Determine the convergence of the d'Alembert series. Numerical series: definitions, properties, convergence criteria, examples, solutions

d'Alembert's convergence criterion Cauchy's radical convergence criterion Cauchy's integral convergence criterion

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

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

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

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

The main prerequisites for applying the d'Alembert sign are as follows:

1) The common member of the series (“stuffing” of the series) includes some number in the degree, for example, , and so on. Moreover, it does not matter at all where this thing is located, in the numerator or in the denominator - it is important that it is present there.

2) The common term of the series includes the factorial. With factorials, we crossed swords in the lesson Number sequence and its limit. However, it does not hurt to spread the self-assembly tablecloth again:

…

…

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

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

Along with powers and (and) factorials, polynomials are often found in the filling of the series, this does not change things - you need to use the d'Alembert test.

In addition, in the general term of the series, both the degree and the factorial can occur at the same time; there may be two factorials, two degrees, it is important that there be at least some of considered points - and this is just a prerequisite for using the d'Alembert sign.

Sign of d'Alembert: Consider positive number series. If there is a limit of the ratio of the next term to the previous one: , then:
a) At a row converges. In particular, the series converges for .
b) At a row diverges. In particular, the series diverges at .
c) When sign does not respond. You need to use another sign. Most often, a unit is obtained when they try to apply the d'Alembert test where it is necessary to use the limit comparison test.

If you still have problems with limits or misunderstanding of limits, please refer to the lesson Limits. Solution examples. Without an understanding of the limit and the ability to reveal the uncertainty further, unfortunately, one cannot move forward.

And now the long-awaited examples.

Example 1

We see that in the common term of the series we have , and this is the correct premise that we need to use the d'Alembert test. First, a complete solution and a design sample, comments below.

We use the d'Alembert test:

converges.

(1) Compose the ratio of the next member of the series to the previous one: . From the condition, we see that the common term of the series . In order to get the next member of the series, it is necessary substitute instead: .
(2) Get rid of the four-story fraction. With some experience in solving this step, you can skip it.
(3) Open the brackets in the numerator. In the denominator we take out the four from the degree.
(4) Reduce by . We take out the constant beyond the sign of the limit. In the numerator, we give like terms in parentheses.
(5) Uncertainty is eliminated in the standard way - by dividing the numerator and denominator by "en" to the highest degree.
(6) Divide the numerators by the denominators term by term, and indicate the terms that tend to zero.
(7) We simplify the answer and make a note that with the conclusion that, according to the d'Alembert criterion, the series under study converges.

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

Example 2

Take a similar series and examine it for convergence

First the full solution, then the comments:

We use the d'Alembert test:

Thus, the series under study converges.

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

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

Example 3

Examine the series for convergence

Complete solution and design sample at the end of the lesson on number sequences.
(4) Reduce everything that can be reduced.
(5) We move the constant beyond the sign of the limit. Open the parentheses in the numerator.
(6) Uncertainty is eliminated in a standard way - by dividing the numerator and denominator by "en" to the highest degree.

Example 5

Examine the series for convergence

Full solution and design sample at the end of the lesson

Example 6

Examine the series for convergence

Sometimes there are rows that contain a “chain” of multipliers in their filling; we have not yet considered this type of row. How to explore a series with a "chain" of factors? Use the sign of d'Alembert. But first, to understand what is happening, we will write a series in detail:

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

An example solution might look like this:

We use the d'Alembert test:

Thus, the series under study converges.

Signs of convergence of series.
Sign of d'Alembert. Signs of Cauchy

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

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

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

d'Alembert convergence test

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

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

The main prerequisites for applying the d'Alembert sign are as follows:

1) The common member of the series ("stuffing" of the series) includes some number in the degree, for example,,, and so on. Moreover, it does not matter at all where this thing is located, in the numerator or in the denominator - it is important that it is present there.

2) The common term of the series includes the factorial. We crossed swords with factorials in the lesson Numerical sequence and its limit. However, it does not hurt to spread the self-assembly tablecloth again:

…

…

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

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

Along with powers and (and) factorials, polynomials are often found in the filling of the series, this does not change things - you need to use the d'Alembert test.

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

Sign of d'Alembert: Consider positive number series. If there is a limit of the ratio of the next term to the previous one: , then:
a) At a row converges
b) At a row diverges
c) When sign does not respond. You need to use another sign. Most often, a unit is obtained when they try to apply the d'Alembert test where it is necessary to use the limit comparison test.

And now the long-awaited examples.

Example 1

We see that in the common term of the series we have , and this is the correct premise that we need to use the d'Alembert test. First, a complete solution and a design sample, comments below.

We use the d'Alembert test:

converges.
(1) Compose the ratio of the next member of the series to the previous one: . From the condition, we see that the common term of the series . In order to get the next member of the series, you need INSTEAD substitute: .
(2) Get rid of the four-story fraction. With some experience in solving this step, you can skip it.
(3) Open the brackets in the numerator. In the denominator we take out the four from the degree.
(4) Reduce by . We take out the constant beyond the sign of the limit. In the numerator, we give like terms in parentheses.
(5) Uncertainty is eliminated in the standard way - by dividing the numerator and denominator by "en" to the highest degree.
(6) Divide the numerators by the denominators term by term, and indicate the terms that tend to zero.
(7) We simplify the answer and make a note that with the conclusion that, according to the d'Alembert criterion, the series under study converges.

Example 2

Take a similar series and examine it for convergence

First the full solution, then the comments:

We use the d'Alembert test:

Thus, the series under study converges.

(1) Compose the ratio .

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

Example 3

Examine the series for convergence

Consider typical examples with factorials:

Example 4

Examine the series for convergence

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

Thus, the series under study diverges.
(1) Compose the ratio . We repeat again. By condition, the common term of the series: . To get the next member of the series, should be substituted instead, thus: .
(2) Get rid of the four-story fraction.
(3) We pinch off the seven from the degree. Factorials are described in detail. How to do this - see the beginning of the lesson or the article on number sequences.
(4) Reduce everything that can be reduced.
(5) We move the constant beyond the sign of the limit. Open the parentheses in the numerator.
(6) Uncertainty is eliminated in a standard way - by dividing the numerator and denominator by "en" to the highest degree.

Example 5

Examine the series for convergence

Full solution and design sample at the end of the lesson

Example 6

Examine the series for convergence

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

An example solution might look like this:

We use the d'Alembert test:

Thus, the series under study converges.

Cauchy's radical sign

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

The Cauchy convergence test for positive numerical series is somewhat similar to the d'Alembert test just considered.

Cauchy's radical sign: Consider positive number series. If there is a limit: , then:
a) At a row converges. In particular, the series converges for .
b) At a row diverges. In particular, the series diverges at .
c) When sign does not respond. You need to use another sign. It is interesting to note that if the Cauchy test does not give us an answer to the question of the convergence of the series, then the d'Alembert test will also not give an answer. But if d'Alembert's sign does not give an answer, then Cauchy's sign may well "work". That is, the Cauchy sign is in this sense a stronger sign.

When should you use the Cauchy radical sign? Cauchy's radical sign is usually used in cases where the root "good" is extracted from a common term in the series. As a rule, this pepper is in the degree, which depends on. There are still exotic cases, but we will not hammer our heads with them.

Example 7

Examine the series for convergence

We see that the fraction is completely under the degree depending on "en", which means that we need to use the radical Cauchy criterion:

Thus, the series under study diverges.

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

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

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

And here is a simpler example for an independent solution:

Example 8

Examine the series for convergence

And a couple more typical examples.

Full solution and design sample at the end of the lesson

Example 9

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

Thus, the series under study converges.

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

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

(5) We actually perform term-by-term division and indicate which terms tend to zero in our case.
(6) Uncertainty is eliminated, we are left with the simplest limit: . Why in infinitely large degree tends to zero? Because the base of the degree satisfies the inequality . If anyone has doubts about the validity of the limit , then I'm not too lazy, I'll pick up a calculator:
If , then
If , then
If , then
If , then
If , then
… etc. to infinity - that is, in the limit:

Just the same infinitely decreasing geometric progression on fingers =)
! Never use this technique as evidence! For if something is obvious, then this does not mean that it is correct.

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

Example 10

Examine the series for convergence

This is a do-it-yourself example.

Sometimes a provocative example is offered for a solution, for example:. Here in the exponent no "en", only a constant. Here you need to square the numerator and denominator (polynomials will turn out), and then follow the algorithm from the article Rows for teapots. In such an example, either the necessary criterion for the convergence of the series or the limit criterion for comparison should work.

Integral Cauchy test

Or just an integral feature. I will disappoint those who poorly learned the material of the first course. In order to apply the integral Cauchy criterion, it is necessary to more or less confidently be able to find derivatives, integrals, and also have the skill of calculating improper integral first kind.

In calculus textbooks integral Cauchy sign given mathematically rigorously, but too confused, so I will formulate the feature not too strictly, but clearly:

Consider positive number series. If there is an improper integral, then the series converges or diverges together with this integral.

And here are some examples for clarification:

Example 11

Examine the series for convergence

Almost a classic. Natural logarithm and some bullshit.

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

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

The main prerequisites for applying the d'Alembert sign are as follows:

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

…

…

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

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

Along with powers and (and) factorials, polynomials are often found in the filling of the series, this does not change things - you need to use the d'Alembert test.

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

For those who still have problems with limits or misunderstanding of limits, please refer to the topic Limits. Solution examples. Without an understanding of the limit and the ability to reveal the uncertainty further, unfortunately, one cannot move forward. And now the long-awaited examples.

Example 1
We see that in the common term of the series we have , and this is the correct premise that we need to use the d'Alembert test. First, a complete solution and a design sample, comments below.

We use the d'Alembert test:

converges.

Example 2 Take a similar series and examine it for convergence
First the full solution, then the comments:

We use the d'Alembert test:

Thus, the series under study converges.

Example 3 .

Consider typical examples with factorials:

Example 4 Examine the series for convergence

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

Thus, the series under study diverges.

(1) Compose the ratio . We repeat again. By condition, the common term of the series: . To get the next member of the series, should be substituted instead, thus: .
(2) Get rid of the four-story fraction.
(3) We pinch off the seven from the degree. Factorials are described in detail. How to do this - see the beginning of the lesson.
(4) Reduce everything that can be reduced.
(5) We move the constant beyond the sign of the limit. Open the parentheses in the numerator.
(6) Uncertainty is eliminated in a standard way - by dividing the numerator and denominator by "en" to the highest degree.

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

Example 6 Examine the series for convergence

An example solution might look like this: Using the d'Alembert test:
Thus, the series under study converges.
RADICAL CAUCHY SIGN

The Cauchy convergence test for positive numerical series is somewhat similar to the d'Alembert test just considered.

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

When should you use the Cauchy radical sign? The radical Cauchy test is usually used in cases where the common term of the series FULLY is in the degree dependent on "en". Or when the root "good" is extracted from the common member of the series. There are still exotic cases, but we will not hammer our heads with them.

Example 7 Examine the series for convergence

We see that the common term of the series is completely under the degree depending on , which means that we need to use the radical Cauchy criterion:

Thus, the series under study diverges.

(1) We make out the common term of the series under the root.
(2) We rewrite the same thing, only without the root, using the property of degrees.
(3) In the exponent, we divide the numerator by the denominator term by term, indicating that
(4) As a result, we have the uncertainty . Here you could go a long way: cube, cube, then divide the numerator and denominator by "en" in the highest degree. But in this case, there is a more efficient solution: you can divide the numerator and denominator term-by-term right below the degree-constant. To eliminate uncertainty, we divide the numerator and denominator by (highest power).
(5) We actually perform the division by term, and indicate the terms that tend to zero.
(6) We bring the answer to mind, mark that and conclude that the series diverges.

And here is a simpler example for an independent solution:

Example 8 Examine the series for convergence

And a couple more typical examples.

The full solution and sample design is below.

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

Thus, the series under study converges.

(1) Place the common term of the series under the root.
(2) We rewrite the same thing, but without the root, while opening the brackets using the abbreviated multiplication formula: .
(3) In the exponent, we divide the numerator by the denominator term by term and indicate that .
(4) An uncertainty of the form is obtained. Here you can divide the numerator by the denominator by "en" to the highest degree right in the bracket. We encountered something similar when studying second remarkable limit. But here the situation is different. If the coefficients at higher powers were the same, for example: , then the trick with term-by-term division would not have passed, and the second wonderful limit would have to be used. But we have these coefficients various(5 and 6), therefore it is possible (and necessary) to divide term by term (by the way, on the contrary - the second wonderful limit for different coefficients at higher powers does not work anymore).
(5) We actually perform term-by-term division and indicate which terms tend to zero in our case.
(6) Uncertainty is eliminated, the simplest limit remains: Why in infinitely large degree tends to zero? Because the base of the degree satisfies the inequality . If anyone has doubts about the fairness of the limit, then I will not be lazy, I will pick up a calculator:
If , then
If , then
If , then
If , then
If , then
… etc. to infinity - that is, in the limit:
(7) We indicate that and we conclude that the series converges.

Example 10 Examine the series for convergence

This is a do-it-yourself example.

I will disappoint those who poorly learned the material of the first course. In order to apply the integral Cauchy criterion, it is necessary to more or less confidently be able to find derivatives, integrals, and also have the skill of calculating improper integral first kind. In textbooks on mathematical analysis, the Cauchy integral criterion is given mathematically rigorously; let us formulate the criterion in a very primitive, but understandable way. And immediately examples for clarification.

Integral Cauchy test: Consider positive number series. This series converges or diverges

Example 11 Examine the series for convergence

Almost a classic. Natural logarithm and some bullshit.

The main prerequisite for using the integral Cauchy test is the fact that the common member of the series contains some function and its derivative. From the topic Derivative you probably remember the simplest tabular thing: , and we have just such a canonical case.

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

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

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

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

I repeat, if the material is running, then reading the paragraph will be difficult and obscure, since the application of the feature essentially boils down to calculating improper integral first kind.

The complete solution and design of the example should look something like this:

We use the integral feature:

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

Example 12 Examine the series for convergence

Solution and sample design at the end of the lesson

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

And two more examples for a snack

Example 13 Examine the series for convergence

According to the common "parameters", the common term of the series seems to be suitable for using the limit comparison criterion. You just need to open the brackets and immediately hand over to the candidate to compare this series with the convergent series as much as possible. However, I was a little cunning, the brackets may not be opened, but all the same, the solution through the limit comparison criterion will look rather pretentious.

Therefore, we use the integral Cauchy test:

The integrand is continuous on

converges together with the corresponding improper integral.

! Note:received number -is not the sum of the series!

Example 14 Examine the series for convergence

Solution and design template at the end of the section that comes to an end.

For the purpose of final and irrevocable assimilation of the topic of number series, visit the topics.

Solutions and answers:

Example 3:We use the d'Alembert test:

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

Example 5: Using the d'Alembert test: Thus, the series under study converges.

Example 8:

Thus, the series under study converges.

Example 10:
We use the radical Cauchy criterion.

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

Example 12: We use the integral sign.

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

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

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

Alternating rows. Leibniz sign. Solution examples

In order to understand the examples of this lesson, it is necessary to be well versed in positive numerical series: to understand what a series is, to know the necessary sign of convergence of the series, to be able to apply comparison signs, d'Alembert's sign, Cauchy's signs. The topic can be raised almost from scratch by sequentially studying the articles Rows for teapots and Sign of d'Alembert. Signs of Cauchy. Logically, this lesson is the third in a row, and it will allow not only to understand the alternating rows, but also to consolidate the material already covered! There will be little novelty, and it will not be difficult to master alternating rows. Everything is simple and affordable.

What is an alternating series? This is clear or almost clear already from the name itself. Immediately the simplest example. Consider the series and write it in more detail:

Now for the killer comment. The members of an alternating series alternate signs: plus, minus, plus, minus, plus, minus, etc. to infinity.
The interleaving provides a multiplier: if even, then there will be a plus sign, if odd, a minus sign. In mathematical jargon, this contraption is called a flasher. Thus, the alternating series is "identified" by minus one to the power of "en".

In practical examples, the alternation of the terms of the series can provide not only the factor , but also its brothers: , , , …. For example:

The pitfall is "tricks":,, etc. are such multipliers do not provide sign change. It is quite clear that for any natural : , , . Rows with tricks are slipped not only to especially gifted students, they occasionally appear “by themselves” in the course of solving functional rows.

How to examine an alternating series for convergence? Use the Leibniz sign. I don’t want to talk about the German giant of thought Gottfried Wilhelm Leibniz, because in addition to mathematical works, he dashed off several volumes on philosophy. Dangerous for the brain.

Leibniz sign: If the members of the alternating series monotonously decrease modulo, then the series converges. Or in two paragraphs:

2) The terms of the series decrease modulo: . Moreover, they decrease monotonically.

If fulfilled both conditions, then the series converges.

Brief information about the module is given in the manualHot School Mathematics Formulas , but again for convenience:

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

– Members of a series without sign decrease.
– The members of the series are decreasing modulo.
– The members of the series are decreasing in absolute value.
– Module the common term of the series tends to zero: End of help

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

Row members strictly monotone decrease modulo if EACH NEXT member of the series modulo LESS than previous: . For the series, a strict monotonicity of decreasing is performed, it can be written in detail:

And we can say in short: each next member of the series modulo less than the previous one: .

Row members not strictly monotone decrease in modulus, if EACH NEXT term of the series modulo IS NOT GREATER THAN the previous one: . Let's consider a series with a factorial: Here, non-strict monotonicity takes place, since the first two terms of the series have the same modulus. That is, each next member of the series modulo no more than the previous one: .

Under the conditions of Leibniz's theorem, the monotonicity of the decrease must be satisfied (it doesn't matter if it is strict or non-strict). In this case, the members of the series can even increase modulo for some time, but the "tail" of the series must necessarily be monotonically decreasing. No need to be afraid of what I piled up, practical examples will put everything in its place:

Example 1 Examine the series for convergence

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

1) Checking the row for alternation. Usually, at this point in the decision, the series is described in detail and the verdict "The series is alternating in sign" is passed.

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

– the terms of the series do not decrease modulo. By the way, there is no need for reasoning about the monotonicity of the decrease. Conclusion: the series diverges.

How to figure out what is equal? Very simple. As you know, the module destroys the minuses, so in order to make up, you just need to remove the flashing beacon from the roof. In this case, the common term of the series is . Stupidly remove the "flasher":.

Example 2 Examine the series for convergence

We use the Leibniz sign:

1) The series is sign-alternating.

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

Conclusion: the series converges.

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

If the series converges according to the Leibniz test, then the series is also said to be converges conditionally.

If the series composed of modules also converges: , then we say that the series converges absolutely.

Therefore, the second stage of solving a typical task is on the agenda - the study of an alternating series for absolute convergence.

I'm not guilty - such a theory of number series =)

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

Thus, our series is not absolutely convergent.
Study Series converges only conditionally.

Note that in Example No. 1 it is not necessary to conduct a study of non-absolute convergence, since at the first step it was concluded that the series diverges.

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

Example 3 Investigate the series for convergence We use the Leibniz test:

1)
This series is sign-alternating.

2) - the terms of the series decrease in absolute value. Each next term of the series is less in modulus than the previous one: , which means that the decrease is monotonous. Conclusion: The series converges.

Analyzing the filling of the series, we come to the conclusion that here it is necessary to use the limit sign of comparison. It is more convenient to open the brackets in the denominator:

Compare this series with the convergent series . We use the limit test of comparison.

A finite number other than zero is obtained, which means that the series converges together with the series . Study Series converges absolutely.

Example 4 Examine the series for convergence

Example 5 Examine the series for convergence

These are self-help examples. A complete solution and sample design at the end of the section.

As you can see, alternating rows are simple and boring! But do not rush to close the page, in just a couple of screens we will consider a case that baffles many. In the meantime, a couple of examples for training and repetition.

Example 6 Examine the series for convergence

We use the Leibniz test.
1) The series is sign-alternating.
2)
The terms of the series decrease modulo. Each next term of the series is less in modulus than the previous one, which means that the decrease is monotonous. Conclusion: the series converges.

Please note that I did not describe in detail the members of the series. It is always desirable to paint them, but from insurmountable laziness in "severe" cases, one can confine oneself to the phrase "The series is alternating in sign." By the way, you do not need to take this point formally, always check(at least mentally) that the series really takes turns. A cursory glance fails, and a mistake is made “on the machine”. Remember about the "tricks" , , , if they exist, then you need to get rid of them by getting a "normal" series with positive members.

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

We examine the series for absolute convergence:

Obviously, you need to use the radical Cauchy test:

Thus, the series converges. Study Series converges absolutely.

Example 7 Examine the series for convergence

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

Example 8 Examine the series for convergence

We use the Leibniz sign:
1) The series is sign-alternating.

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

NOTE: the concept of the growth order of a function is covered in detail in the articleLimit Solving Methods . We have sequence limits, but that doesn't change the point.

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

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

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

– Any exponential sequence grows faster than any power sequence, for example: , . exponential sequence higher order of growth than any power sequence. Similarly to the factorial, the exponential sequence "pulls" the product of any number of any power sequences or polynomials: .

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

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

Conclusion: the series converges.

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

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

And finally, a couple of examples for an independent decision. One of the same opera (re-read the help), but simpler. Another for gourmets is to fix the integral sign of convergence.

Example 9 Examine the series for convergence

Example 10 Examine the series for convergence

After a qualitative study of numerical positive and alternating series with a clear conscience, you can go to functional rows, which are no less monotonous and uniform, are interesting.

Solutions and answers:

Example 4: We use the Leibniz sign:

1) This series is alternating.
2)
The terms of the series do not decrease modulo. Conclusion: The series diverges.. , at the same time, each next term of the series is less in absolute value than the previous one, thus, the decrease is monotonous.

Thus, the series diverges together with the corresponding improper integral. Study Series converges only conditionally.

This article has collected and structured the information necessary to solve almost any example on the topic of number series, from finding the sum of a series to examining its convergence.

Article review.

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

Page navigation.

Basic definitions and concepts.

Let we have a numerical sequence , where .

Here is an example of a numerical sequence: .

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

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

are called common member of the number series or the kth member of the series.

For the previous example, the common term of the number series is .

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

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

Partial sums form an infinite sequence of partial sums of a numerical series.

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

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

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

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

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

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

Sum view called harmonic number series.

Sum view , where s is some real number, is called generalized harmonic number series.

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

THE HARMONIC SERIES IS Divergent.

Let us prove the divergence of the harmonic series.

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

On the other side,

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

Conclusion: the harmonic series diverges.

THE SUMMATION OF A GEOMETRIC PROGRESSION OF THE TYPE WITH A DENOMINATOR q IS A CONVERGENT NUMERICAL SERIES IF , AND A DIVERGENT SERIES AT .

Let's prove it.

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

When fair

which indicates the convergence of the numerical series.

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

If q \u003d -1, then the number series will take the form . Partial sums take on a value for odd n , and for even n . From this we can conclude that the limit of partial sums does not exist and the series diverges.

When fair

which indicates the divergence of the numerical series.

GENERALIZED HARMONIC SERIES CONVERGES FOR s > 1 AND DIVERS FOR .

Proof.

For s = 1 we get the harmonic series , and above we have established its divergence.

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

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

Let's write the difference:

Obviously, then

Let's write the resulting inequality for n = 2, 4, 8, 16, …

Using these results, the following actions can be performed with the original numerical series:

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

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

The number line is called alternating if the signs of its neighboring terms are different. An alternating number series can be written as or , where .

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

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

ranks

are sign-positive, sign-alternating, and sign-alternating, respectively.

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

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

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

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

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

Properties of convergent numerical series.

Example.

Prove the convergence of the numerical series.

Decision.

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

Example.

Does the number series converge?

Decision.

Let's transform the original series: . Thus, we have obtained the sum of two numerical series and , and each of them converges (see the previous example). Therefore, due to the third property of convergent numerical series, the original series also converges.

Example.

Prove the convergence of the number series and calculate its sum.

Decision.

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

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

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

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

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

A necessary condition for the convergence of a series.

If the number series converges, then the limit of its k-th term is equal to zero: .

In the study of any numerical series for convergence, first of all, it is necessary to check the fulfillment of the necessary condition for convergence. Failure to comply with this condition indicates the divergence of the numerical series, that is, if , then the series diverges.

On the other hand, it must be understood that this condition is not sufficient. That is, the fulfillment of equality does not indicate the convergence of the numerical series. For example, for a harmonic series, the necessary convergence condition is satisfied, and the series diverges.

Example.

Examine the number series for convergence.

Decision.

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

Limit n-th member of the numerical series is not equal to zero, therefore, the series diverges.

Sufficient conditions for the convergence of a positive sign series.

When using sufficient features to study numerical series for convergence, you constantly have to deal with , so we recommend that you refer to this section in case of difficulty.

A necessary and sufficient condition for the convergence of a positive-sign number series.

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

Let's start with series comparison features. Their essence lies in comparing the studied numerical series with a series whose convergence or divergence is known.

First, second and third signs of comparison.

The first sign of comparison of rows.

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

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

Example.

Set the convergence or divergence of the series.

Decision.

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

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

Example.

Examine the number series for convergence.

Decision.

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

Example.

Determine the convergence or divergence of the number series.

Decision.

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

The second sign of comparison.

Let and be sign-positive numerical series. If , then the convergence of the series implies the convergence of . If , then the divergence of the numerical series implies the divergence of .

Consequence.

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

We examine the series for convergence using the second comparison criterion. Let's take a convergent series as a series. Let's find the limit of the ratio of the k-th members of the numerical series:

Thus, according to the second criterion of comparison, the convergence of the numerical series implies the convergence of the original series.

Example.

Investigate the convergence of a number series.

Decision.

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

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

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

The third sign of comparison.

Let and be sign-positive numerical series. If the condition is satisfied from a certain number N, then the convergence of the series implies the convergence, and the divergence of the series implies the divergence.

Sign of d'Alembert.

Comment.

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

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

Example.

Examine the number series for convergence on the basis of d'Alembert.

Decision.

Let's check the fulfillment of the necessary condition for the convergence of the numerical series, we calculate the limit by:

The condition is met.

Let's use d'Alembert's sign:

Thus, the series converges.

Cauchy's radical sign.

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

Comment.

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

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

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

Example.

Investigate a positive-sign number series for convergence using the radical Cauchy test.

Decision.

. By the radical Cauchy test, we get .

Therefore, the series converges.

Example.

Does the number series converge? .

Decision.

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

Integral Cauchy test.

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

When checking the decay of a function y = f(x) over an interval, you may find the theory in the section useful.

Example.

Examine the number series with positive terms for convergence.

Decision.

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

Before starting work with this topic, I advise you to look at the section with terminology for number series. It is especially worth paying attention to the concept of a common term of a series. If you have doubts about the correct choice of the sign of convergence, I advise you to look at the topic "Choosing the sign of convergence of numerical series".

The D'Alembert test (or d'Alembert test) is used to study the convergence of series whose common term is strictly greater than zero, i.e. $u_n > 0$. Such series are called strictly positive. In standard examples, the sign of D "Alembert is used in the limiting form.

Sign of D "Alamber (in the limiting form)

If the series $\sum\limits_(n=1)^(\infty)u_n$ is strictly positive and $$ \lim_(n\to\infty)\frac(u_(n+1))(u_n)=L, $ $ then for $L<1$ ряд сходится, а при $L>1$ (and for $L=\infty$) the series diverges.

The formulation is quite simple, but the following question remains open: what happens if $L=1$? The sign of D "Alembert is not able to answer this question. If $L \u003d 1 $, then the series can both converge and diverge.

Most often, in standard examples, the sign of D "Alembert is used if the expression of the common term of the series contains a polynomial in $n$ (the polynomial can also be under the root) and a degree of the form $a^n$ or $n!$. For example, $u_n= \frac(5^n\cdot(3n+7))(2n^3-1)$ (see example #1) or $u_n=\frac(\sqrt(4n+5))((3n-2)$ (см. пример №2). Вообще, для стандартного примера наличие $n!$ - это своеобразная "визитная карточка" признака Д"Аламбера.!}

What does the expression "n!" stand for? show/hide

Recording "n!" (read "en factorial") denotes the product of all natural numbers from 1 to n, i.e.

$$ n!=1\cdot2\cdot 3\cdot \ldots\cdot n $$

By definition, it is assumed that $0!=1!=1$. For example, let's find 5!:

$$ 5!=1\cdot 2\cdot 3\cdot 4\cdot 5=120. $$

In addition, the D "Alembert test is often used to determine the convergence of a series whose common term contains the product of the following structure: $u_n=\frac(3\cdot 5\cdot 7\cdot\ldots\cdot(2n+1))(2\ cdot 5\cdot 8\cdot\ldots\cdot(3n-1))$.

Example #1

Examine the series $\sum\limits_(n=1)^(\infty)\frac(5^n\cdot(3n+7))(2n^3-1)$ for convergence.

Since the lower summation limit is 1, the common term of the series is written under the sum sign: $u_n=\frac(5^n\cdot(3n+7))(2n^3-1)$. Since for $n≥ 1$ we have $3n+7 > 0$, $5^n>0$ and $2n^3-1 > 0$, then $u_n > 0$. Therefore, our series is strictly positive.

$$ 5\cdot\lim_(n\to\infty)\frac((3n+10)\left(2n^3-1\right))(\left(2(n+1)^3-1\right )(3n+7))=\left|\frac(\infty)(\infty)\right|= 5\cdot\lim_(n\to\infty)\frac(\frac((3n+10)\left (2n^3-1\right))(n^4))(\frac(\left(2(n+1)^3-1\right)(3n+7))(n^4))= 5 \cdot\lim_(n\to\infty)\frac(\frac(3n+10)(n)\cdot\frac(2n^3-1)(n^3))(\frac(\left(2( n+1)^3-1\right))(n^3)\cdot\frac(3n+7)(n))=\\ =5\cdot\lim_(n\to\infty)\frac(\ left(\frac(3n)(n)+\frac(10)(n)\right)\cdot\left(\frac(2n^3)(n^3)-\frac(1)(n^3) \right))(\left(2\left(\frac(n)(n)+\frac(1)(n)\right)^3-\frac(1)(n^3)\right)\cdot \left(\frac(3n)(n)+\frac(7)(n)\right))=5\cdot\lim_(n\to\infty)\frac(\left(3+\frac(10) (n)\right)\cdot\left(2-\frac(1)(n^3)\right))(\left(2\left(1+\frac(1)(n)\right)^3 -\frac(1)(n^3)\right)\cdot\left(3+\frac(7)(n)\right))=5\cdot\frac(3\cdot 2)(2\cdot 3 )=5. $$

Since $\lim_(n\to\infty)\frac(u_(n+1))(u_n)=5>1$, then according to the given series diverges.

To be honest, the sign of D "Alembert is not the only option in this situation. You can use, for example, the radical Cauchy sign. However, the use of the radical Cauchy sign will require knowledge (or proof) of additional formulas. Therefore, using the sign of D" Alembert in this situation is more convenient.

Answer: the series diverges.

Example #2

Explore the series $\sum\limits_(n=1)^(\infty)\frac(\sqrt(4n+5))((3n-2)$ на сходимость.!}

Since the lower summation limit is 1, the common term of the series is written under the sum sign: $u_n=\frac(\sqrt(4n+5))((3n-2)$. Заданный ряд является строго положительным, т.е. $u_n>0$.!}

The common term of the series contains a polynomial under the root, i.e. $\sqrt(4n+5)$, and factorial $(3n-2)!$. The presence of a factorial in a standard example is an almost one hundred percent guarantee of the application of the D "Alembert sign.

To apply this feature, we have to find the limit of the relation $\frac(u_(n+1))(u_n)$. To write $u_(n+1)$, you need to use the formula $u_n=\frac(\sqrt(4n+5))((3n-2)$ вместо $n$ подставить $n+1$:!}

$$ u_(n+1)=\frac(\sqrt(4(n+1)+5))((3(n+1)-2)=\frac{\sqrt{4n+9}}{(3n+1)!}. $$ !}

Since $(3n+1)!=(3n-2)!\cdot (3n-1)\cdot 3n\cdot(3n+1)$, the formula for $u_(n+1)$ can be written as otherwise:

$$ u_(n+1)=\frac(\sqrt(4n+9))((3n+1)=\frac{\sqrt{4n+9}}{(3n-2)!\cdot (3n-1)\cdot 3n\cdot(3n+1)}. $$ !}

This entry is convenient for further solution when we have to reduce the fraction under the limit. If equality with factorials requires clarification, then please expand the note below.

How did we get $(3n+1)!=(3n-2)!\cdot (3n-1)\cdot 3n\cdot(3n+1)$? show/hide

The notation $(3n+1)!$ means the product of all natural numbers from 1 to $3n+1$. Those. this expression can be written like this:

$$ (3n+1)!=1\cdot 2\cdot\ldots\cdot(3n+1). $$

Immediately before the number $3n+1$ there is a number one less, i.e. number $3n+1-1=3n$. And immediately before the number $3n$ is the number $3n-1$. Well, just before the number $3n-1$ we have the number $3n-1-1=3n-2$. Let's rewrite the formula for $(3n+1)!$:

$$ (3n+1)!=1\cdot2\cdot\ldots\cdot(3n-2)\cdot(3n-1)\cdot 3n\cdot (3n+1) $$

What is the product of $1\cdot2\cdot\ldots\cdot(3n-2)$? This product is equal to $(3n-2)!$. Therefore, the expression for $(3n+1)!$ can be rewritten in this form:

$$(3n+1)!=(3n-2)!\cdot (3n-1)\cdot 3n\cdot(3n+1)$$

This entry is convenient for further solution when we have to reduce the fraction under the limit.

Calculate the value of $\lim_(n\to\infty)\frac(u_(n+1))(u_n)$:

$$ \lim_(n\to\infty)\frac(u_(n+1))(u_n)=\lim_(n\to\infty)\frac(\frac(\sqrt(4n+9))(( 3n-2)!\cdot (3n-1)\cdot 3n\cdot(3n+1)))(\frac(\sqrt(4n+5))((3n-2)}= \lim_{n\to\infty}\left(\frac{\sqrt{4n+9}}{\sqrt{4n+5}}\cdot\frac{(3n-2)!}{(3n-2)!\cdot (3n-1)\cdot 3n\cdot(3n+1)}\right)=\\ =\lim_{n\to\infty}\frac{\sqrt{4n+9}}{\sqrt{4n+5}}\cdot\lim_{n\to\infty}\frac{1}{(3n-1)\cdot 3n\cdot(3n+1)}= \lim_{n\to\infty}\frac{\sqrt{4+\frac{9}{n}}}{\sqrt{4+\frac{5}{n}}}\cdot\lim_{n\to\infty}\frac{1}{(3n-1)\cdot 3n\cdot(3n+1)}=1\cdot 0=0. $$ !}

Since $\lim_(n\to\infty)\frac(u_(n+1))(u_n)=0<1$, то согласно

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