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 such series converges. For example, because
then the series converges. If we failed to find the sum of the series, then other methods should be used to check the convergence of the series.
One of these methods is sign of d'Alembert
here and are the n-th and (n + 1)-th terms of the series, respectively, and the convergence is determined by the value of D: If D< 1 - ряд сходится, если D >
As an example, we examine the convergence of a series using the d'Alembert test. First, let's write expressions for and . Now let's find the corresponding limit:
Since , according to d'Alembert's test, the series converges.
Another method to check the convergence of a series is radical sign of Cauchy, which is written as follows:
here is the nth term of the series, and convergence, as in the case of the d'Alembert test, is determined by the value of D: If D< 1 - ряд сходится, если D >1 - diverges. When D = 1 - this sign does not give an answer and additional research is needed.
As an example, we study the convergence of a series using the radical Cauchy test. First, let's write an expression for . Now let's find the corresponding limit:
Because title="15625/64>1"> , according to the radical Cauchy test, the series diverges.
It should be noted that along with the above, there are other signs of convergence of the series, such as the integral Cauchy test, the Raabe test, etc.
Our online calculator, built on the basis of the Wolfram Alpha system, allows you to test the convergence of the series. In this case, if the calculator gives a specific number as the sum of the series, then the series converges. Otherwise, it is necessary to pay attention to the item "Test of series convergence". If the phrase “series converges” is present there, then the series converges. If the phrase "series diverges" is present, then the series diverges.
Below is a translation of all possible values of the item "Test of convergence of a series":
| Text on English language | Text in Russian |
|---|---|
| By the harmonic series test, the series diverges. | When comparing the studied series with the harmonic series, the original series diverges. |
| The ratio test is inconclusive. | d'Alembert's test cannot give an answer about the convergence of a series. |
| The root test is inconclusive. | Cauchy's radical 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 divergences. | Based on the fact that title="The limit of the nth member of the series when 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- the sum composed of an infinite number of terms reciprocal of successive numbers of the natural series:
∑ 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 #1. Mat. analysis. Fourier series in the trigonometric system. The simplest properties
✪ ROWS. Review
Subtitles
The sum of the first n terms of the series
The individual terms of the series tend to zero, but its sum diverges. The n-th partial sum of the s n harmonic series is the n-th 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 values of partial sums
| 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 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 precise 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 calculation error on it 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) )
Series Divergence
S n → ∞ (\displaystyle s_(n)\rightarrow \infty ) at n → ∞ (\displaystyle n\rightarrow \infty )
The harmonic series diverges very slowly (in order 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 the telescopic series:
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))),whose partial sum 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)).Orem's proof
The proof of the 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 belongs to the medieval scholar 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 the Euler - Mascheroni constant.
The difference between different harmonic numbers is never an integer and no harmonic number except H 1 = 1 (\displaystyle H_(1)=1), is not an integer.
Related rows
Dirichlet row
The generalized harmonic series (or the Dirichlet series) is called the 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) .
The sum of the generalized harmonic order series α (\displaystyle \alpha ) is equal to the value of the Riemann zeta function:
∑ k = 1 ∞ 1 k α = ζ (α) (\displaystyle \sum _(k=1)^(\infty )(\frac (1)(k^(\alpha )))=\zeta (\alpha ))For even numbers, this value is explicitly expressed in terms of 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, such a series is said to have with probability 1 , and the sum of the series is a random quantity with interesting properties. For example, the function density probability 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, the denominators of which 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, in the end, the vast majority of the terms that form the sum of the harmonic series are discarded so as not to exceed the geometric progression limiting from above.
This article is a structured and detailed information that can be useful during the analysis of exercises and problems. We will consider the topic of number series.
This article begins with basic definitions and concepts. Next, we will standard options and study the basic formulas. In order to consolidate the material, the article provides the main examples and tasks.
Basic theses
First, 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
The number series is the sum of the terms ∑ a k k = 1 ∞ = a 1 + a 2 + . . . + a n + . . . .
To better understand the definition, consider this case where q = - 0 . 5: 8 - 4 + 2 - 1 + 1 2 - 1 4 + . . . = ∑ k = 1 ∞ (- 16) · - 1 2 k .
Definition 2
a k is common or k-th a member of the row.
It looks something like this - 16 · - 1 2 k .
Definition 3
Partial sum of a 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 number n-th the sum is found by the formula S n \u003d a 1 (1 - q n) 1 - q \u003d 8 1 - - 1 2 n 1 - - 1 2 \u003d 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 converging 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 the 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 , series ∑ 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 geometric progression with a denominator greater than one: 1 + 2 + 4 + 8 + . . . + 2n - 1 + . . . = ∑ k = 1 ∞ 2 k - 1 .
The nth partial sum is defined by the expression 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 the sum of the form ∑ k = 1 ∞ 5 = 5 + 5 + . . . . In this case, the nth partial sum can be calculated as S n = 5 n . The limit of partial sums is infinite lim n → + ∞ S n = lim n → + ∞ 5 n = + ∞ .
Definition 6
A sum similar to ∑ k = 1 ∞ = 1 + 1 2 + 1 3 + . . . + 1n + . . . - This harmonic number line.
Definition 7
The sum ∑ k = 1 ∞ 1 k s = 1 + 1 2 s + 1 3 s + . . . + 1n s + . . . , Where s is a real number, is a generalized harmonic number series.
The definitions discussed above will help you solve most of the examples and problems.
In order to complete the definitions, it is necessary to prove certain equations.
- ∑ k = 1 ∞ 1 k is divergent.
We act in reverse. 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 + . . . + 1n + 1n + 1 + 1n + 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 reached. 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 the sequence of numbers converges for q< 1 , и расходится при q ≥ 1 .
According to the above definitions, the sum n members 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 proved 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 row 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 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 proved 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 get ∑ k = 1 ∞ 1 k , the series diverges.
For 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 chosen series diverges at s< 1 .
It is necessary to provide evidence that the series ∑ k = 1 ∞ 1 k s converges for s > 1.
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(2n - 1)s
Let's say 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
Imagine an 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 + . . . + 17s + 18s + . . . + 1 15 s + . . . \u003d \u003d 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 1 + 1 2 s - 1 + 1 2 s - 1 2 + 1 2 s - 1 3 + . . . is the sum of a geometric progression q = 1 2 s - 1 . According to the initial data 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 . Imagine that there is a limit and the series is convergent ∑ k = 1 ∞ 1 k s .
Definition 8
Series ∑ k = 1 ∞ a k positive in that case, if its members > 0 a k > 0 , k = 1 , 2 , . . . .
Series ∑ k = 1 ∞ b k alternating if the signs of the numbers are different. This example is represented 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 variant of the series is a special case of the third variant.
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 define absolute and conditional convergence.
Definition 9
The alternating series ∑ k = 1 ∞ b k is absolutely convergent if ∑ k = 1 ∞ b k is also considered to be convergent.
Let's take a look at some of the typical options.
Example 2
If rows 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 variant is considered convergent because 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 recognized as convergent. It can be noted that the series m members is also considered convergent. If we add several numbers to ∑ k = m + 1 ∞ a k, then the resulting result will also converge.
- If ∑ k = 1 ∞ a k converges and the sum = S, then the series ∑ k = 1 ∞ A ak , ∑ k = 1 ∞ A ak = A S , 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 A+B And A-B respectively.
Determine that the series converges ∑ k = 1 ∞ 2 3 k · k 3 .
Let us 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 for s > 1. According to the second property, ∑ k = 1 ∞ 2 3 · 1 k 4 3 .
Example 5
Determine if the series converges ∑ n = 1 ∞ 3 + n n 5 2 .
We 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 recognized as convergent according to the property. Since the series converge, so does the original version.
Example 6
Calculate if the series 1 - 6 + 1 2 - 2 + 1 4 - 2 3 + 1 8 - 2 9 + converges. . . and calculate the amount.
Let's break down the original:
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 the numerical 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 , 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 decreasing numerical sequence = 1 3 . We get: ∑ k = 1 ∞ 1 3 k - 2 = 3 1 - 1 3 = 9 2 .
We use the expressions obtained above in order 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 k-th term = 0: lim k → + ∞ a k = 0 .
If we check any option, then we must not forget about the indispensable condition. If it is not satisfied, 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 take 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 convergence ∑ n = 1 ∞ n 2 1 + n .
Check the original expression for 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 0 . We have proved that this series diverges.
How to determine the convergence of a positive sign series.
If you constantly use these features, you will have to constantly calculate the limits. This section will help you avoid difficulties while solving examples and problems. In order to determine the convergence of a positive sign series, there is a certain condition.
For convergence of positive sign ∑ k = 1 ∞ a k , a k > 0 ∀ k = 1 , 2 , 3 , . . . need to be determined limited sequence amounts.
How to compare rows
There are several signs of series comparison. 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 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 get ∑ k = 1 ∞ a k . Since ∑ k = 1 ∞ a k diverges, 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. Difficulties may lie in the fact that it is far from possible to find a suitable example for comparison in every case. Quite often, a series is chosen according to the principle that the indicator k-th term will be equal to the result of subtracting the exponents of the numerator and denominator k-th row member. Let's say that a k = k 2 + 3 4 k 2 + 5 , the difference will be equal to 2 – 3 = - 1 . In this case, it can be determined that a series with k-th term b k = k - 1 = 1 k , which is harmonic.
In order to consolidate the material received, we will consider in detail a couple of typical options.
Example 8
Determine what the series is ∑ k = 1 ∞ 1 k - 1 2 .
Since the 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 proved 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 met, since lim k → + ∞ 1 k 3 + 3 k - 1 = 0 . We represent in the form of 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 feature, we can conclude that the number series is convergent.
Example 10
Determine what the series is ∑ k = 3 ∞ 1 k ln (ln k) . lim k → + ∞ 1 k ln (ln k) = 1 + ∞ + ∞ = 0 .
In this variant, it is possible to mark the fulfillment of the desired condition. Let's define a series for comparison. For example, ∑ k = 1 ∞ 1 k s . To determine what a degree is, consider the sequence ( ln (ln k) ) , k = 3 , 4 , 5 . . . . The members of the sequence ln (ln 3) , ln (ln 4) , ln (ln 5) , . . . increases to infinity. After analyzing the equation, it can be noted that, taking N = 1619 as a value, then the terms of the sequence > 2 . For this sequence, the inequality 1 k ln (ln k)< 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 sign 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 with the help of the second feature. For comparison ∑ k = 1 ∞ b k we take the convergent series ∑ k = 1 ∞ 1 k 3 . 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 is ∑ n = 1 ∞ k 2 + 3 4 k 3 + 5 .
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, the divergent series entails the divergence of the original series.
Third sign
Consider the third sign of comparison.
Assume that ∑ k = 1 ∞ a k and _ ∑ k = 1 ∞ b k are positive-sign numerical series. If the condition is satisfied for some 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 .
Sign of d'Alembert
Imagine that ∑ k = 1 ∞ a k is a positive sign 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.
Remark 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 test 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 by the d'Alembert 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 + ∞ log 2 = 0
We can see that the condition is met. Let's use the d'Alembert 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 if the series is divergent ∑ k = 1 ∞ k k k ! .
Let's use the d'Alembert 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.
Cauchy's radical sign
Assume that ∑ k = 1 ∞ a k is a series with positive sign. If lim k → + ∞ a k k< 1 , то ряд является сходящимся, если lim k → + ∞ a k k >1 , then divergent.
Remark 2
If lim k → + ∞ a k k = 1 , then this feature 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 characteristic when a member of the number series is an exponential exponential expression.
In order to consolidate the information received, consider a few typical examples.
Example 14
Determine whether the positive series ∑ k = 1 ∞ 1 (2 k + 1) k converges on the series.
The necessary condition is considered to be satisfied, since lim k → + ∞ 1 (2 k + 1) k = 1 + ∞ + ∞ = 0 .
According to the test considered above, we get 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 converge ∑ k = 1 ∞ 1 3 k · 1 + 1 k k 2 .
We use the sign 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
Assume that ∑ k = 1 ∞ a k is a series with positive sign. It is necessary to designate a function of a continuous argument y=f(x), which matches a n = f (n) . If y=f(x) greater than zero, does not break and decreases on [ a ; + ∞) , where a ≥ 1
Then, if the improper integral ∫ a + ∞ f (x) d x is convergent, then the series under consideration also converges. If it diverges, then in the example under consideration the series also diverges.
When checking the decay of a function, you can use the material discussed in previous lessons.
Example 16
Consider the example ∑ k = 2 ∞ 1 k ln k for convergence.
The convergence condition 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 features of the principle that we considered above. We 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 because 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 , the condition is considered to be 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. Thus, we can determine that the original expression is also convergent.
Let us proceed 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 does not terminate and decreases on [ 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.
Sign of Raabe
Let us assume that ∑ k = 1 ∞ a k is a positive-sign 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 method of determination can be used if the techniques described above do not give visible results.
Study for absolute convergence
For research we take ∑ k = 1 ∞ b k . We use the positive sign ∑ k = 1 ∞ b k . We can use any of the suitable features we have 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 the d'Alembert test and the radical Cauchy test will help to draw conclusions about ∑ k = 1 ∞ b k from the divergence from the modules ∑ k = 1 ∞ b k . The series ∑ k = 1 ∞ b k also diverges if the necessary convergence condition is not met, 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 k-th term is represented as b k = k ! 7k.
Let's explore the series ∑ k = 1 ∞ b k = ∑ k = 1 ∞ k ! 7 k for convergence by the d'Alembert criterion: lim k → + ∞ b k + 1 b k = lim k → + ∞ (k + 1) ! 7k + 1k ! 7 k = 1 7 limk → + ∞ (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.
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 satisfied, therefore ∑ k = 1 ∞ (- 1) k · k 2 + 1 ln (k + 1) the series is divergent. The limit was calculated according to L'Hospital's rule.
Conditional convergence criteria
Leibniz sign
Definition 12If the values of the members of the alternating series decrease b 1 > b 2 > b 3 > . . . > . . . and 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 get 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: the 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 conditionally converges.
Abel-Dirichlet sign
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.
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 , . . . - 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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Expressions can consist of functions (notations are given in alphabetical order): absolute(x) Absolute value x
(module x or |x|)
arccos(x) Function - arc cosine of x arccosh(x) Arc cosine hyperbolic from x arcsin(x) Arcsine from x arcsinh(x) Arcsine hyperbolic from x arctg(x) Function - arc tangent from x arctgh(x) The arc tangent is hyperbolic from x e e a number that is approximately equal to 2.7 exp(x) Function - exponent from x(which is e^x)
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(To obtain log7(x), you need to enter log(x)/log(7) (or, for example, for log10(x)=log(x)/log(10)) pi The number is "Pi", which is approximately equal to 3.14 sin(x) Function - Sine of x cos(x) Function - Cosine of x sinh(x) Function - Hyperbolic sine of x cash(x) Function - Hyperbolic cosine of x sqrt(x) Function - Square root from x sqr(x) or x^2 Function - Square x tg(x) Function - Tangent from x tgh(x) Function - Hyperbolic tangent of x cbrt(x) The function is the cube root of x
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