Denis Valentinovich Manturov Minister of Industry and Trade of the Russian ...
There are several ways to check the convergence of the series. First, you can simply find the sum of the series. If as a result we get a finite number, then such the series converges... For example, since
then the series converges. If we could not 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 sign
here and, respectively, the n-th 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, let us investigate the convergence of a series using the d'Alembert test. First, let's write expressions for and. Now let's find the corresponding limit:
Because, in accordance with the d'Alembert sign, the series converges.
Another method to check the convergence of the series is radical Cauchy sign which is written as follows:
here is the nth term of the series, and the 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, let us investigate the convergence of a series using the radical Cauchy test. First, let's write the expression for. Now let's find the corresponding limit:
Since title = "15625/64> 1">, in accordance with the radical Cauchy sign, the series diverges.
It is worth noting that along with the above, there are other criteria for the convergence of series, such as the Cauchy integral 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. Moreover, if the calculator gives a specific number as the sum of a series, then the series converges. Otherwise, it is necessary to pay attention to the item "Series convergence test". 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 values of the "Series convergence test" item:
| Text on English language | Russian text |
|---|---|
| By the harmonic series test, the series diverges. | When comparing the investigated series with the harmonic series, the original series diverges. |
| The ratio test is inconclusive. | The d'Alembert test cannot give an answer about the convergence of the series. |
| The root test is inconclusive. | Cauchy's radical criterion cannot give an answer about the convergence of the series. |
| By the comparison test, the series converges. | On the basis of comparison, the series converges |
| By the ratio test, the series converges. | On the basis of d'Alembert, the series converges |
| By the limit test, the series diverges. | Based on the fact that title = "(! LANG: The limit of the n-th member of the series at n-> oo is not equal to zero or does not exist"> , или указанный предел не существует, сделан вывод о том, что ряд расходится. !} |
This article is a structured and detailed information, which can come in handy during the analysis of exercises and tasks. We will look at the topic of number series.
This article starts with basic definitions and concepts. Next, we will standard options and explore 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, 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 in which q = - 0. 5: 8 - 4 + 2 - 1 + 1 2 - 1 4 +. ... ... = ∑ k = 1 ∞ (- 16) - 1 2 k.
Definition 2
a k is generic or k th a member of the series.
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 endless sequence of numbers.
For a number 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 converging if 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 converging 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 nn → + ∞ = lim 16 3 t → + ∞ 1 - 1 2 n = 16 3 lim n → + ∞ 1 - - 1 2 n = 16 3, the 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 diverging 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 determined by the expression S n = a 1 (1 - qn) 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 diverging numerical series is a sum of the form ∑ k = 1 ∞ 5 = 5 + 5 +. ... ... ... In this case, the n-th 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 that ∑ k = 1 ∞ = 1 + 1 2 + 1 3 +. ... ... + 1 n +. ... ... - this is harmonic number series.
Definition 7
The 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 with 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 by the method from the opposite. If it converges, then the limit is finite. The equation can be written 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 hold: 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 row 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 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 qn - 1 q - 1 = b 1 lim n → + ∞ qnq - 1 - lim n → + ∞ 1 q - 1 = = b 1 0 - 1 q - 1 = b 1 q - 1
We 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 row 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 will make sure that there is no limit and the series is divergent.
For q> 1, lim n → + ∞ S n = lim n → + ∞ b 1 (qn - 1) q - 1 = b 1 lim n → + ∞ qnq - 1 - lim n → + ∞ 1 q - 1 = = b 1 ∞ - 1 q - 1 = ∞
We 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 obtain ∑ 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 selected row diverges at s< 1 .
It is necessary to provide evidence that the series ∑ k = 1 ∞ 1 k s converges for s> 1.
Let's represent 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
Suppose 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 represent 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 + . . .
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 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. 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 terms> 0 a k> 0, k = 1, 2,. ... ... ...
Series ∑ k = 1 ∞ b k alternating if the signs of the numbers are different. This example is presented as ∑ k = 1 ∞ bk = ∑ k = 1 ∞ (- 1) k ak or ∑ k = 1 ∞ bk = ∑ k = 1 ∞ (- 1) k + 1 ak, where ak> 0 , k = 1, 2,. ... ... ...
Series ∑ k = 1 ∞ b k alternating, since it contains a lot of numbers, negative and positive.
The second option row is a special case of the third option.
Let's give 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 converges absolutely if ∑ k = 1 ∞ b k is also assumed to be convergent.
Let's take a closer look at several typical options.
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 converging, then it is true to assume that 6 + 3 + 3 2 + 3 4 + 3 8 + 3 16 +. ... ...
Definition 10
An alternating series ∑ k = 1 ∞ b k is considered to be conditionally convergent if ∑ k = 1 ∞ b k is divergent, and the series ∑ k = 1 ∞ b k is considered to be convergent.
Example 3
Let us analyze 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 to be 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 converging 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 to be convergent. It can be noted that the row without m members are also considered to be convergent. If we add several numbers to ∑ k = m + 1 ∞ a k, then the result will also converge.
- 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 also, then the series ∑ k = 1 ∞ a k + b k and ∑ k = 1 ∞ a k - b k also converge. The amounts will equal A + B and A - B respectively.
Determine that the series converges ∑ k = 1 ∞ 2 3 k · k 3.
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 to be 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 ∑ n = 1 ∞ 3 + n n 5 2 converges.
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 row is recognized as converging according to its property. Since the rows converge, then the original version too.
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 row converges, since it is one of the members of the numerical sequence. According to the third property, we can calculate that the original variant 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 a decreasing number 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
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 sine qua non condition. If it is not met, 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 a 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 us 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 equal to 0. We have proved that the given series diverges.
How to determine the convergence of a positive series.
If you constantly use these features, you will have to constantly calculate the limits. This section will help you avoid complications while solving examples and problems. In order to determine the convergence of a positive series, there is a certain condition.
For the convergence of a positive ∑ k = 1 ∞ a k, a k> 0 ∀ k = 1, 2, 3,. ... ... it is necessary to define a limited sequence of amounts.
How to compare ranks
There are several indications of a series comparison. We compare the series, the convergence of which is proposed to be determined, with the series, the convergence of which is known.
The 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 obtain ∑ 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 strong argument that will help determine convergence. Difficulties may lie in the fact that it is not possible to find a suitable example for comparison in every case. Quite often, the series is selected according to the principle that the indicator k-th term will be equal to the result of subtracting exponents of the numerator and denominator k-th member of the series. Suppose that a k = k 2 + 3 4 k 2 + 5, the difference will be 2 – 3 = - 1 ... 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 resulting material, let's take a closer look at a couple of typical options.
Example 8
Determine what the series ∑ k = 1 ∞ 1 k - 1 2 is.
Since the limit = 0 lim k → + ∞ 1 k - 1 2 = 0, we performed necessary condition... The inequality will be valid 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 variant 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 as 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 sign, 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.
V 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. After analyzing the equation, it can be noted that taking N = 1619 as the value, then the members of the sequence are> 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
Suppose that ∑ k = 1 ∞ a k and ∑ k = 1 ∞ b k are positive numerical series.
If lim k → + ∞ a k b k ≠ ∞, then the series ∑ k = 1 ∞ b k converges, and ∑ k = 1 ∞ a k converges as well.
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 feature. For comparison, ∑ k = 1 ∞ b k take a converging series ∑ k = 1 ∞ 1 k 3. We 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 converging 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 fulfilled in this variant. 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 diverging series entails a divergence of the original series.
Third sign
Consider the third feature of comparison.
Suppose that ∑ k = 1 ∞ a k and _ ∑ k = 1 ∞ b k are positive 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 implies the divergence ∑ k = 1 ∞ b k.
D'Alembert sign
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.
Remark 1
The d'Alembert 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 the d'Alembert 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 us calculate the limit using L'Hôpital'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. We 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 whether the series is divergent ∑ k = 1 ∞ k k k! ...
We will 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 kk (k + 1) = = lim k → + ∞ (k + 1) kkk = lim k → + ∞ k + 1 kk = lim k → + ∞ 1 + 1 kk = e> 1
Therefore, the series is divergent.
Cauchy's radical sign
Suppose that ∑ k = 1 ∞ a k is a positive series. 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 attribute does not provide any information - additional analysis is required.
This feature can be used in examples that are easy to define. The case will be typical when a member of a numerical series is an exponential exponential expression.
In order to consolidate the information received, we will consider several typical examples.
Example 14
Determine if the positive series ∑ k = 1 ∞ 1 (2 k + 1) k is convergent.
The required condition is considered to be satisfied, since lim k → + ∞ 1 (2 k + 1) k = 1 + ∞ + ∞ = 0.
According to the criterion considered 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
Suppose that ∑ k = 1 ∞ a k is a positive series. It is necessary to denote the function of continuous argument y = f (x) which matches a n = f (n). If y = f (x) is 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 also converges. If it diverges, then in the example under consideration the series also diverges.
When checking the decreasing function, you can use the material discussed in the previous lessons.
Example 16
Consider the example ∑ k = 2 ∞ 1 k · ln k for convergence.
The condition of 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 one should dwell in more detail. Find the derivative: y "= 1 x ln x" = x ln x "x ln x 2 = ln x + x 1 xx 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 + ∞ dxx 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, 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 to be 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.
We 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 is not interrupted and decreases by [4; + ∞). We use the feature described in the previous paragraph:
∫ 4 + ∞ dx (5 x + 8) (ln (5 x + 8)) 3 = lim A → + ∞ ∫ 4 A dx (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 + ∞ dx (5 x + 8) (ln (5 x + 8)) 3, we can define that ∑ k = 4 ∞ 1 (5 k + 8) (ln (5 k + 8 )) 3 also converges.
Raabe's sign
Suppose that ∑ k = 1 ∞ a k is a sign-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 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. Use the positive ∑ k = 1 ∞ b k. We can use any of the appropriate traits 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 feature: 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 satisfied, that is, if lim k → ∞ + b k ≠ 0.
Example 19
Check the 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! 7 k.
Let us examine the series ∑ k = 1 ∞ b k = ∑ k = 1 ∞ k! 7 k for the d'Alembert convergence: 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) converging.
Consider the necessary condition lim k → + ∞ bk = 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) is a divergent series. The limit was calculated according to L'Hôpital's rule.
Tests for conditional convergence
Leibniz's sign
Definition 12If the values of the members 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 required 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 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.
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,. ... ... - nonincreasing, and the sequence (v k) = 1, - 3, 2, 1, - 3, 2,. ... ... bounded (S k) = 1, - 2, 0, 1, - 2, 0,. ... ... ... The series converges.
If you notice an error in the text, please select it and press Ctrl + Enter
Answer: the row diverges.
Example No. 3
Find the sum of the series $ \ sum \ limits_ (n = 1) ^ (\ infty) \ frac (2) ((2n + 1) (2n + 3)) $.
Since the lower limit of summation is 1, the common term of the series is written under the sum sign: $ u_n = \ frac (2) ((2n + 1) (2n + 3)) $. Let's compose the n-th partial sum of the series, i.e. Let's sum up the first $ n $ members of a given numerical series:
$$ S_n = u_1 + u_2 + u_3 + u_4 + \ ldots + u_n = \ frac (2) (3 \ cdot 5) + \ frac (2) (5 \ cdot 7) + \ frac (2) (7 \ cdot 9 ) + \ frac (2) (9 \ cdot 11) + \ ldots + \ frac (2) ((2n + 1) (2n + 3)). $$
Why I write exactly $ \ frac (2) (3 \ cdot 5) $, and not $ \ frac (2) (15) $, it will be clear from the further narration. However, recording a partial amount did not bring us one iota closer to our goal. We need to find $ \ lim_ (n \ to \ infty) S_n $, but if we just write:
$$ \ lim_ (n \ to \ infty) S_n = \ lim_ (n \ to \ infty) \ left (\ frac (2) (3 \ cdot 5) + \ frac (2) (5 \ cdot 7) + \ frac (2) (7 \ cdot 9) + \ frac (2) (9 \ cdot 11) + \ ldots + \ frac (2) ((2n + 1) (2n + 3)) \ right), $$
then this record, completely correct in form, will give us nothing in essence. To find the limit, the expression for the partial sum must first be simplified.
To do this, there is a standard transformation, which consists in expanding the fraction $ \ frac (2) ((2n + 1) (2n + 3)) $, which represents the common term of the series, into elementary fractions. Decomposition rational fractions a separate topic is devoted to elementary ones (see, for example, example # 3 on this page). Expanding the fraction $ \ frac (2) ((2n + 1) (2n + 3)) $ into elementary fractions, we will have:
$$ \ frac (2) ((2n + 1) (2n + 3)) = \ frac (A) (2n + 1) + \ frac (B) (2n + 3) = \ frac (A \ cdot (2n +3) + B \ cdot (2n + 1)) ((2n + 1) (2n + 3)). $$
We equate the numerators of the fractions on the left and right sides of the resulting equality:
$$ 2 = A \ cdot (2n + 3) + B \ cdot (2n + 1). $$
There are two ways to find the values of $ A $ and $ B $. You can expand the parentheses and rearrange the terms, or you can simply substitute some suitable values for $ n $. Strictly for a change, in this example we will go the first way, and the next - we will substitute the private values of $ n $. Expanding the brackets and rearranging the terms, we get:
$$ 2 = 2An + 3A + 2Bn + B; \\ 2 = (2A + 2B) n + 3A + B. $$
There is a zero on the left-hand side of the equality before $ n $. If you like, the left-hand side of the equality can be represented as $ 0 \ cdot n + 2 $ for clarity. Since on the left side of the equality before $ n $ there is zero, and on the right side of the equality before $ n $ there is $ 2A + 2B $, we have the first equation: $ 2A + 2B = 0 $. We immediately divide both sides of this equation by 2, after which we get $ A + B = 0 $.
Since the free term is equal to 2 on the left side of the equality, and the free term is equal to $ 3A + B $ on the right side of the equality, then $ 3A + B = 2 $. So, we have a system:
$$ \ left \ (\ begin (aligned) & A + B = 0; \\ & 3A + B = 2. \ end (aligned) \ right. $$
The proof will be carried out by the method of mathematical induction. At the first step, it is necessary to check whether the equality being proved holds: $ S_n = \ frac (1) (3) - \ frac (1) (2n + 3) $ for $ n = 1 $. We know that $ S_1 = u_1 = \ frac (2) (15) $, but will the expression $ \ frac (1) (3) - \ frac (1) (2n + 3) $ give the value $ \ frac (2 ) (15) $, if you substitute $ n = 1 $? Let's check:
$$ \ frac (1) (3) - \ frac (1) (2n + 3) = \ frac (1) (3) - \ frac (1) (2 \ cdot 1 + 3) = \ frac (1) (3) - \ frac (1) (5) = \ frac (5-3) (15) = \ frac (2) (15). $$
So, for $ n = 1 $, the equality $ S_n = \ frac (1) (3) - \ frac (1) (2n + 3) $ holds. This completes the first step of the method of mathematical induction.
Suppose that for $ n = k $ the equality holds, that is, $ S_k = \ frac (1) (3) - \ frac (1) (2k + 3) $. Let us prove that the same equality will hold for $ n = k + 1 $. To do this, consider $ S_ (k + 1) $:
$$ S_ (k + 1) = S_k + u_ (k + 1). $$
Since $ u_n = \ frac (1) (2n + 1) - \ frac (1) (2n + 3) $, then $ u_ (k + 1) = \ frac (1) (2 (k + 1) + 1) - \ frac (1) (2 (k + 1) +3) = \ frac (1) (2k + 3) - \ frac (1) (2 (k + 1) +3) $. According to the above assumption, $ S_k = \ frac (1) (3) - \ frac (1) (2k + 3) $, therefore the formula $ S_ (k + 1) = S_k + u_ (k + 1) $ takes the form:
$$ S_ (k + 1) = S_k + u_ (k + 1) = \ frac (1) (3) - \ frac (1) (2k + 3) + \ frac (1) (2k + 3) - \ frac (1) (2 (k + 1) +3) = \ frac (1) (3) - \ frac (1) (2 (k + 1) +3). $$
Conclusion: the formula $ S_n = \ frac (1) (3) - \ frac (1) (2n + 3) $ is correct for $ n = k + 1 $. Therefore, according to the method of mathematical induction, the formula $ S_n = \ frac (1) (3) - \ frac (1) (2n + 3) $ is true for any $ n \ in N $. Equality is proven.
In the standard course of higher mathematics, they are usually satisfied with "crossing out" the canceling terms without requiring any proof. So, we got the expression for nth partial sums: $ S_n = \ frac (1) (3) - \ frac (1) (2n + 3) $. Find the value of $ \ lim_ (n \ to \ infty) S_n $:
Conclusion: the given series converges and its sum is $ S = \ frac (1) (3) $.
The second way to simplify the formula for the partial sum.
Honestly, I myself prefer this method :) Let's write down the partial sum in an abbreviated form:
$$ S_n = \ sum \ limits_ (k = 1) ^ (n) u_k = \ sum \ limits_ (k = 1) ^ (n) \ frac (2) ((2k + 1) (2k + 3)). $$
We got earlier that $ u_k = \ frac (1) (2k + 1) - \ frac (1) (2k + 3) $, so:
$$ S_n = \ sum \ limits_ (k = 1) ^ (n) \ frac (2) ((2k + 1) (2k + 3)) = \ sum \ limits_ (k = 1) ^ (n) \ left (\ frac (1) (2k + 1) - \ frac (1) (2k + 3) \ right). $$
The sum of $ S_n $ contains a finite number of terms, so we can rearrange them as we please. I want to first add all the terms of the form $ \ frac (1) (2k + 1) $, and only then go to the terms of the form $ \ frac (1) (2k + 3) $. This means that we will represent the partial amount in the following form:
$$ S_n = \ frac (1) (3) - \ frac (1) (5) + \ frac (1) (5) - \ frac (1) (7) + \ frac (1) (7) - \ frac (1) (9) + \ frac (1) (9) - \ frac (1) (11) + \ ldots + \ frac (1) (2n + 1) - \ frac (1) (2n + 3) = \\ = \ frac (1) (3) + \ frac (1) (5) + \ frac (1) (7) + \ frac (1) (9) + \ ldots + \ frac (1) (2n + 1 ) - \ left (\ frac (1) (5) + \ frac (1) (7) + \ frac (1) (9) + \ ldots + \ frac (1) (2n + 3) \ right). $$
Of course, the expanded notation is extremely inconvenient, so the equality presented above can be formatted more compactly:
$$ S_n = \ sum \ limits_ (k = 1) ^ (n) \ left (\ frac (1) (2k + 1) - \ frac (1) (2k + 3) \ right) = \ sum \ limits_ ( k = 1) ^ (n) \ frac (1) (2k + 1) - \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2k + 3). $$
Now we transform the expressions $ \ frac (1) (2k + 1) $ and $ \ frac (1) (2k + 3) $ to the same form. I think it is convenient to bring to the form of a larger fraction (although it is possible to reduce it, this is a matter of taste). Since $ \ frac (1) (2k + 1)> \ frac (1) (2k + 3) $ (the larger the denominator, the smaller the fraction), we will reduce the fraction $ \ frac (1) (2k + 3) $ to the form $ \ frac (1) (2k + 1) $.
I will represent the expression in the denominator of the fraction $ \ frac (1) (2k + 3) $ as follows:
$$ \ frac (1) (2k + 3) = \ frac (1) (2k + 2 + 1) = \ frac (1) (2 (k + 1) +1). $$
And the sum $ \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2k + 3) $ can now be written like this:
$$ \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2k + 3) = \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2 (k + 1 ) +1) = \ sum \ limits_ (k = 2) ^ (n + 1) \ frac (1) (2k + 1). $$
If equality $ \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2k + 3) = \ sum \ limits_ (k = 2) ^ (n + 1) \ frac (1) (2k + 1) $ does not raise questions, then let's go further. If you have any questions, please expand the note.
How did we get the converted amount? show \ hide
We had a series $ \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2k + 3) = \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2 ( k + 1) +1) $. Let's introduce a new variable instead of $ k + 1 $ - for example, $ t $. So, $ t = k + 1 $.
How did the old $ k $ variable change? And it changed from 1 to $ n $. Let's find out how the new $ t $ variable will change. If $ k = 1 $, then $ t = 1 + 1 = 2 $. If $ k = n $, then $ t = n + 1 $. So the expression $ \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2 (k + 1) +1) $ is now $ \ sum \ limits_ (t = 2) ^ (n +1) \ frac (1) (2t + 1) $.
$$ \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2 (k + 1) +1) = \ sum \ limits_ (t = 2) ^ (n + 1) \ frac (1 ) (2t + 1). $$
We have the sum $ \ sum \ limits_ (t = 2) ^ (n + 1) \ frac (1) (2t + 1) $. The question is: does it really matter which letter to use in this amount? :) Trite writing the letter $ k $ instead of $ t $, we get the following:
$$ \ sum \ limits_ (t = 2) ^ (n + 1) \ frac (1) (2t + 1) = \ sum \ limits_ (k = 2) ^ (n + 1) \ frac (1) (2k +1). $$
This is how we get the equality $ \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2 (k + 1) +1) = \ sum \ limits_ (k = 2) ^ (n + 1) \ frac (1) (2k + 1) $.
Thus, the partial amount can be represented as follows:
$$ S_n = \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2k + 1) - \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2k + 3 ) = \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2k + 1) - \ sum \ limits_ (k = 2) ^ (n + 1) \ frac (1) (2k + 1 ). $$
Note that the sums $ \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2k + 1) $ and $ \ sum \ limits_ (k = 2) ^ (n + 1) \ frac (1 ) (2k + 1) $ differ only in the limits of summation. Let's make these limits the same. Taking the first element from the sum $ \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2k + 1) $ we will have:
$$ \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2k + 1) = \ frac (1) (2 \ cdot 1 + 1) + \ sum \ limits_ (k = 2) ^ (n) \ frac (1) (2k + 1) = \ frac (1) (3) + \ sum \ limits_ (k = 2) ^ (n) \ frac (1) (2k + 1). $$
Taking the last element from the sum $ \ sum \ limits_ (k = 2) ^ (n + 1) \ frac (1) (2k + 1) $, we get:
$$ \ sum \ limits_ (k = 2) ^ (n + 1) \ frac (1) (2k + 1) = \ sum \ limits_ (k = 2) ^ (n) \ frac (1) (2k + 1 ) + \ frac (1) (2 (n + 1) +1) = \ sum \ limits_ (k = 2) ^ (n) \ frac (1) (2k + 1) + \ frac (1) (2n + 3). $$
Then the expression for the partial sum will take the form:
$$ S_n = \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2k + 1) - \ sum \ limits_ (k = 2) ^ (n + 1) \ frac (1) (2k +1) = \ frac (1) (3) + \ sum \ limits_ (k = 2) ^ (n) \ frac (1) (2k + 1) - \ left (\ sum \ limits_ (k = 2) ^ (n) \ frac (1) (2k + 1) + \ frac (1) (2n + 3) \ right) = \\ = \ frac (1) (3) + \ sum \ limits_ (k = 2) ^ (n) \ frac (1) (2k + 1) - \ sum \ limits_ (k = 2) ^ (n) \ frac (1) (2k + 1) - \ frac (1) (2n + 3) = \ frac (1) (3) - \ frac (1) (2n + 3). $$
If we skip all the explanations, then the process of finding an abbreviated formula for the n-th partial sum will take the following form:
$$ S_n = \ sum \ limits_ (k = 1) ^ (n) u_k = \ sum \ limits_ (k = 1) ^ (n) \ frac (2) ((2k + 1) (2k + 3)) = \ sum \ limits_ (k = 1) ^ (n) \ left (\ frac (1) (2k + 1) - \ frac (1) (2k + 3) \ right) = \\ = \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2k + 1) - \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2k + 3) = \ frac (1) (3) + \ sum \ limits_ (k = 2) ^ (n) \ frac (1) (2k + 1) - \ left (\ sum \ limits_ (k = 2) ^ (n) \ frac (1) (2k + 1 ) + \ frac (1) (2n + 3) \ right) = \ frac (1) (3) - \ frac (1) (2n + 3). $$
Let me remind you that we reduced the fraction $ \ frac (1) (2k + 3) $ to the form $ \ frac (1) (2k + 1) $. Of course, you can do the opposite, i.e. represent the fraction $ \ frac (1) (2k + 1) $ as $ \ frac (1) (2k + 3) $. The final expression for the partial amount will not change. In this case, I will hide the process of finding a partial sum under a note.
How to find $ S_n $ if we reduce it to another fraction? show \ hide
$$ S_n = \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2k + 1) - \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2k + 3 ) = \ sum \ limits_ (k = 0) ^ (n-1) \ frac (1) (2k + 3) - \ sum \ limits_ (k = 1) ^ (n) \ frac (1) (2k + 3 ) = \\ = \ frac (1) (3) + \ sum \ limits_ (k = 1) ^ (n-1) \ frac (1) (2k + 3) - \ left (\ sum \ limits_ (k = 1) ^ (n-1) \ frac (1) (2k + 3) + \ frac (1) (2n + 3) \ right) = \ frac (1) (3) - \ frac (1) (2n + 3). $$
So $ S_n = \ frac (1) (3) - \ frac (1) (2n + 3) $. Find the limit $ \ lim_ (n \ to \ infty) S_n $:
$$ \ lim_ (n \ to \ infty) S_n = \ lim_ (n \ to \ infty) \ left (\ frac (1) (3) - \ frac (1) (2n + 3) \ right) = \ frac (1) (3) -0 = \ frac (1) (3). $$
The given series converges and its sum is $ S = \ frac (1) (3) $.
Answer: $ S = \ frac (1) (3) $.
The continuation of the topic of finding the sum of a series will be considered in the second and third parts.
Harmonic series- a sum made up of an infinite number of terms, reciprocal 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).Collegiate 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. Simplest properties
✪ SERIES. Overview
Subtitles
The sum of the first n members of the series
Individual members of the series tend to zero, but its sum diverges. n-th partial sum s n of a harmonic series is called the n-th harmonic number:
sn = ∑ 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's formula
When the 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 a harmonic series:
sn ≍ 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 kn 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 for 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))
Divergence of the series
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 telescopic row :
vn = 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
Divergence can be proved 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 is obviously diverging. This proof belongs to a medieval scholar Nikolay Orem(about 1350).
Alternative proof of divergence
we invite the reader to be convinced of 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 an integer and no harmonic number other than H 1 = 1 (\ displaystyle H_ (1) = 1) is not whole.
Related rows
Dirichlet series
Generalized harmonic series (or near Dirichlet) call 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 series of order α (\ displaystyle \ alpha) equal to the value the Riemann zeta function :
∑ k = 1 ∞ 1 k α = ζ (α) (\ displaystyle \ sum _ (k = 1) ^ (\ infty) (\ frac (1) (k ^ (\ alpha))) = \ zeta (\ alpha ))For even, 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 is the relation ζ (1 + 1 n) ∼ n (\ displaystyle \ zeta (1 + (\ frac (1) (n))) \ sim n). Therefore, they say that such a series possesses with 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 the harmonic series, in which only the 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), less and less terms are taken for the sum of the "thinned" series. That is, in the end, the overwhelming majority of the members forming the sum of the harmonic series are discarded so as not to exceed the geometric progression bounding from above.
