Why does the series 1 n diverge? Sum of the series. Some partial sum values
There are several ways to check the convergence of a series. First, you can simply find the sum of the series. If as a result we get a finite number, then this series converges. For example, because
then the series converges. If we were unable to find the sum of the series, then we should use other methods to check the convergence of the series.
One such method is d'Alembert's sign
here and respectively are the nth and (n+1)th terms of the series, and the convergence is determined by the value of D: If D< 1 - ряд сходится, если D >
As an example, we study the convergence of a series using d'Alembert's test. First, let's write down expressions for and . Now let's find the corresponding limit:
Since, in accordance with d'Alembert's test, the series converges.
Another method to check the convergence of a series is radical Cauchy's sign, which is written as follows:
here is the nth term of the series, and convergence, as in the case of d'Alembert's test, is determined by the value of D: If D< 1 - ряд сходится, если D >1 - diverges. When D = 1, this sign does not provide an answer and additional research needs to be carried out.
As an example, we study the convergence of a series using the radical Cauchy test. First, let's write down the expression for . Now let's find the corresponding limit:
Since title="15625/64>1"> , in accordance with the radical Cauchy test, the series diverges.
It is worth noting that, along with those listed, there are other signs of convergence of series, such as the integral Cauchy test, Raabe test, etc.
Our online calculator, built on the basis of the Wolfram Alpha system, allows you to test the convergence of the series. Moreover, if the calculator produces a specific number as the sum of a series, then the series converges. Otherwise, you need to pay attention to the “Series convergence test” item. If the phrase “series converges” is present, then the series converges. If the phrase “series diverges” is present, then the series diverges.
Below is a translation of all possible meanings of the “Series Convergence Test” item:
| Text on English language | Text in Russian |
|---|---|
| By the harmonic series test, the series diverges. | When comparing the series under study with the harmonic series, the original series diverges. |
| The ratio test is inclusive. | D'Alembert's test cannot give an answer about the convergence of a series. |
| The root test is inclusive. | The radical Cauchy test cannot give an answer about the convergence of the series. |
| By the comparison test, the series converges. | By comparison, the series converges |
| By the ratio test, the series converges. | According to d'Alembert's test, the series converges |
| By the limit test, the series diverges. | Based on the fact that title="The limit of the nth term of the series for n->oo is not equal to zero or does not exist"> , или указанный предел не существует, сделан вывод о том, что ряд расходится. !} |
Harmonic series- a sum made up of an infinite number of terms, the reciprocals of consecutive numbers natural range :
∑ k = 1 ∞ 1 k = 1 + 1 2 + 1 3 + 1 4 + ⋯ + 1 k + ⋯ (\displaystyle \sum _(k=1)^(\mathcal (\infty ))(\frac (1 )(k))=1+(\frac (1)(2))+(\frac (1)(3))+(\frac (1)(4))+\cdots +(\frac (1) (k))+\cdots ).Encyclopedic YouTube
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✪ Number series. Basic concepts - bezbotvy
✪ Proof of the divergence of the harmonic series
✪ Number series-9. Convergence and divergence of the Dirichlet series
✪ Consultation No. 1. Mat. analysis. Fourier series in the trigonometric system. The simplest properties
✪ RANKS. Review
Subtitles
Sum of the first n terms of the series
Individual members of the series tend to zero, but their sum diverges. nth partial amount s n of a harmonic series is called the nth harmonic number:
s n = ∑ k = 1 n 1 k = 1 + 1 2 + 1 3 + 1 4 + ⋯ + 1 n (\displaystyle s_(n)=\sum _(k=1)^(n)(\frac (1 )(k))=1+(\frac (1)(2))+(\frac (1)(3))+(\frac (1)(4))+\cdots +(\frac (1) (n)))Some partial sum values
| s 1 = 1 s 2 = 3 2 = 1 , 5 s 3 = 11 6 ≈ 1.833 s 4 = 25 12 ≈ 2.083 s 5 = 137 60 ≈ 2.283 (\displaystyle (\begin(matrix)s_(1)&=&1 \\\\s_(2)&=&(\frac (3)(2))&=&1(,)5\\\\s_(3)&=&(\frac (11)(6))& \approx &1(,)833\\\\s_(4)&=&(\frac (25)(12))&\approx &2(,)083\\\\s_(5)&=&(\frac (137)(60))&\approx &2(,)283\end(matrix))) | s 6 = 49 20 = 2.45 s 7 = 363,140 ≈ 2.593 s 8 = 761,280 ≈ 2.718 s 10 3 ≈ 7.484 s 10 6 ≈ 14.393 (\displaystyle (\begin(matrix)s_(6)&=&( \frac (49)(20))&=&2(,)45\\\\s_(7)&=&(\frac (363)(140))&\approx &2(,)593\\\\s_ (8)&=&(\frac (761)(280))&\approx &2(,)718\\\\s_(10^(3))&\approx &7(,)484\\\\s_( 10^(6))&\approx &14(,)393\end(matrix))) |
Euler's formula
When value ε n → 0 (\displaystyle \varepsilon _(n)\rightarrow 0), therefore, for large n (\displaystyle n):
s n ≈ ln (n) + γ (\displaystyle s_(n)\approx \ln(n)+\gamma )- Euler's formula for the sum of the first n (\displaystyle n) members of the harmonic series.| n (\displaystyle n) | s n = ∑ k = 1 n 1 k (\displaystyle s_(n)=\sum _(k=1)^(n)(\frac (1)(k))) | ln (n) + γ (\displaystyle \ln(n)+\gamma ) | ε n (\displaystyle \varepsilon _(n)), (%) |
| 10 | 2,93 | 2,88 | 1,7 |
| 25 | 3,82 | 3,80 | 0,5 |
A more accurate asymptotic formula for the partial sum of the harmonic series:
s n ≍ ln (n) + γ + 1 2 n − 1 12 n 2 + 1 120 n 4 − 1 252 n 6 ⋯ = ln (n) + γ + 1 2 n − ∑ k = 1 ∞ B 2 k 2 k n 2 k (\displaystyle s_(n)\asymp \ln(n)+\gamma +(\frac (1)(2n))-(\frac (1)(12n^(2)))+(\ frac (1)(120n^(4)))-(\frac (1)(252n^(6)))\dots =\ln(n)+\gamma +(\frac (1)(2n))- \sum _(k=1)^(\infty )(\frac (B_(2k))(2k\,n^(2k)))), Where B 2 k (\displaystyle B_(2k)) - Bernoulli numbers.This series diverges, but the error in its calculations never exceeds half of the first discarded term.
Number-theoretic properties of partial sums
∀ n > 1 s n ∉ N (\displaystyle \forall n>1\;\;\;\;s_(n)\notin \mathbb (N) )
Divergence of series
S n → ∞ (\displaystyle s_(n)\rightarrow \infty ) at n → ∞ (\displaystyle n\rightarrow \infty )
The harmonic series diverges very slowly (for the partial sum to exceed 100, about 10 43 elements of the series are needed).
The divergence of the harmonic series can be demonstrated by comparing it with telescopic row :
v n = ln (n + 1) − ln n = ln (1 + 1 n) ∼ + ∞ 1 n (\displaystyle v_(n)=\ln(n+1)-\ln n=\ln \ left(1+(\frac (1)(n))\right)(\underset (+\infty )(\sim ))(\frac (1)(n))),the partial sum of which is obviously equal to:
∑ i = 1 n − 1 v i = ln n ∼ s n (\displaystyle \sum _(i=1)^(n-1)v_(i)=\ln n\sim s_(n)).Oresme's proof
The proof of divergence can be constructed by grouping the terms as follows:
∑ k = 1 ∞ 1 k = 1 + [ 1 2 ] + [ 1 3 + 1 4 ] + [ 1 5 + 1 6 + 1 7 + 1 8 ] + [ 1 9 + ⋯ ] + ⋯ > 1 + [ 1 2 ] + [ 1 4 + 1 4 ] + [ 1 8 + 1 8 + 1 8 + 1 8 ] + [ 1 16 + ⋯ ] + ⋯ = 1 + 1 2 + 1 2 + 1 2 + 1 2 + ⋯ . (\displaystyle (\begin(aligned)\sum _(k=1)^(\infty )(\frac (1)(k))&()=1+\left[(\frac (1)(2) )\right]+\left[(\frac (1)(3))+(\frac (1)(4))\right]+\left[(\frac (1)(5))+(\frac (1)(6))+(\frac (1)(7))+(\frac (1)(8))\right]+\left[(\frac (1)(9))+\cdots \ right]+\cdots \\&()>1+\left[(\frac (1)(2))\right]+\left[(\frac (1)(4))+(\frac (1) (4))\right]+\left[(\frac (1)(8))+(\frac (1)(8))+(\frac (1)(8))+(\frac (1) (8))\right]+\left[(\frac (1)(16))+\cdots \right]+\cdots \\&()=1+\ (\frac (1)(2))\ \ \ +\quad (\frac (1)(2))\ \quad +\ \qquad \quad (\frac (1)(2))\qquad \ \quad \ +\quad \ \ (\frac (1 )(2))\ \quad +\ \cdots .\end(aligned)))The last row obviously diverges. This proof comes from a medieval scientist Nicholas Orem(c. 1350).
Alternative proof of divergence
We invite the reader to verify the fallacy of this proof
Difference between n (\displaystyle n) th harmonic number and natural logarithm n (\displaystyle n) converges to Euler-Mascheroni constant.
The difference between different harmonic numbers is never equal to a whole number and no harmonic number except H 1 = 1 (\displaystyle H_(1)=1), is not an integer.
Related series
Dirichlet series
Generalized harmonic series (or next to Dirichlet) is called a series
∑ k = 1 ∞ 1 k α = 1 + 1 2 α + 1 3 α + 1 4 α + ⋯ + 1 k α + ⋯ (\displaystyle \sum _(k=1)^(\infty )(\frac ( 1)(k^(\alpha )))=1+(\frac (1)(2^(\alpha )))+(\frac (1)(3^(\alpha )))+(\frac ( 1)(4^(\alpha )))+\cdots +(\frac (1)(k^(\alpha )))+\cdots ).The generalized harmonic series diverges at α ⩽ 1 (\displaystyle \alpha \leqslant 1) and converges at α > 1 (\displaystyle \alpha >1) .
Sum of generalized harmonic series of order α (\displaystyle \alpha ) equal to the value Riemann zeta functions :
∑ k = 1 ∞ 1 k α = ζ (α) (\displaystyle \sum _(k=1)^(\infty )(\frac (1)(k^(\alpha )))=\zeta (\alpha ))For even numbers this value is expressed explicitly through Pi, For example, ζ (2) = π 2 6 (\displaystyle \zeta (2)=(\frac (\pi ^(2))(6))), and already for α=3 its value is analytically unknown.
Another illustration of the divergence of the harmonic series can be the relation ζ (1 + 1 n) ∼ n (\displaystyle \zeta (1+(\frac (1)(n)))\sim n) . Therefore, they say that such a series has probability 1, and the sum of the series is random value with interesting properties. For example, probability density function, calculated at points +2 or −2 has the value:
0,124 999 999 999 999 999 999 999 999 999 999 999 999 999 7 642 …,differing from ⅛ by less than 10 −42.
“Thinned” harmonic series
Kempner series (English)If we consider a harmonic series in which only terms are left whose denominators do not contain the number 9, then it turns out that the remaining sum converges to the number<80 . Более того, доказано, что если оставить слагаемые, не содержащие любой заранее выбранной последовательности цифр, то полученный ряд будет сходиться. Однако из этого будет ошибочно заключать о сходимости исходного гармонического ряда, так как с ростом разрядов в числе n (\displaystyle n), fewer and fewer terms are taken for the sum of the “thinned” series. That is, ultimately, the overwhelming majority of terms forming the sum of the harmonic series are discarded so as not to exceed the geometric progression limiting from above.
This article provides structured and detailed information that may be useful when analyzing exercises and tasks. We will look at the topic of number series.
This article begins with basic definitions and concepts. Next, we will use standard options and study the basic formulas. In order to consolidate the material, the article provides basic examples and tasks.
Basic theses
First, let's imagine the system: a 1 , a 2 . . . , a n , . . . , where a k ∈ R, k = 1, 2. . . .
For example, let's take numbers such as: 6, 3, - 3 2, 3 4, 3 8, - 3 16, . . . .
Definition 1
A number series is the sum of terms ∑ a k k = 1 ∞ = a 1 + a 2 + . . . + a n + . . . .
To better understand the definition, consider the given case in which q = - 0. 5: 8 - 4 + 2 - 1 + 1 2 - 1 4 + . . . = ∑ k = 1 ∞ (- 16) · - 1 2 k .
Definition 2
a k is general or k –th member of the series.
It looks something like this - 16 · - 1 2 k.
Definition 3
Partial sum of series looks something like this S n = a 1 + a 2 + . . . + a n , in which n– any number. S n is nth the sum of the series.
For example, ∑ k = 1 ∞ (- 16) · - 1 2 k is S 4 = 8 - 4 + 2 - 1 = 5.
S 1 , S 2 , . . . , S n , . . . form an infinite sequence of numbers.
For a row nth the sum is found by the formula S n = a 1 · (1 - q n) 1 - q = 8 · 1 - - 1 2 n 1 - - 1 2 = 16 3 · 1 - - 1 2 n. We use the following sequence of partial sums: 8, 4, 6, 5, . . . , 16 3 · 1 - - 1 2 n , . . . .
Definition 4
The series ∑ k = 1 ∞ a k is convergent when the sequence has a finite limit S = lim S n n → + ∞ . If there is no limit or the sequence is infinite, then the series ∑ k = 1 ∞ a k is called divergent.
Definition 5
The sum of a convergent series∑ k = 1 ∞ a k is the limit of the sequence ∑ k = 1 ∞ a k = lim S n n → + ∞ = S .
In this example, lim S n n → + ∞ = lim 16 3 t → + ∞ · 1 - 1 2 n = 16 3 · lim n → + ∞ 1 - - 1 2 n = 16 3 , row ∑ k = 1 ∞ (- 16) · - 1 2 k converges. The sum is 16 3: ∑ k = 1 ∞ (- 16) · - 1 2 k = 16 3 .
Example 1
An example of a divergent series is the sum of a geometric progression with a denominator greater than one: 1 + 2 + 4 + 8 +. . . + 2 n - 1 + . . . = ∑ k = 1 ∞ 2 k - 1 .
The nth partial sum is given by S n = a 1 (1 - q n) 1 - q = 1 (1 - 2 n) 1 - 2 = 2 n - 1, and the limit of partial sums is infinite: lim n → + ∞ S n = lim n → + ∞ (2 n - 1) = + ∞ .
Another example of a divergent number series is a sum of the form ∑ k = 1 ∞ 5 = 5 + 5 + . . . . In this case, the nth partial sum can be calculated as Sn = 5n. The limit of partial sums is infinite lim n → + ∞ S n = lim n → + ∞ 5 n = + ∞ .
Definition 6
A sum of the same form as ∑ k = 1 ∞ = 1 + 1 2 + 1 3 + . . . + 1 n + . . . - This harmonic number series.
Definition 7
Sum ∑ k = 1 ∞ 1 k s = 1 + 1 2 s + 1 3 s + . . . + 1 n s + . . . , Where s– a real number, is a generalized harmonic number series.
The definitions discussed above will help you solve most examples and problems.
In order to complete the definitions, it is necessary to prove certain equations.
- ∑ k = 1 ∞ 1 k – divergent.
We use the reverse method. If it converges, then the limit is finite. We can write the equation as lim n → + ∞ S n = S and lim n → + ∞ S 2 n = S . After certain actions we get the equality l i m n → + ∞ (S 2 n - S n) = 0.
Against,
S 2 n - S n = 1 + 1 2 + 1 3 + . . . + 1 n + 1 n + 1 + 1 n + 2 + . . . + 1 2 n - - 1 + 1 2 + 1 3 + . . . + 1 n = 1 n + 1 + 1 n + 2 + . . . + 1 2 n
The following inequalities are valid: 1 n + 1 > 1 2 n, 1 n + 1 > 1 2 n, . . . , 1 2 n - 1 > 1 2 n . We get that S 2 n - S n = 1 n + 1 + 1 n + 2 + . . . + 1 2 n > 1 2 n + 1 2 n + . . . + 1 2 n = n 2 n = 1 2 . The expression S 2 n - S n > 1 2 indicates that lim n → + ∞ (S 2 n - S n) = 0 is not achieved. The series is divergent.
- b 1 + b 1 q + b 1 q 2 + . . . + b 1 q n + . . . = ∑ k = 1 ∞ b 1 q k - 1
It is necessary to confirm that the sum of a sequence of numbers converges at q< 1 , и расходится при q ≥ 1 .
According to the above definitions, the amount n terms is determined according to the formula S n = b 1 · (q n - 1) q - 1 .
If q< 1 верно
lim n → + ∞ S n = lim n → + ∞ b 1 · q n - 1 q - 1 = b 1 · lim n → + ∞ q n q - 1 - lim n → + ∞ 1 q - 1 = = b 1 · 0 - 1 q - 1 = b 1 q - 1
We have proven that the number series converges.
For q = 1 b 1 + b 1 + b 1 + . . . ∑ k = 1 ∞ b 1 . The sums can be found using the formula S n = b 1 · n, the limit is infinite lim n → + ∞ S n = lim n → + ∞ b 1 · n = ∞. In the presented version, the series diverges.
If q = - 1, then the series looks like b 1 - b 1 + b 1 - . . . = ∑ k = 1 ∞ b 1 (- 1) k + 1 . Partial sums look like S n = b 1 for odd n, and S n = 0 for even n. Having considered this case, we will make sure that there is no limit and the series is divergent.
For q > 1, lim n → + ∞ S n = lim n → + ∞ b 1 · (q n - 1) q - 1 = b 1 · lim n → + ∞ q n q - 1 - lim n → + ∞ 1 q - 1 = = b 1 · ∞ - 1 q - 1 = ∞
We have proven that the number series diverges.
- The series ∑ k = 1 ∞ 1 k s converges if s > 1 and diverges if s ≤ 1.
For s = 1 we obtain ∑ k = 1 ∞ 1 k , the series diverges.
When s< 1 получаем 1 k s ≥ 1 k для k,natural number. Since the series is divergent ∑ k = 1 ∞ 1 k , there is no limit. Following this, the sequence ∑ k = 1 ∞ 1 k s is unbounded. We conclude that the selected series diverges when s< 1 .
It is necessary to provide evidence that the series ∑ k = 1 ∞ 1 k s converges for s > 1.
Let's imagine S 2 n - 1 - S n - 1:
S 2 n - 1 - S n - 1 = 1 + 1 2 s + 1 3 s + . . . + 1 (n - 1) s + 1 n s + 1 (n + 1) s + . . . + 1 (2 n - 1) s - - 1 + 1 2 s + 1 3 s + . . . + 1 (n - 1) s = 1 n s + 1 (n + 1) s + . . . + 1 (2 n - 1) s
Let us assume that 1 (n + 1) s< 1 n s , 1 (n + 2) s < 1 n s , . . . , 1 (2 n - 1) s < 1 n s , тогда S 2 n - 1 - S n - 1 = 1 n s + 1 (n + 1) s + . . . + 1 (2 n - 1) s < < 1 n s + 1 n s + . . . + 1 n s = n n s = 1 n s - 1
Let's imagine the equation for numbers that are natural and even n = 2: S 2 n - 1 - S n - 1 = S 3 - S 1 = 1 2 s + 1 3 s< 1 2 s - 1 n = 4: S 2 n - 1 - S n - 1 = S 7 - S 3 = 1 4 s + 1 5 s + 1 6 s + 1 7 s < 1 4 s - 1 = 1 2 s - 1 2 n = 8: S 2 n - 1 - S n - 1 = S 15 - S 7 = 1 8 s + 1 9 s + . . . + 1 15 s < 1 8 s - 1 = 1 2 s - 1 3 . . .
We get:
∑ k = 1 ∞ 1 k s = 1 + 1 2 s + 1 3 s + 1 4 s + . . . + 1 7 s + 1 8 s + . . . + 1 15 s + . . . = = 1 + S 3 - S 1 + S 7 - S 3 + S 15 + S 7 + . . .< < 1 + 1 2 s - 1 + 1 2 s - 1 2 + 1 2 s - 1 3 + . . .
The expression is 1 + 1 2 s - 1 + 1 2 s - 1 2 + 1 2 s - 1 3 + . . . is the sum of the geometric progression q = 1 2 s - 1. According to the initial data at s > 1, then 0< q < 1 . Получаем, ∑ k = 1 ∞ < 1 + 1 2 s - 1 + 1 2 s - 1 2 + 1 2 s - 1 3 + . . . = 1 1 - q = 1 1 - 1 2 s - 1 . Последовательность ряда при s > 1 increases and is limited from above 1 1 - 1 2 s - 1 . Let's imagine that there is a limit and the series is convergent ∑ k = 1 ∞ 1 k s .
Definition 8
Series ∑ k = 1 ∞ a k is positive in that case, if its members > 0 a k > 0 , k = 1 , 2 , . . . .
Series ∑ k = 1 ∞ b k signalternating, if the signs of the numbers are different. This example is presented as ∑ k = 1 ∞ b k = ∑ k = 1 ∞ (- 1) k · a k or ∑ k = 1 ∞ b k = ∑ k = 1 ∞ (- 1) k + 1 · a k , where a k > 0 , k = 1 , 2 , . . . .
Series ∑ k = 1 ∞ b k alternating, since it contains many numbers, negative and positive.
The second option series is a special case of the third option.
Here are examples for each case, respectively:
6 + 3 + 3 2 + 3 4 + 3 8 + 3 16 + . . . 6 - 3 + 3 2 - 3 4 + 3 8 - 3 16 + . . . 6 + 3 - 3 2 + 3 4 + 3 8 - 3 16 + . . .
For the third option, you can also determine absolute and conditional convergence.
Definition 9
The alternating series ∑ k = 1 ∞ b k is absolutely convergent in the case when ∑ k = 1 ∞ b k is also considered convergent.
Let's look at several typical options in detail.
Example 2
If the rows are 6 - 3 + 3 2 - 3 4 + 3 8 - 3 16 +. . . and 6 + 3 - 3 2 + 3 4 + 3 8 - 3 16 + . . . are defined as convergent, then it is correct to assume that 6 + 3 + 3 2 + 3 4 + 3 8 + 3 16 + . . .
Definition 10
An alternating series ∑ k = 1 ∞ b k is considered conditionally convergent if ∑ k = 1 ∞ b k is divergent, and the series ∑ k = 1 ∞ b k is considered convergent.
Example 3
Let us examine in detail the option ∑ k = 1 ∞ (- 1) k + 1 k = 1 - 1 2 + 1 3 - 1 4 + . . . . The series ∑ k = 1 ∞ (- 1) k + 1 k = ∑ k = 1 ∞ 1 k, which consists of absolute values, is defined as divergent. This option is considered convergent since it is easy to determine. From this example we learn that the series ∑ k = 1 ∞ (- 1) k + 1 k = 1 - 1 2 + 1 3 - 1 4 + . . . will be considered conditionally convergent.
Features of convergent series
Let's analyze the properties for certain cases
- If ∑ k = 1 ∞ a k converges, then the series ∑ k = m + 1 ∞ a k is also considered convergent. It can be noted that the row without m terms is also considered convergent. If we add several numbers to ∑ k = m + 1 ∞ a k, then the resulting result will also be convergent.
- If ∑ k = 1 ∞ a k converges and the sum = S, then the series ∑ k = 1 ∞ A · a k , ∑ k = 1 ∞ A · a k = A · S also converges, where A-constant.
- If ∑ k = 1 ∞ a k and ∑ k = 1 ∞ b k are convergent, the sums A And B too, then the series ∑ k = 1 ∞ a k + b k and ∑ k = 1 ∞ a k - b k also converge. The amounts will be equal A+B And A - B respectively.
Determine that the series converges ∑ k = 1 ∞ 2 3 k · k 3 .
Let's change the expression ∑ k = 1 ∞ 2 3 k · k 3 = ∑ k = 1 ∞ 2 3 · 1 k 4 3 . The series ∑ k = 1 ∞ 1 k 4 3 is considered convergent, since the series ∑ k = 1 ∞ 1 k s converges when s > 1. According to the second property, ∑ k = 1 ∞ 2 3 · 1 k 4 3 .
Example 5
Determine whether the series ∑ n = 1 ∞ 3 + n n 5 2 converges.
Let's transform the original version ∑ n = 1 ∞ 3 + n n 5 2 = ∑ n = 1 ∞ 3 n 5 2 + n n 2 = ∑ n = 1 ∞ 3 n 5 2 + ∑ n = 1 ∞ 1 n 2 .
We get the sum ∑ n = 1 ∞ 3 n 5 2 and ∑ n = 1 ∞ 1 n 2 . Each series is considered convergent according to the property. So, as the series converge, so does the original version.
Example 6
Calculate whether the series 1 - 6 + 1 2 - 2 + 1 4 - 2 3 + 1 8 - 2 9 + converges. . . and calculate the amount.
Let's expand the original version:
1 - 6 + 1 2 - 2 + 1 4 - 2 3 + 1 8 - 2 9 + . . . = = 1 + 1 2 + 1 4 + 1 8 + . . . - 2 · 3 + 1 + 1 3 + 1 9 + . . . = = ∑ k = 1 ∞ 1 2 k - 1 - 2 · ∑ k = 1 ∞ 1 3 k - 2
Each series converges because it is one of the members of a number sequence. According to the third property, we can calculate that the original version is also convergent. We calculate the sum: The first term of the series ∑ k = 1 ∞ 1 2 k - 1 = 1, and the denominator = 0. 5, this is followed by, ∑ k = 1 ∞ 1 2 k - 1 = 1 1 - 0 . 5 = 2. The first term is ∑ k = 1 ∞ 1 3 k - 2 = 3 , and the denominator of the descending number sequence = 1 3 . We get: ∑ k = 1 ∞ 1 3 k - 2 = 3 1 - 1 3 = 9 2 .
We use the expressions obtained above to determine the sum 1 - 6 + 1 2 - 2 + 1 4 - 2 3 + 1 8 - 2 9 +. . . = ∑ k = 1 ∞ 1 2 k - 1 - 2 ∑ k = 1 ∞ 1 3 k - 2 = 2 - 2 9 2 = - 7
A necessary condition for determining whether a series is convergent
Definition 11If the series ∑ k = 1 ∞ a k is convergent, then its limit kth term = 0: lim k → + ∞ a k = 0 .
If we check any option, we must not forget about the indispensable condition. If it is not fulfilled, then the series diverges. If lim k → + ∞ a k ≠ 0, then the series is divergent.
It should be clarified that the condition is important, but not sufficient. If the equality lim k → + ∞ a k = 0 holds, then this does not guarantee that ∑ k = 1 ∞ a k is convergent.
Let's give an example. For the harmonic series ∑ k = 1 ∞ 1 k the condition is satisfied lim k → + ∞ 1 k = 0 , but the series still diverges.
Example 7
Determine the convergence ∑ n = 1 ∞ n 2 1 + n .
Let's check the original expression for the fulfillment of the condition lim n → + ∞ n 2 1 + n = lim n → + ∞ n 2 n 2 1 n 2 + 1 n = lim n → + ∞ 1 1 n 2 + 1 n = 1 + 0 + 0 = + ∞ ≠ 0
Limit nth member is not equal to 0. We have proven that this series diverges.
How to determine the convergence of a positive series.
If you constantly use these characteristics, you will have to constantly calculate the limits. This section will help you avoid difficulties when solving examples and problems. In order to determine the convergence of a positive series, there is a certain condition.
For convergence of positive sign ∑ k = 1 ∞ a k , a k > 0 ∀ k = 1 , 2 , 3 , . . . it is necessary to determine a limited sequence of sums.
How to compare series
There are several signs of comparing series. We compare the series whose convergence is proposed to be determined with the series whose convergence is known.
First sign
∑ k = 1 ∞ a k and ∑ k = 1 ∞ b k are positive sign series. The inequality a k ≤ b k is valid for k = 1, 2, 3, ... It follows from this that from the series ∑ k = 1 ∞ b k we can obtain ∑ k = 1 ∞ a k . Since ∑ k = 1 ∞ a k is divergent, the series ∑ k = 1 ∞ b k can be defined as divergent.
This rule is constantly used to solve equations and is a serious argument that will help determine convergence. The difficulty may lie in the fact that it is not possible to find a suitable example for comparison in every case. Quite often, a series is selected according to the principle that the indicator kth term will be equal to the result of subtracting the exponents of the numerator and denominator kth member of the series. Let's assume that a k = k 2 + 3 4 k 2 + 5 , the difference will be equal to 2 – 3 = - 1 . IN in this case it can be determined that for comparison a series with k-th term b k = k - 1 = 1 k , which is harmonic.
In order to consolidate the obtained material, we will consider in detail a couple of typical options.
Example 8
Determine what the series ∑ k = 1 ∞ 1 k - 1 2 is.
Since limit = 0 lim k → + ∞ 1 k - 1 2 = 0 , we have done necessary condition. The inequality will be fair 1 k< 1 k - 1 2 для k, which are natural. From the previous paragraphs we learned that the harmonic series ∑ k = 1 ∞ 1 k is divergent. According to the first criterion, it can be proven that the original version is divergent.
Example 9
Determine whether the series is convergent or divergent ∑ k = 1 ∞ 1 k 3 + 3 k - 1 .
In this example, the necessary condition is satisfied, since lim k → + ∞ 1 k 3 + 3 k - 1 = 0. We represent it as the inequality 1 k 3 + 3 k - 1< 1 k 3 для любого значения k. The series ∑ k = 1 ∞ 1 k 3 is convergent, since the harmonic series ∑ k = 1 ∞ 1 k s converges for s > 1. According to the first criterion, we can conclude that the number series is convergent.
Example 10
Determine what the series ∑ k = 3 ∞ 1 k ln (ln k) is. lim k → + ∞ 1 k ln (ln k) = 1 + ∞ + ∞ = 0 .
In this option, you can mark the fulfillment of the desired condition. Let's define a series for comparison. For example, ∑ k = 1 ∞ 1 k s . To determine what the degree is, consider the sequence (ln (ln k)), k = 3, 4, 5. . . . Members of the sequence ln (ln 3) , ln (ln 4) , ln (ln 5) , . . . increases to infinity. Having analyzed the equation, we can note that, taking N = 1619 as the value, then the terms of the sequence > 2. For this sequence the inequality 1 k ln (ln k) will be true< 1 k 2 . Ряд ∑ k = N ∞ 1 k 2 сходится согласно первому признаку, так как ряд ∑ k = 1 ∞ 1 k 2 тоже сходящийся. Отметим, что согласно первому признаку ряд ∑ k = N ∞ 1 k ln (ln k) сходящийся. Можно сделать вывод, что ряд ∑ k = 3 ∞ 1 k ln (ln k) также сходящийся.
Second sign
Let us assume that ∑ k = 1 ∞ a k and ∑ k = 1 ∞ b k are positive number series.
If lim k → + ∞ a k b k ≠ ∞ , then the series ∑ k = 1 ∞ b k converges, and ∑ k = 1 ∞ a k also converges.
If lim k → + ∞ a k b k ≠ 0, then since the series ∑ k = 1 ∞ b k diverges, then ∑ k = 1 ∞ a k also diverges.
If lim k → + ∞ a k b k ≠ ∞ and lim k → + ∞ a k b k ≠ 0, then the convergence or divergence of a series means the convergence or divergence of another.
Consider ∑ k = 1 ∞ 1 k 3 + 3 k - 1 using the second sign. For comparison ∑ k = 1 ∞ b k we take the convergent series ∑ k = 1 ∞ 1 k 3 . Let's define the limit: lim k → + ∞ a k b k = lim k → + ∞ 1 k 3 + 3 k - 1 1 k 3 = lim k → + ∞ k 3 k 3 + 3 k - 1 = 1
According to the second criterion, it can be determined that the convergent series ∑ k = 1 ∞ 1 k 3 means that the original version also converges.
Example 11
Determine what the series ∑ n = 1 ∞ k 2 + 3 4 k 3 + 5 is.
Let us analyze the necessary condition lim k → ∞ k 2 + 3 4 k 3 + 5 = 0, which is satisfied in this version. According to the second criterion, take the series ∑ k = 1 ∞ 1 k . We are looking for the limit: lim k → + ∞ k 2 + 3 4 k 3 + 5 1 k = lim k → + ∞ k 3 + 3 k 4 k 3 + 5 = 1 4
According to the above theses, a divergent series entails the divergence of the original series.
Third sign
Let's consider the third sign of comparison.
Let us assume that ∑ k = 1 ∞ a k and _ ∑ k = 1 ∞ b k are positive number series. If the condition is satisfied for a certain number a k + 1 a k ≤ b k + 1 b k , then the convergence of this series ∑ k = 1 ∞ b k means that the series ∑ k = 1 ∞ a k is also convergent. The divergent series ∑ k = 1 ∞ a k entails the divergence ∑ k = 1 ∞ b k .
D'Alembert's sign
Let's imagine that ∑ k = 1 ∞ a k is a positive number series. If lim k → + ∞ a k + 1 a k< 1 , то ряд является сходящимся, если lim k → + ∞ a k + 1 a k >1, then divergent.
Note 1
D'Alembert's test is valid if the limit is infinite.
If lim k → + ∞ a k + 1 a k = - ∞ , then the series is convergent, if lim k → ∞ a k + 1 a k = + ∞ , then it is divergent.
If lim k → + ∞ a k + 1 a k = 1, then d'Alembert's sign will not help and several more studies will be required.
Example 12
Determine whether the series is convergent or divergent ∑ k = 1 ∞ 2 k + 1 2 k using d’Alembert’s criterion.
It is necessary to check whether the necessary convergence condition is satisfied. Let's calculate the limit using L'Hopital's rule: lim k → + ∞ 2 k + 1 2 k = ∞ ∞ = lim k → + ∞ 2 k + 1 " 2 k " = lim k → + ∞ 2 2 k ln 2 = 2 + ∞ ln 2 = 0
We can see that the condition is met. Let's use d'Alembert's test: lim k → + ∞ = lim k → + ∞ 2 (k + 1) + 1 2 k + 1 2 k + 1 2 k = 1 2 lim k → + ∞ 2 k + 3 2 k + 1 = 12< 1
The series is convergent.
Example 13
Determine whether the series is divergent ∑ k = 1 ∞ k k k ! .
Let's use d'Alembert's test to determine the divergence of the series: lim k → + ∞ a k + 1 a k = lim k → + ∞ (k + 1) k + 1 (k + 1) ! k k k! = lim k → + ∞ (k + 1) k + 1 · k ! k k · (k + 1) ! = lim k → + ∞ (k + 1) k + 1 k k · (k + 1) = = lim k → + ∞ (k + 1) k k k = lim k → + ∞ k + 1 k k = lim k → + ∞ 1 + 1 k k = e > 1
Therefore, the series is divergent.
Radical Cauchy's sign
Let us assume that ∑ k = 1 ∞ a k is a series with a positive sign. If lim k → + ∞ a k k< 1 , то ряд является сходящимся, если lim k → + ∞ a k k >1, then divergent.
Note 2
If lim k → + ∞ a k k = 1, then this sign does not provide any information - additional analysis is required.
This feature can be used in examples that are easy to identify. The case will be typical when a member of a number series is an exponential power expression.
In order to consolidate the information received, let's consider several typical examples.
Example 14
Determine whether the positive sign series ∑ k = 1 ∞ 1 (2 k + 1) k is convergent.
The necessary condition is considered satisfied, since lim k → + ∞ 1 (2 k + 1) k = 1 + ∞ + ∞ = 0 .
According to the criterion discussed above, we obtain lim k → + ∞ a k k = lim k → + ∞ 1 (2 k + 1) k k = lim k → + ∞ 1 2 k + 1 = 0< 1 . Данный ряд является сходимым.
Example 15
Does the number series ∑ k = 1 ∞ 1 3 k · 1 + 1 k k 2 converge?
We use the feature described in the previous paragraph lim k → + ∞ 1 3 k 1 + 1 k k 2 k = 1 3 lim k → + ∞ 1 + 1 k k = e 3< 1 , следовательно, числовой ряд сходится.
Integral Cauchy test
Let us assume that ∑ k = 1 ∞ a k is a series with positive sign. It is necessary to denote the function of a continuous argument y = f(x), which coincides with a n = f (n) . If y = f(x) greater than zero, is not interrupted and decreases by [ a ; + ∞) , where a ≥ 1
Then in case improper integral∫ a + ∞ f (x) d x is convergent, then the series under consideration is also convergent. If it diverges, then in the example under consideration the series also diverges.
When checking whether a function is decreasing, you can use the material covered in previous lessons.
Example 16
Consider the example ∑ k = 2 ∞ 1 k · ln k for convergence.
The condition for the convergence of the series is considered to be satisfied, since lim k → + ∞ 1 k · ln k = 1 + ∞ = 0 . Consider y = 1 x ln x. It is greater than zero, is not interrupted and decreases by [ 2 ; + ∞) . The first two points are known for certain, but the third should be discussed in more detail. Find the derivative: y " = 1 x · ln x " = x · ln x " x · ln x 2 = ln x + x · 1 x x · ln x 2 = - ln x + 1 x · ln x 2. It is less than zero on [ 2 ; + ∞).This proves the thesis that the function is decreasing.
Actually, the function y = 1 x ln x corresponds to the characteristics of the principle that we considered above. Let's use it: ∫ 2 + ∞ d x x · ln x = lim A → + ∞ ∫ 2 A d (ln x) ln x = lim A → + ∞ ln (ln x) 2 A = = lim A → + ∞ (ln ( ln A) - ln (ln 2)) = ln (ln (+ ∞)) - ln (ln 2) = + ∞
According to the results obtained, the original example diverges, since the improper integral is divergent.
Example 17
Prove the convergence of the series ∑ k = 1 ∞ 1 (10 k - 9) (ln (5 k + 8)) 3 .
Since lim k → + ∞ 1 (10 k - 9) (ln (5 k + 8)) 3 = 1 + ∞ = 0, then the condition is considered satisfied.
Starting with k = 4, the correct expression is 1 (10 k - 9) (ln (5 k + 8)) 3< 1 (5 k + 8) (ln (5 k + 8)) 3 .
If the series ∑ k = 4 ∞ 1 (5 k + 8) (ln (5 k + 8)) 3 is considered convergent, then, according to one of the principles of comparison, the series ∑ k = 4 ∞ 1 (10 k - 9) ( ln (5 k + 8)) 3 will also be considered convergent. This way we can determine that the original expression is also convergent.
Let's move on to the proof: ∑ k = 4 ∞ 1 (5 k + 8) (ln (5 k + 8)) 3 .
Since the function y = 1 5 x + 8 (ln (5 x + 8)) 3 is greater than zero, it is not interrupted and decreases by [ 4 ; + ∞) . We use the feature described in the previous paragraph:
∫ 4 + ∞ d x (5 x + 8) (l n (5 x + 8)) 3 = lim A → + ∞ ∫ 4 A d x (5 x + 8) (ln (5 x + 8)) 3 = = 1 5 lim A → + ∞ ∫ 4 A d (ln (5 x + 8) (ln (5 x + 8)) 3 = - 1 10 lim A → + ∞ 1 (ln (5 x + 8)) 2 | 4 A = = - 1 10 lim A → + ∞ 1 (ln (5 A + 8)) 2 - 1 (ln (5 4 + 8)) 2 = = - 1 10 1 + ∞ - 1 (ln 28) 2 = 1 10 · ln 28 2
In the resulting convergent series, ∫ 4 + ∞ d x (5 x + 8) (ln (5 x + 8)) 3, we can determine that ∑ k = 4 ∞ 1 (5 k + 8) (ln (5 k + 8 )) 3 also converges.
Raabe's sign
Let us assume that ∑ k = 1 ∞ a k is a positive number series.
If lim k → + ∞ k · a k a k + 1< 1 , то ряд расходится, если lim k → + ∞ k · a k a k + 1 - 1 >1, then it converges.
This determination method can be used if the techniques described above do not give visible results.
Absolute Convergence Study
For the study we take ∑ k = 1 ∞ b k . We use positive sign ∑ k = 1 ∞ b k . We can use any of the suitable features that we described above. If the series ∑ k = 1 ∞ b k converges, then the original series is absolutely convergent.
Example 18
Investigate the series ∑ k = 1 ∞ (- 1) k 3 k 3 + 2 k - 1 for convergence ∑ k = 1 ∞ (- 1) k 3 k 3 + 2 k - 1 = ∑ k = 1 ∞ 1 3 k 3 + 2 k - 1 .
The condition is satisfied lim k → + ∞ 1 3 k 3 + 2 k - 1 = 1 + ∞ = 0 . We use ∑ k = 1 ∞ 1 k 3 2 and use the second sign: lim k → + ∞ 1 3 k 3 + 2 k - 1 1 k 3 2 = 1 3 .
The series ∑ k = 1 ∞ (- 1) k 3 k 3 + 2 k - 1 converges. The original series is also absolutely convergent.
Divergence of alternating series
If the series ∑ k = 1 ∞ b k is divergent, then the corresponding alternating series ∑ k = 1 ∞ b k is either divergent or conditionally convergent.
Only d'Alembert's test and the radical Cauchy test will help to draw conclusions about ∑ k = 1 ∞ b k from the divergence from the moduli ∑ k = 1 ∞ b k . The series ∑ k = 1 ∞ b k also diverges if the necessary convergence condition is not satisfied, that is, if lim k → ∞ + b k ≠ 0.
Example 19
Check divergence 1 7, 2 7 2, - 6 7 3, 24 7 4, 120 7 5 - 720 7 6, . . . .
Module kth term is represented as b k = k ! 7 k.
Let us examine the series ∑ k = 1 ∞ b k = ∑ k = 1 ∞ k ! 7 k for convergence using d'Alembert's criterion: lim k → + ∞ b k + 1 b k = lim k → + ∞ (k + 1) ! 7 k + 1 k ! 7 k = 1 7 · lim k → + ∞ (k + 1) = + ∞ .
∑ k = 1 ∞ b k = ∑ k = 1 ∞ k ! 7 k diverges in the same way as the original version.
Example 20
Is ∑ k = 1 ∞ (- 1) k · k 2 + 1 ln (k + 1) convergent.
Let's consider the necessary condition lim k → + ∞ b k = lim k → + ∞ k 2 + 1 ln (k + 1) = ∞ ∞ = lim k → + ∞ = k 2 + 1 " (ln (k + 1)) " = = lim k → + ∞ 2 k 1 k + 1 = lim k → + ∞ 2 k (k + 1) = + ∞ . The condition is not met, therefore ∑ k = 1 ∞ (- 1) k · k 2 + 1 ln (k + 1) the series is divergent. The limit was calculated using L'Hopital's rule.
Conditional convergence criteria
Leibniz's test
Definition 12If the values of the terms of the alternating series decrease b 1 > b 2 > b 3 > . . . > . . . and the modulus limit = 0 as k → + ∞, then the series ∑ k = 1 ∞ b k converges.
Example 17
Consider ∑ k = 1 ∞ (- 1) k 2 k + 1 5 k (k + 1) for convergence.
The series is represented as ∑ k = 1 ∞ (- 1) k 2 k + 1 5 k (k + 1) = ∑ k = 1 ∞ 2 k + 1 5 k (k + 1) . The necessary condition is satisfied: lim k → + ∞ = 2 k + 1 5 k (k + 1) = 0 . Consider ∑ k = 1 ∞ 1 k by the second comparison criterion lim k → + ∞ 2 k + 1 5 k (k + 1) 1 k = lim k → + ∞ 2 k + 1 5 (k + 1) = 2 5
We find that ∑ k = 1 ∞ (- 1) k 2 k + 1 5 k (k + 1) = ∑ k = 1 ∞ 2 k + 1 5 k (k + 1) diverges. The series ∑ k = 1 ∞ (- 1) k 2 k + 1 5 k (k + 1) converges according to the Leibniz criterion: sequence 2 1 + 1 5 1 1 1 + 1 = 3 10, 2 2 + 1 5 · 2 · (2 + 1) = 5 30 , 2 · 3 + 1 5 · 3 · 3 + 1 , . . . decreases and lim k → + ∞ = 2 k + 1 5 k (k + 1) = 0 .
The series converges conditionally.
Abel-Dirichlet test
Definition 13∑ k = 1 + ∞ u k · v k converges if ( u k ) does not increase and the sequence ∑ k = 1 + ∞ v k is bounded.
Example 17
Explore 1 - 3 2 + 2 3 + 1 4 - 3 5 + 1 3 + 1 7 - 3 8 + 2 9 + . . . for convergence.
Let's imagine
1 - 3 2 + 2 3 + 1 4 - 3 5 + 1 3 + 1 7 - 3 8 + 2 9 + . . . = 1 1 + 1 2 (- 3) + 1 3 2 + 1 4 1 + 1 5 (- 3) + 1 6 = ∑ k = 1 ∞ u k v k
where (u k) = 1, 1 2, 1 3, . . . is non-increasing, and the sequence (v k) = 1, - 3, 2, 1, - 3, 2, . . . limited (S k) = 1, - 2, 0, 1, - 2, 0, . . . . The series converges.
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