Before starting to 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 member of a series. If you have doubts about the correct choice of a convergence criterion, I advise you to look at the topic “Choosing a convergence criterion for number series”.

D'Alembert's test (or D'Alembert's 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 D'Alembert sign is used in its extreme form.

D'Alembert's sign (in its 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 $L=1$? D'Alembert's test 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 D'Alembert criterion is used if the expression of the general term of the series contains a polynomial of $n$ (the polynomial can 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 No. 1) or $u_n=\frac(\sqrt(4n+5))((3n-2)$ (см. пример №2). Вообще, для стандартного примера наличие $n!$ - это своеобразная "!} business card"D"Alembert's sign.

What does the expression "n!" mean? 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 No. 1

Examine 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 general 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 D'Alembert test 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 D'Alembert test in this situation is more convenient.

Answer: the series diverges.

Example No. 2

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

Since the lower limit of summation is 1, the general 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 under the root, i.e. $\sqrt(4n+5)$, and the factorial $(3n-2)!$. Availability of factorial in standard example- almost one hundred percent guarantee of the use of the D'Alembert sign.

To apply this criterion, we will have to find the limit of the ratio $\frac(u_(n+1))(u_n)$. To write $u_(n+1)$, you need in 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 $(3n+1)!=(3n-2)!\cdot (3n-1)\cdot 3n\cdot(3n+1)$, then 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 solutions when we have to reduce the fraction under the limit. If equality with factorials requires explanation, please open 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 as follows:

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

Directly before the number $3n+1$ there is a number that is 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 in the following form:

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

This notation is convenient for further solutions when we have to reduce 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

This article collects and structures 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 positive and alternating series and the concept of convergence. Next, we will consider standard series, such as the harmonic series, the generalized harmonic series, and recall the formula for finding the sum of an infinitely decreasing geometric progression. After this, we will move on to the properties of convergent series, dwell on the necessary condition for the convergence of the series and state sufficient criteria for the convergence of the series. We will dilute the theory with solutions to typical examples with detailed explanations.

Page navigation.

Basic definitions and concepts.

Let us have a number sequence where .

Here is an example of a number sequence: .

Number series is the sum of the terms 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 denominator q = -0.5: .

Called common member of the number series or the kth member of the series.

For the previous example, the general term of the number series has the form .

Partial sum of a number series is a sum of the form , where n is some natural number. also called the nth 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 nth partial sum is found using 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 series is called convergent, if there is a finite limit to the sequence of partial sums. If the limit of the sequence of partial sums of a number 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 determined by the expression , and the limit of partial sums is infinite: .

Another example of a divergent number series is a sum of the form . In this case, the nth 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 by harmonic number series.

The above definitions are sufficient to justify 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 to 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 GEOMETRIC PROGRESSION OF THE KIND WITH DENOMINATOR q IS A CONVERGING NUMERIC SERIES IF , AND A DIVERGING SERIES FOR .

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 number series.

For q = 1 we have the 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 value for odd n, and for even n. From this we can conclude that there is no limit on partial sums and the series diverges.

When fair


which indicates the divergence of the number series.

GENERALLY, THE HARMONIC SERIES CONVERGES AT s > 1 AND DIVERGES AT .

Proof.

For s = 1 we obtain 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 unlimited (since there is no finite limit). Then the sequence of partial sums of a 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 as s.

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

Let's write down 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 whose denominator is . Since we are considering the case for s > 1, then. That's why
. Thus, the sequence of partial sums of a generalized harmonic series for s > 1 is increasing and at the same time limited 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 sign, if all its terms are positive, that is, .

The number series is called signalternating, 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 sign, if it contains an infinite number of both positive and negative terms.

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

Rows

are positive, alternating and 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 number series converges.

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

An 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 terms 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 is an alternating series conditionally convergent.

Properties of convergent number series.

Example.

Prove the convergence of the number series.

Solution.

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?

Solution.

Let's transform the original series: . Thus, we have obtained the sum of two number series and , and each of them converges (see the previous example). Consequently, by virtue of the third property of convergent number series, the original series also converges.

Example.

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

Solution.

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

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

The first term of the series is one, and the denominator of the corresponding geometric progression is equal to 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 results obtained to find the sum of the original number series:

A necessary condition for the convergence of a series.

If a number series converges, then the limit of its kth term is equal to zero: .

When examining any number series for convergence, the first thing to check is the fulfillment of the necessary convergence condition. Failure to fulfill this condition indicates the divergence of the number 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 indicate the convergence of the number series. For example, for a harmonic series the necessary condition for convergence is satisfied, and the series diverges.

Example.

Examine a number series for convergence.

Solution.

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

Limit The nth term 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 number series for convergence, you constantly encounter problems, so we recommend turning 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 bounded.

Let's start with the signs of comparing series. Their essence lies in comparing the numerical series under study with a series whose convergence or divergence is known.

The first, second and third signs of comparison.

The first sign of comparison of series.

Let and be two positive number 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 of .

The first comparison criterion is used very often and is a very powerful tool for studying number series for convergence. The main problem is selecting a suitable series for comparison. A series for comparison is usually (but not always) chosen so that the exponent of its kth term is equal to the difference between the exponents of the numerator and denominator of the kth term of the numerical series under study. For example, let the difference between the exponents of the numerator and the denominator be equal to 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 convergence or divergence of a series.

Solution.

Since the limit of the general 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 is divergent; therefore, by the first criterion of comparison, the original series is also divergent.

Example.

Examine the number series for convergence.

Solution.

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

Example.

Determine the convergence or divergence of a number series.

Solution.

, therefore, the necessary condition for the convergence of the number series is satisfied. Which row should I choose for comparison? A number series suggests itself, and in order to decide on s, we carefully examine the number sequence. The terms of a number 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 true. A 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, by the first criterion of comparison, the series is convergent, and by virtue of the first property of convergent number series, the series will also converge.

The second sign of comparison.

Let and be positive number series. If , then the convergence of the series implies the convergence of . If , then the divergence of the number 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. As a series we take a convergent series. Let's find the limit of the ratio of the kth terms of the number series:

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

Example.

Examine the convergence of a number series.

Solution.

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

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

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

The third sign of comparison.

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

D'Alembert's sign.

Comment.

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

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

Example.

Examine a number series for convergence using d'Alembert's test.

Solution.

Let's check the fulfillment of the necessary condition for the convergence of a number series; calculate the limit using:

The condition is met.

Let's use d'Alembert's sign:

Thus, the series converges.

Radical Cauchy sign.

Let be a positive number series. If , then the number 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 fairly easy to discern cases where it is best to use the radical Cauchy test. A typical case is when the general term of a number series is an exponential power expression. Let's look at a few examples.

Example.

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

Solution.

. Using the radical Cauchy test we obtain .

Therefore, the series converges.

Example.

Does the number series converge? .

Solution.

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

Integral Cauchy test.

Let be a positive number series. Let's create 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 the number series under study converges. If 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 section may be useful to you.

Example.

Examine a number series with positive terms for convergence.

Solution.

The necessary condition for the convergence of the series is satisfied, since . Let's consider the 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.

D'Alembert's convergence test Radical Cauchy convergence test Integral Cauchy convergence test

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

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

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

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

The main prerequisites for the application of d'Alembert's test are as follows:

1) The common term of the series (“stuffing” of the series) includes some number to a degree, for example, , and so on. Moreover, it doesn’t matter at all where this thing is located, in the numerator or in the denominator - what matters is that it is present there.

2) The common term of the series includes the factorial. We crossed swords with factorials back in class Number sequence and its limit. However, it won’t hurt to spread out the self-assembled tablecloth again:





…


…

! When using d'Alembert's test, we will 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 general term of the series there is a “chain of factors”, for example, . This case is rare, but! When studying such a series, a mistake is often made - see Example 6.

Along with powers and/or factorials, polynomials are often found in the filling of a series; this does not change the situation - you need to use D'Alembert's sign.

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

D'Alembert's sign: Let's consider positive number series. If there is a limit on the ratio of the subsequent term to the previous one: , then:
a) When row converges. In particular, the series converges at .
b) When row diverges. In particular, the series diverges at .
c) When the sign does not give an answer. You need to use another sign. Most often, one is obtained in the case when they try to apply the d'Alembert test where it is necessary to use the limiting comparison test.

For those who still have problems with limits or misunderstandings of limits, refer to the lesson Limits. Examples of solutions. Without an understanding of the limit and the ability to reveal uncertainty, unfortunately, one cannot advance further.

And now the long-awaited examples.

Example 1

We see that in the general term of the series we have , and this is a sure prerequisite for using d'Alembert's test. First, the full solution and sample design, comments below.

We use d'Alembert's 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 general term of the series is . In order to get the next member of the series it is necessary instead of substituting: .
(2) We get rid of the four-story fraction. If you have some experience with the solution, you can skip this step.
(3) Open the parentheses in the numerator. In the denominator we take the four out of the power.
(4) Reduce by . We take the constant beyond the limit sign. In the numerator in brackets we give similar terms.
(5) Uncertainty is eliminated in the standard way - by dividing the numerator and denominator by “en” to the highest power.
(6) We divide the numerators term by term by the denominators, and indicate the terms that tend to zero.
(7) We simplify the answer and make a note that with the conclusion that, according to D’Alembert’s criterion, the series under study converges.

In the example considered, in the general term of the series we encountered a polynomial of the 2nd degree. What to do 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 use the “turbo” solution method.

Example 2

Let's take a similar series and examine it for convergence

First the complete solution, then comments:

We use d'Alembert's sign:

Thus, the series under study converges.

(1) We create the relation .
(2) We 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 we need to open the brackets and raise them to the fourth power: , which we absolutely don’t 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 higher 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 dividing the numerator and denominator term by term, we end up with 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 the same way: and , they too same order of growth, and their ratio tends to unity.

In fact, such a “hack” could have been pulled off 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 3rd or more high degrees, I use the “turbo” method similar to Example 2.

Example 3

Examine the series for convergence

Full solution and sample design at the end of the lesson number sequences.
(4) We cut everything that can be cut.
(5) We move the constant beyond the limit sign. Open the parentheses in the numerator.
(6) We eliminate uncertainty in the standard way - by dividing the numerator and denominator by “en” to the highest power.

Example 5

Examine the series for convergence

Full solution and sample design at the end of the lesson

Example 6

Examine the series for convergence

Sometimes there are series that contain a “chain” of factors in their filling; we have not yet considered this type of series. How to study a series with a “chain” of factors? Use d'Alembert's sign. But first, to understand what is happening, let’s describe the series in detail:

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

A sample solution might look like this:

We use d'Alembert's sign:

Thus, the series under study converges.

Signs of series convergence.
D'Alembert's sign. Cauchy's signs

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

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

For those who are just starting to study 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 the topics:

One of the common comparison signs that is found 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 simply, accessible and understandable. The topic is not the most difficult, and all tasks are to a certain extent standard.

D'Alembert's convergence test

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

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

Let's start with a review first. Let's remember the cases when you need to use the most popular limit of comparison . The limiting criterion for comparison is applied when in the general term of the series:

1) The denominator contains a polynomial.
2) Polynomials are in both the numerator and 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 the application of d'Alembert's test are as follows:

1) The common term of the series (“filling” of the series) includes some number to a degree, for example, , , and so on. Moreover, it doesn’t matter at all where this thing is located, in the numerator or in the denominator - what matters is that it is present there.

2) The common term of the series includes the factorial. We crossed swords with factorials back in class Number sequence and its limit. However, it won’t hurt to spread out the self-assembled tablecloth again:





…


…

! When using d'Alembert's test, we will 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 general term of the series there is a “chain of factors”, for example, . This case is rare, but! When studying such a series, a mistake is often made - see Example 6.

Along with powers and/or factorials, polynomials are often found in the filling of a series; this does not change the situation - you need to use D'Alembert's sign.

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

D'Alembert's sign: Let's consider positive number series. If there is a limit on the ratio of the subsequent term to the previous one: , then:
a) When row converges
b) When row diverges
c) When the sign does not give an answer. You need to use another sign. Most often, a unit is obtained when they try to apply the D’Alembert sign where it should be used limit of comparison.

For those who still have problems with limits or misunderstandings of limits, refer to the lesson Limits. Examples of solutions . Without an understanding of the limit and the ability to reveal uncertainty, unfortunately, one cannot advance further.

And now the long-awaited examples.

Example 1

We see that in the general term of the series we have , and this is a sure prerequisite for using d'Alembert's test. First, the full solution and sample design, comments below.

We use d'Alembert's 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 general term of the series is . In order to get the next member of the series you need INSTEAD to substitute: .
(2) Getting rid of four-story fraction. If you have some experience with the solution, you can skip this step.
(3) Open the parentheses in the numerator. In the denominator we take the four out of the power.
(4) Reduce by . We take the constant beyond the limit sign. In the numerator we present similar terms in parentheses.
(5) Uncertainty is resolved in a standard way - dividing the numerator and denominator to "en" in the senior degree.
(6) We divide the numerators term by term by the denominators, and indicate the terms that tend to zero.
(7) We simplify the answer and make a note that with the conclusion that, according to D’Alembert’s criterion, the series under study converges.

In the example considered, in the general term of the series we encountered a polynomial of the 2nd degree. What to do 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 use the “turbo” solution method.

Example 2

Let's take a similar series and examine it for convergence

First the complete solution, then comments:

We use d'Alembert's sign:


Thus, the series under study converges.

(1) We create the relation .

(3) Consider the expression in the numerator and the expression in the denominator. We see that in the numerator we need to open the brackets and raise them to the fourth power: , which we absolutely don’t 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 open the brackets at the top , then we will get a senior degree. Below we have the same senior degree: . By analogy with the previous example, it is obvious that when dividing the numerator and denominator term by term, we end up with one in the limit. Or, as mathematicians say, polynomials And - same order of growth. Thus, it is quite possible to outline the relation with a simple pencil and immediately indicate that this thing is tending to one. We deal with the second pair of polynomials in the same way: and , they too same order of growth, and their ratio tends to unity.

In fact, such a “hack” could have been pulled off 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 for solving Example 1. If I come across a polynomial of the 3rd or higher degrees, 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 common term of the series includes both the degree and the factorial. It is clear as day that d'Alembert's sign must be used here. Let's decide.

Thus, the series under study diverges.
(1) We create the relation . We repeat again. By condition, the common term of the series is: . In order to get the next term in the series, instead you need to substitute, Thus: .
(2) We get rid of the four-story fraction.
(3) Pinch off the seven from the degree. We describe factorials in detail. How to do this - see the beginning of the lesson or article about number sequences.
(4) We cut everything that can be cut.
(5) We move the constant beyond the limit sign. Open the parentheses in the numerator.
(6) We eliminate uncertainty in the standard way - by dividing the numerator and denominator by “en” to the highest power.

Example 5

Examine the series for convergence

Full solution and sample design at the end of the lesson

Example 6

Examine the series for convergence

Sometimes there are series that contain a “chain” of factors in their filling; we have not yet considered this type of series. How to study a series with a “chain” of factors? Use d'Alembert's sign. But first, to understand what is happening, let’s describe the series in detail:

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

A sample solution might look like this:

We use d'Alembert's sign:

Thus, the series under study converges.

Radical Cauchy's sign

Augustin Louis Cauchy is an even more famous French mathematician. Any student can tell you Cauchy's biography. technical specialty. 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 number series is somewhat similar to D'Alembert's test just discussed.

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

When should you use the radical Cauchy sign? The radical Cauchy test is usually used in cases where the root “good” is extracted from a common member of the series. As a rule, this pepper is in a degree which depends on. There are also exotic cases, but we won’t worry about them.

Example 7

Examine the series for convergence

We see that the fraction is completely under a power depending on “en”, which means we need to use the radical Cauchy test:


Thus, the series under study diverges.

(1) We formulate 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 indicator, we divide the numerator by the denominator term by term, indicating that
(4) As a result, we have uncertainty. Here you could go the long way: cube, cube, then divide numerator and denominator to "en" in cube. But in this case there is a more effective solution: this technique can be used directly under the constant degree. To eliminate uncertainty, divide the numerator and denominator by (the highest power 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 what we have and conclude that the series diverges.

Here is a simpler example for you to solve on your own:

Example 8

Examine the series for convergence

And a couple more typical examples.

Full solution and sample design 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 indicator, 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 under the degree. But with one condition: the coefficients of the higher powers of the polynomials must be different. Ours are 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 such a trick does not work and you need to use second wonderful limit . If you remember, these subtleties were discussed in the last paragraph of the article Methods for solving limits .

(5) We actually perform term-by-term division and indicate which terms tend to zero.
(6) The uncertainty has been eliminated, we are left with the simplest limit: . Why in infinitely large tends to zero? Because the base of the degree satisfies the inequality. If anyone has doubts about the fairness of the limit , then I won’t be 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 like that infinitely decreasing geometric progression on your fingers =)
! Never use this technique as evidence! Because just because something is obvious, that doesn’t mean it’s right.

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

Example 10

Examine the series for convergence

This is an example for you to solve on your own.

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

Integral Cauchy test

Or just an integral sign. I will disappoint those who did not understand the first course material well. In order to apply the Cauchy integral test, you must be more or less confident in finding derivatives, integrals, and also have the skill of calculation improper integral first kind.

In textbooks on mathematical analysis integral Cauchy test given mathematically strictly, but too confusingly, so I will formulate the sign not too strictly, but clearly:

Let's consider positive number series. If there is an improper integral, then the series converges or diverges along with this integral.

And just 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 Cauchy integral test is is the fact that the general term of the series contains factors similar to a certain function and its derivative. From topic