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 it for convergence.

Review of the article.

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

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Basic definitions and concepts.

Suppose we have a numerical sequence, where .

Let's give an example of a numerical sequence: .

Number series Is the sum of the 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 k-th 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 a number series.

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

Partial amounts form an infinite sequence of partial sums of a number series.

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

The number series is called 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 converging 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 diverging series is the sum of a geometric progression with a denominator greater than one: ... n-th partial sum is determined by the expression , and the limit of partial sums is infinite: .

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

Sum of the form called harmonic number series.

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

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

THE HARMONIC SERIES ARE DISPENSING.

Let us prove the divergence of the harmonic series.

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

On the other side,


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

Conclusion: the harmonic series diverges.

THE SUM OF THE GEOMETRIC PROGRESSION OF THE VIEW WITH THE DENOMINATOR q IS A CONVERGING NUMBER SERIES, IF, AND DIVIDING 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 it is true


which indicates the convergence of the number 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 = -1, then the number series will take the form ... Partial sums take on values ​​for odd n and even n. From this we can conclude that the limit of partial sums does not exist and the series diverges.

When it is true


which indicates the divergence of the number series.

GENERALIZED, THE HARMONIC SERIES CONVERGES FOR s> 1 AND DIVERSES FOR.

Proof.

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

At s the inequality holds for all natural k. Due to the divergence of the harmonic series, it can be argued that the sequence of its partial sums is unbounded (since there is no finite limit). Then the sequence of partial sums of the numerical series is all the more unlimited (each term of this series is larger than the corresponding term 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 us write down the resulting inequality for n = 2, 4, 8, 16, ...



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



Expression is the sum of a geometric progression, the denominator of which is. Since we are considering the case for s> 1, then. That's why
... 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 series is called positive if all its members are positive, that is, .

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

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

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

The 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 sign-positive number series converges.

For example, number series and converge absolutely, since the series , 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.

As an example of a conventionally converging number series, we can give 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 alternating series conditionally convergent.

Properties of converging numerical series.

Example.

Prove the convergence of a number series.

Solution.

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

Example.

Whether the number series converges.

Solution.

Let's transform the original row: ... Thus, we got the sum of two numerical series and, each of them converges (see the previous example). Consequently, by virtue of the third property of converging numerical series, the original series also converges.

Example.

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

Solution.

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

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

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

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

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

A necessary condition for the convergence of the series.

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

When examining any numerical series for convergence, first of all, one should check the fulfillment of the necessary convergence condition. Failure to meet this condition indicates the divergence of the numerical series, that is, if, then the series diverges.

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

Example.

Investigate a series of numbers for convergence.

Solution.

Let us check the necessary condition for the convergence of a numerical series:

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

Sufficient signs of convergence of a positive series.

When using sufficient features to study numerical series for convergence, you constantly have to deal with, so we recommend that you refer to this section if you have any difficulties.

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

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

Let's start with the series comparison signs. Their essence lies in comparing the studied numerical series with a series, the convergence or divergence of which is known.

The first, second and third signs of comparison.

The first sign of comparing the rows.

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

The first comparison criterion is used very often and is a very powerful tool for investigating 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 term is equal to the difference between the exponents of the numerator and denominator of the k-th term of the numerical series under study. For example, suppose the difference between the exponents of the numerator and denominator is 2 - 3 = -1, therefore, for comparison, we select a series with the kth term, that is, a harmonic series. Let's look at a few examples.

Example.

Establish the convergence or divergence of the series.

Solution.

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

It is easy to see that the inequality holds for all natural numbers 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.

Solution.

The necessary condition for the convergence of a numerical series is satisfied, since ... Obviously, the inequality for any natural value k. The series converges since the generalized harmonic series is convergent for s> 1. Thus, the first sign of comparing the series allows us to state the convergence of the original numerical series.

Example.

Determine the convergence or divergence of the number series.

Solution.

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

Second sign of comparison.

Let and be positive numerical series. If, then convergence follows from the convergence of the series. If, then the divergence follows from the divergence of the numerical series.

Consequence.

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

Let us investigate the series for convergence using the second comparison criterion. Take a converging series as a series. Let us find the limit of the ratio of the k-th terms of the numerical series:

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

Example.

Investigate the convergence of a number series.

Solution.

Let us check the necessary condition for the convergence of the series ... The condition is fulfilled. To apply the second comparison criterion, we take a harmonic series. Let us find the limit of the ratio of the k-th terms:

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

For information, we will give the third sign of comparing the series.

The third sign of comparison.

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

D'Alembert sign.

Comment.

The d'Alembert test 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 on the convergence or divergence of the series and additional research is required.

Example.

Examine the number series for d'Alembert convergence.

Solution.

Let us check the fulfillment of the necessary condition for the convergence of a numerical series, the limit is calculated by:

The condition is fulfilled.

Let's use the d'Alembert test:

Thus, the series converges.

Cauchy's radical sign.

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

Comment.

Cauchy's radical criterion 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 on the convergence or divergence of the series and additional research is required.

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

Example.

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

Solution.

... By the Cauchy radical criterion, we obtain .

Consequently, the series converges.

Example.

Does the number series converge .

Solution.

We use the radical Cauchy criterion , therefore, the number series converges.

Integral Cauchy test.

Let be a positive number series. Let's 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 the case of convergence improper integral the numerical series under study converges. If the improper integral diverges, then the original series also diverges.

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

Example.

Examine a series of numbers with positive terms for convergence.

Solution.

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

D'Alembert convergence criterion Radical Cauchy convergence criterion Integral Cauchy convergence criterion

One of the common signs of comparison, which is found in practical examples, is the d'Alembert sign. Cauchy 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 the tasks are to a certain extent stencilable.

Jean Leron 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 could fly better cannonballs. At the same time, I did not forget about the numerical ranks, it was not for nothing that the ranks of Napoleon's troops converged and diverged so clearly.

Before formulating the feature itself, consider an important question:
When should the d'Alembert convergence criterion be applied?

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

The main prerequisites for the use of the d'Alembert feature are as follows:

1) The common term of the series ("stuffing" of the series) includes some number in the power, 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 factorial is included in the general term of the series. We crossed swords with factorials during the lesson Numerical sequence and its limit... However, it does not hurt to spread the self-assembled tablecloth again:





…


…

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

3) If there is a “chain of factors” in the common term of the series, for example,. This case is rare, but! When examining such a series, mistakes are 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 the matter - you need to use the d'Alembert sign.

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

D'Alembert sign: Consider positive number series... If there is a limit on the relationship of the next member to the previous one:, then:
a) For a series converges... In particular, the series converges for.
b) For a series diverges... In particular, the series diverges at.
c) When the sign does not give an answer... Another sign should be used. Most often, the unit is obtained when the d'Alembert test is tried to be applied where it is necessary to use the limiting comparison feature.

Anyone who still has problems with limits or a misunderstanding of limits, refer to the lesson Limits. Examples of solutions... Unfortunately, without understanding the limit and the ability to disclose uncertainty, one cannot advance further.

And now the long-awaited examples.

Example 1

We see that we have in the common term of the series, and this is a correct precondition for using the d'Alembert sign. First, a complete solution and a sample design, comments below.

We use the d'Alembert sign:

converges.

(1) We compose the ratio of the next member of the series to the previous one:. From the condition we see that the common term of the series. In order to get the next member of the series, it is necessary instead of substituting: .
(2) Getting rid of the four-story fraction. With some experience with the solution, this step can be skipped.
(3) Expand the parentheses in the numerator. In the denominator, we take out the four from the degree.
(4) Reduce by. The constant is taken out of the limit sign. We give similar terms in the numerator in parentheses.
(5) Uncertainty is eliminated in the standard way - dividing the numerator and denominator by "en" in the highest power.
(6) We divide the numerators by 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 test, the series under study converges.

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

Example 2

Take a similar series and examine it for convergence

First the complete solution, then the comments:

We use the d'Alembert sign:

Thus, the series under study converges.

(1) Composing the relation.
(2) Getting rid of the four-story fraction.
(3) Consider an expression in the numerator and an expression in the denominator. We see that in the numerator you need to open parentheses and raise to the fourth power: which you absolutely do not want to do. In addition, for those who are not familiar with Newton's binomial, this task may not be feasible at all. Let's analyze the 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 when the numerator and denominator are divided by term by, we get one in the limit. Or, as mathematicians say, polynomials and - same order of growth... Thus, it is quite possible to circle the ratio with a simple pencil and immediately indicate that this thing tends to one. We deal with the second pair of polynomials in a similar way: and, they are also same order of growth, and their ratio tends to unity.

In fact, such a "hack" could have been done in Example # 1, but for a polynomial of the 2nd degree, such a solution still looks somehow undignified. Personally, I do this: if there is a polynomial (or polynomials) of the first or second degree, I use the "long" way to solve Example 1. If I come across a polynomial of the third or higher degrees, I use the "turbo" -method similar to Example 2.

Example 3

Examine the series for convergence

Complete solution and sample design at the end of the number sequences lesson.
(4) Reducing everything that can be reduced.
(5) The constant is taken out of the limit sign. Expand the parentheses in the numerator.
(6) Uncertainty is eliminated in the standard way - by dividing the numerator and denominator by "en" in the highest power.

Example 5

Examine the series for convergence

Complete solution and sample design at the end of the lesson

Example 6

Examine the series for convergence

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

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

A rough example of a solution might look like this:

We use the d'Alembert sign:

Thus, the series under study converges.

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

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

Sign D "Alamber" (in extreme 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 will happen if $ L = 1 $? Alambert's sign D is not able to give an answer to this question. If $ L = 1 $, then the series can both converge and diverge.

Most often, in standard examples, the Alamber D sign is used if the expression for the general term of the series contains a polynomial of $ n $ (the polynomial may 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!" Mean? show \ hide

The entry "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, Alambert's D "feature 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

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

Since the lower limit of summation 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.

Honestly, the Alambert D "sign is not the only option in this situation. You can use, for example, the radical Cauchy test. However, the use of the radical Cauchy test will require knowledge (or proof) of additional formulas. Therefore, the use of the Alamber D" feature in this situation is more convenient.

Answer: the row diverges.

Example No. 2

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

Since the lower limit of summation 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 the polynomial at the root, i.e. $ \ sqrt (4n + 5) $, and the factorial $ (3n-2)! $. The presence of a factorial in a standard example is an almost one hundred percent guarantee of the use of Alamber's characteristic D.

To apply this feature, we have to find the limit of the ratio $ \ frac (u_ (n + 1)) (u_n) $. To write $ u_ (n + 1) $, you need $ 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 to another:

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

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

How did we get the equality $ (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:

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

Directly before the number $ 3n + 1 $ there is a number one less, i.e. number $ 3n + 1-1 = 3n $. And immediately before the number $ 3n $ there is the number $ 3n-1 $. Well, immediately 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 $ 1 \ cdot2 \ cdot \ ldots \ cdot (3n-2) $? This product is equal to $ (3n-2)! $. Therefore, the expression for $ (3n + 1)! $ Can be rewritten as follows:

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

This notation is convenient for further solution when we have to cancel the fraction under the limit.

Let's 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

Convergence criteria for series.
D'Alembert sign. Cauchy signs

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 the fact that you have been looking forward to and longing to know for a long time - convergence criteria for positive numerical series... The holiday of September 1 and my congratulations are always relevant, it's okay if it's actually summer outside, you are now retaking the exam for the third time, if you go to this page!

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

One of the common signs of comparison, which is found in practical examples, is the d'Alembert sign. Cauchy 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 the tasks are to a certain extent stencilable.

The d'Alembert convergence test

Jean Leron 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 could fly better cannonballs. At the same time, I did not forget about the numerical ranks, it was not for nothing that the ranks of Napoleon's troops converged and diverged so clearly.

Before formulating the feature itself, consider an important question:
When should the d'Alembert convergence criterion be applied?

Let's start with repetition. Let us recall the cases when you need to use the most popular limit comparison criterion... The limiting comparison criterion is applied when in the common term of the series:

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

The main prerequisites for the use of the d'Alembert feature are as follows:

1) The common term of the series ("stuffing" of the series) includes some number in the power, 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 factorial is included in the general term of the series. We crossed swords with factorials in the lesson Numerical sequence and its limit. However, it does not hurt to spread the self-assembled tablecloth again:





…


…

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

3) If in the common term of the series there is a "chain of factors", for example, ... This case is rare, but! When examining such a series, mistakes are 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 the matter - you need to use the d'Alembert sign.

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

D'Alembert sign: Consider positive number series... If there is a limit on the relationship of the next member to the previous one:, then:
a) For a series converges
b) For a series diverges
c) When the sign does not give an answer... Another sign should be used. Most often, the unit is obtained when the d'Alembert test is tried to be applied where it is necessary to use the limiting comparison feature.

Anyone who still has problems with limits or a misunderstanding of limits, refer to the lesson Limits. Examples of solutions... Unfortunately, without understanding the limit and the ability to disclose uncertainty, one cannot advance further.

And now the long-awaited examples.

Example 1

We see that we have in the common term of the series, and this is a correct precondition for using the d'Alembert sign. First, a complete solution and a sample design, comments below.

We use the d'Alembert sign:


converges.
(1) We compose the ratio of the next member of the series to the previous one:. From the condition we see that the common term of the series. In order to get the next member of the series, you need INSTEADLY substitute: .
(2) Getting rid of the four-story fraction. With some experience with the solution, this step can be skipped.
(3) Expand the parentheses in the numerator. In the denominator, we take out the four from the degree.
(4) Reduce by. The constant is taken out of the limit sign. We give similar terms in the numerator in parentheses.
(5) Uncertainty is eliminated in the standard way - dividing the numerator and denominator by "en" in the highest power.
(6) We divide the numerators by 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 test, the series under study converges.

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

Example 2

Take a similar series and examine it for convergence

First the complete solution, then the comments:

We use the d'Alembert sign:


Thus, the series under study converges.

(1) Composing the relation.

(3) Consider the expression in the numerator and expression in the denominator. We see that in the numerator you need to open parentheses and raise to the fourth power: which you absolutely do not want to do. And for those who are not familiar with Newton's binomial, this task will be even more difficult. Let's analyze the higher degrees: if we expand the parentheses 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 when the numerator and denominator are divided by term by, we get one in the limit. Or, as mathematicians say, polynomials and - same 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. We deal with the second pair of polynomials in a similar way: and, they are also same 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" way to solve Example 1. If I come across a polynomial of the third degree or higher, I use the "turbo" -method similar to Example 2.

Example 3

Examine the series for convergence

Let's consider typical examples with factorials:

Example 4

Examine the series for convergence

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

Thus, the series under study diverges.
(1) Composing the relation. We repeat once more. By condition, the common term of the series: ... To get the next term in the series, instead you need to substitute, thus: .
(2) Getting rid of the four-story fraction.
(3) We pinch off the seven from the degree. We paint factorials in detail... How to do this - see the beginning of the lesson or the article on number sequences.
(4) Reducing everything that can be reduced.
(5) The constant is taken out of the limit sign. Expand the parentheses in the numerator.
(6) Uncertainty is eliminated in the standard way - by dividing the numerator and denominator by "en" in the highest power.

Example 5

Examine the series for convergence

Complete solution and sample design at the end of the lesson

Example 6

Examine the series for convergence

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

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

A rough example of a solution might look like this:

We use the d'Alembert sign:

Thus, the series under study converges.

Cauchy's radical sign

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

Cauchy's convergence test for positive 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) For a series converges... In particular, the series converges for.
b) For a series diverges... In particular, the series diverges at.
c) When the sign does not give an answer... Another sign should be used. 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 also does not give an answer. But if the d'Alembert sign does not give an answer, then the Cauchy sign may well “work”. That is, the Cauchy sign is in this sense a stronger sign.

When should you use the radical Cauchy sign? The Cauchy radical criterion is usually used in cases where the root "good" is extracted from a common member of the series. Typically, this pepper is in the degree which depends on... There are also exotic cases, but we will not bother 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 you need to use the radical Cauchy criterion:


Thus, the series under study diverges.

(1) We form the common term of the series as a root.

(2) We rewrite the same thing, only without the root, using the power property.
(3) In the exponent, divide the numerator by the denominator term by term, indicating that
(4) The result is uncertainty. Here one could go a long way: build to a cube, build to a 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 right under the degree-constant. To eliminate uncertainty, divide the numerator and denominator by (the highest degree of the polynomials).

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

And here is a simpler example for a do-it-yourself solution:

Example 8

Examine the series for convergence

And a couple more typical examples.

Complete solution and sample design at the end of the lesson

Example 9

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


Thus, the series under study converges.

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

(2) We rewrite the same, but without the root, while expanding the parentheses using the formula for abbreviated multiplication: .
(3) In the indicator, divide the numerator by the denominator term by term and indicate that.
(4) The uncertainty of the form is obtained, and here, too, you can perform division directly under the degree. But with one condition: the coefficients at the highest degrees of the 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) Actually, we perform term-by-term division and indicate which terms tend to zero.
(6) Uncertainty is removed, we have 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 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:

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

(7) We point out that we conclude that the series converges.

Example 10

Examine the series for convergence

This is an example for a do-it-yourself solution.

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 (you get polynomials), and then adhere to the algorithm from the article Rows for dummies... In such an example, either the necessary criterion for the convergence of the series or the limiting comparison criterion should work.

Integral Cauchy test

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

In textbooks on calculus integral Cauchy test given mathematically rigorously, but too distorted, so I will formulate the criterion not too strictly, but understandably:

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

And immediately examples for clarification:

Example 11

Examine the series for convergence

Almost a classic. Natural logarithm and some kind of byaka.

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