converges, then our series will also converge.

2) If it turns out that the integral diverges, then our series will also diverge.

We use the integral sign:

The integrand function is continuous on

Thus, the series under study diverges together with the corresponding improper integral.

Example: Investigate the convergence of the series

Solution: first of all, let's check a necessary sign of convergence of a series. This is not a formality, but an excellent chance to deal with the example with “little bloodshed.”

Number sequence higher growth order, than , therefore , that is, the necessary sign of convergence is satisfied, and the series can either converge or diverge.

Thus, you need to use some kind of sign. But which one? Limit of comparison clearly does not fit, since a logarithm has been squeezed into the common term of the series, d'Alembert's and Cauchy's signs also do not lead to results. If we had, then at the very least we could get out through integral feature.

“Inspection of the scene” suggests a divergent series (the case of a generalized harmonic series), but again the question arises, how to take into account the logarithm in the numerator?

What remains is the very first sign of comparison, based on inequalities, which is often not taken into account and collects dust on a distant shelf. Let's describe the series in more detail:

Let me remind you that – unlimitedly growing number sequence:

And, starting from the number, the inequality will be satisfied:

that is, the members of the series will be even more relevant members divergent row.

As a result, the series has no choice but to disperse.

The convergence or divergence of a number series depends on its “infinite tail” (remainder). In our case, we can ignore the fact that the inequality is not true for the first two numbers - this does not affect the conclusion.

The finished example should look something like this:

Let's compare this series with a divergent series.
For all numbers, starting with , the inequality is satisfied, therefore, according to the comparison criterion, the series under study diverges.

Alternating rows. Leibniz's sign. Examples of solutions.

What is an alternating series? This is clear or almost clear from the name itself. Just a simple example.

Let's look at the series and describe it in more detail:

Alignment provides a multiplier: if even, there will be a plus sign, if odd, there will be a minus sign.

In practical examples, the alternation of the terms of the series can be provided not only by the multiplier, but also by its siblings: , , , …. For example:

The pitfall is “deceptions”: , , etc. - such multipliers do not provide sign change. It is absolutely clear that for any natural: , , .

How to examine an alternating series for convergence? Use Leibniz's test.

Leibniz's test: If in an alternating series two conditions are met: 1) the terms of the series monotonically decrease in absolute value. 2) the limit of the common term in modulus is equal to zero, then the series converges, and the modulus of the sum of this series does not exceed the modulus of the first term.

Brief information about the module:

What does “modulo” mean? The module, as we remember from school, “eats” the minus sign. Let's go back to the row . Mentally erase all the signs with an eraser and let's look at the numbers. We will see that every next series member less than the previous one.

Now a little about monotony.

Members of the series strictly monotonous decrease in modulus if EVERY NEXT member of the series modulo LESS than previous: . For a row The strict monotonicity of decreasing is fulfilled; it can be described in detail:

Or we can say briefly: each next member of the series modulo less than the previous one: .

Members of the series not strictly monotonous decrease in modulo if EACH FOLLOWING member of the series modulo is NOT GREATER than the previous one: . Consider a series with factorial: Here there is a loose monotonicity, since the first two terms of the series are identical in modulus. That is, each next member of the series modulo no more than the previous one: .

Example: Examine the series for convergence

Solution: The common term of the series includes a factor, which means that you need to use the Leibniz criterion

1) Checking the series for monotonic decrease.

1<2<3<…, т.е. n+1>n – the first condition is not met

2) – the second condition is also not met.

Conclusion: the series diverges.

Definition: If a series converges according to the Leibniz criterion and a series composed of modules also converges, then they say that the series converges absolutely.

If a series converges according to the Leibniz criterion, and a series composed of modules diverges, then the series is said to be converges conditionally.

If a series composed of modules converges, then this series also converges.

Therefore, an alternating convergent series must be examined for absolute or conditional convergence.

Example:

Solution: We use Leibniz's criterion:

1) Each next member of the series is less in absolute value than the previous one: – the first condition is met.

2) – the second condition is also satisfied.

Conclusion: the series converges.

Let's check for conditional or absolute convergence.

Let's make a series of modules - again we simply remove the multiplier, which ensures sign alternation:
– diverges (harmonic series).

Thus our series is not absolutely convergent.
Series under study converges conditionally.

Example: Examine a series for conditional or absolute convergence

Solution: We use Leibniz's criterion:
1) Let's try to write down the first few terms of the series:

…?!

If the numerator at grows faster than the factorial, then . If, at infinity, the factorial grows faster than the numerator, then it, on the contrary, will “pull” the limit to zero: . Or maybe this limit is equal to some non-zero number? or . Instead, you can substitute some polynomial of the thousandth degree, this again will not change the situation - sooner or later the factorial will still “overtake” such a terrible polynomial. Factorial higher order of growth.

– The factorial is growing faster than product of any quantity exponential and power sequences(our case).

– Any an exponential sequence grows faster than any power sequence, for example: , . Exponential sequence higher order of growth than any power sequence. Similar to the factorial, an exponential sequence “drags” the product of any number of any power sequences or polynomials: .

– Is there anything “stronger” than factorial? Eat! A power exponential sequence (“en” to the power of “en”) grows faster than the factorial. In practice it is rare, but the information will not be superfluous.

End of help

Thus, the second point of the study can be written as follows:
2) , since the growth order is higher than .
The terms of the series decrease in modulus, starting from some number, in this case, each next member of the series is less in absolute value than the previous one, thus the decrease is monotonous.

Conclusion: the series converges.

Here is exactly the curious case when the terms of the series first increase in absolute value, which is why we had an erroneous initial opinion about the limit. But, starting from some number "en", the factorial overtakes the numerator, and the “tail” of the series becomes monotonically decreasing, which is fundamentally important for fulfilling the conditions of Leibniz’s theorem. What exactly this “en” equals is quite difficult to find out..

We examine the series for absolute or conditional convergence:

And here D’Alembert’s sign already works:

We use d'Alembert's sign:

Thus, the series converges.

Series under study converges absolutely.

The analyzed example can be solved in another way (we use a sufficient criterion for the convergence of an alternating series).

A sufficient sign of convergence of an alternating series: If a series composed of the absolute values of the terms of a given series converges, then the given series also converges.

Second way:

Examine a series for conditional or absolute convergence

Solution : We examine the series for absolute convergence:

We use d'Alembert's sign:

Thus, the series converges.
Based on a sufficient criterion for the convergence of an alternating series, the series itself converges.

Conclusion: Study series converges absolutely.

To calculate the sum of a series with a given accuracy We will use the following theorem:

Let the alternating series sign satisfies the conditions of Leibniz's criterion and let - his n th partial amount. Then the series converges and the error in approximate calculation of its sum S in absolute value does not exceed the modulus of the first discarded term: