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 criteria for the convergence of the series. We will dilute the theory by solving typical examples with detailed explanations.

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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 numerical 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 hand,


The following inequalities are beyond doubt. In this way, . 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 . 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 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.

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 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.

Solution.

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: .

When examining any number series for convergence, first of all, one should check the fulfillment necessary condition 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.

Solution.

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 examining numerical series for convergence. The main problem is the selection of a suitable series for comparison. The comparison series is usually (but not always) chosen so that the exponent of its kth term is equal to the difference exponents of the numerator and denominator of the k-th member of the studied number series. 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.

Solution.

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.

Solution.

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.

Solution.

, 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.

Solution.

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 members:

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.

Solution.

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 characteristic case is when the common term of the number series is exponentially power expression. Let's look at a few examples.

Example.

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

Solution.

. By the radical Cauchy test, we get .

Therefore, the series converges.

Example.

Does the number series converge? .

Solution.

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.

Solution.

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.

Standard methods, but reached a dead end with another example.

What is the difficulty and where can there be a snag? Let's put aside the soapy rope, calmly analyze the reasons and get acquainted with the practical methods of solution.

First and most important: in the vast majority of cases, to study the convergence of a series, it is necessary to apply some familiar method, but the common term of the series is filled with such tricky stuffing that it is not at all obvious what to do with it. And you go around in circles: the first sign does not work, the second does not work, the third, fourth, fifth method does not work, then the drafts are thrown aside and everything starts anew. This is usually due to a lack of experience or gaps in other sections of calculus. In particular, if running sequence limits and superficially disassembled function limits, then it will be difficult.

In other words, a person simply does not see the necessary solution due to a lack of knowledge or experience.

Sometimes “eclipse” is also to blame, when, for example, the necessary criterion for the convergence of a series is simply not fulfilled, but due to ignorance, inattention or negligence, this falls out of sight. And it turns out like in that bike where the professor of mathematics solved a children's problem with the help of wild recurrent sequences and number series =)

In the best traditions, immediately living examples: rows and their relatives - diverge, since in theory it is proved sequence limits. Most likely, in the first semester you will be beaten out of your soul for a proof of 1-2-3 pages, but for now it is quite enough to show that the necessary condition for the convergence of the series is not met, referring to known facts. Famous? If the student does not know that the root of the nth degree is an extremely powerful thing, then, say, the series put him in a rut. Although the solution is like two and two: , i.e. for obvious reasons, both series diverge. A modest comment “these limits have been proven in theory” (or even its absence at all) is quite enough for offset, after all, the calculations are quite heavy and they definitely do not belong to the section of numerical series.

And after studying the next examples, you will only be surprised at the brevity and transparency of many solutions:

Example 1

Investigate the convergence of a series

Solution: first of all, check the execution necessary criterion for convergence. This is not a formality, but a great chance to deal with the example of "little bloodshed".

"Inspection of the scene" suggests a divergent series (the case of a generalized harmonic series), but again the question arises, how to take into account the logarithm in the numerator?

Approximate examples of tasks at the end of the lesson.

It is not uncommon when you have to carry out a two-way (or even three-way) reasoning:

Example 6

Investigate the convergence of a series

Solution: first, carefully deal with the gibberish of the numerator. The sequence is limited: . Then:

Let's compare our series with the series . By virtue of the double inequality just obtained, for all "en" it will be true:

Now let's compare the series with the divergent harmonic series.

Fraction denominator less the denominator of the fraction, so the fraction itself – more fractions (write down the first few terms, if not clear). Thus, for any "en":

So, by comparison, the series diverges along with the harmonic series.

If we change the denominator a little: , then the first part of the reasoning will be similar: . But to prove the divergence of the series, only the limit test of comparison is already applicable, since the inequality is false.

The situation with converging series is “mirror”, that is, for example, both comparison criteria can be used for a series (the inequality is true), and for a series, only the limiting criterion (the inequality is false).

We continue our safari through the wild, where a herd of graceful and succulent antelopes loomed on the horizon:

Example 7

Investigate the convergence of a series

Solution: the necessary convergence criterion is satisfied, and we again ask the classic question: what to do? Before us is something resembling a convergent series, however, there is no clear rule here - such associations are often deceptive.

Often, but not this time. By using Limit comparison criterion Let's compare our series with the convergent series . When calculating the limit, we use wonderful limit , where as infinitesimal stands:

converges together with next to .

Instead of using the standard artificial method of multiplication and division by a "three", it was possible to initially compare with a convergent series.
But here a caveat is desirable that the constant-multiplier of the general term does not affect the convergence of the series. And just in this style the solution of the following example is designed:

Example 8

Investigate the convergence of a series

Sample at the end of the lesson.

Example 9

Investigate the convergence of a series

Solution: in the previous examples, we used the boundedness of the sine, but now this property is out of play. The denominator of a fraction of a higher order of growth than the numerator, so when the sine argument and the entire common term infinitely small. The necessary condition for convergence, as you understand, is satisfied, which does not allow us to shirk from work.

We will conduct reconnaissance: in accordance with remarkable equivalence , mentally discard the sine and get a series. Well, something like that….

Making a decision:

Let us compare the series under study with the divergent series . We use the limit comparison criterion:

Let us replace the infinitesimal with the equivalent one: for .

A finite number other than zero is obtained, which means that the series under study diverges along with the harmonic series.

Example 10

Investigate the convergence of a series

This is a do-it-yourself example.

To plan next steps in similar examples mental rejection of the sine, arcsine, tangent, arctangent helps a lot. But remember, this possibility exists only when infinitesimal argument, not so long ago I came across a provocative series:

Example 11

Investigate the convergence of a series
.

Solution: it is useless to use the limitedness of the arc tangent here, and the equivalence does not work either. The output is surprisingly simple:


Study Series diverges, since the necessary criterion for the convergence of the series is not satisfied.

The second reason"Gag on the job" consists in a decent sophistication of the common member, which causes difficulties of a technical nature. Roughly speaking, if the series discussed above belong to the category of “figures you guess”, then these ones belong to the category of “you decide”. Actually, this is called complexity in the "usual" sense. Not everyone will correctly resolve several factorials, degrees, roots and other inhabitants of the savannah. Of course, factorials cause the most problems:

Example 12

Investigate the convergence of a series

How to raise a factorial to a power? Easily. According to the rule of operations with powers, it is necessary to raise each factor of the product to a power:

And, of course, attention and once again attention, the d'Alembert sign itself works traditionally:

Thus, the series under study converges.

I remind you of a rational technique for eliminating uncertainty: when it is clear order of growth numerator and denominator - it is not at all necessary to suffer and open the brackets.

Example 13

Investigate the convergence of a series

The beast is very rare, but it is found, and it would be unfair to bypass it with a camera lens.

What is double exclamation point factorial? The factorial "winds" the product of positive even numbers:

Similarly, the factorial “winds up” the product of positive odd numbers:

Analyze what is the difference between

Example 14

Investigate the convergence of a series

And in this task, try not to get confused with the degrees, wonderful equivalences and wonderful limits.

Sample solutions and answers at the end of the lesson.

But the student gets to feed not only tigers - cunning leopards also track down their prey:

Example 15

Investigate the convergence of a series

Solution: the necessary criterion of convergence, the limiting criterion, the d'Alembert and Cauchy criteria disappear almost instantly. But worst of all, the feature with inequalities, which has repeatedly rescued us, is powerless. Indeed, comparison with a divergent series is impossible, since the inequality incorrect - the multiplier-logarithm only increases the denominator, reducing the fraction itself in relation to the fraction. And another global question: why are we initially sure that our series is bound to diverge and must be compared with some divergent series? Does he fit in at all?

Integral feature? Improper integral evokes a mournful mood. Now, if we had a row … then yes. Stop! This is how ideas are born. We make a decision in two steps:

1) First, we study the convergence of the series . We use integral feature:

Integrand continuous on the

Thus, a number diverges together with the corresponding improper integral.

2) Compare our series with the divergent series . We use the limit comparison criterion:

A finite number other than zero is obtained, which means that the series under study diverges along with side by side .

And there is nothing unusual or creative in such a decision - that's how it should be decided!

I propose to independently draw up the following two-move:

Example 16

Investigate the convergence of a series

A student with some experience in most cases immediately sees whether the series converges or diverges, but it happens that a predator cleverly disguises itself in the bushes:

Example 17

Investigate the convergence of a series

Solution: at first glance, it is not at all clear how this series behaves. And if we have fog in front of us, then it is logical to start with a rough check of the necessary condition for the convergence of the series. In order to eliminate uncertainty, we use an unsinkable multiplication and division method by adjoint expression:

The necessary sign of convergence did not work, but brought our Tambov comrade to light. As a result of the performed transformations, an equivalent series was obtained , which in turn strongly resembles a convergent series .

We write a clean solution:

Let's compare this series with converging next to . We use the limit comparison criterion:

Multiply and divide by the adjoint expression:

A finite number other than zero is obtained, which means that the series under study converges together with next to .

Perhaps some have a question, where did the wolves come from on our African safari? Don't know. They probably brought it. You will get the following trophy skin:

Example 18

Investigate the convergence of a series

An example solution at the end of the lesson

And, finally, one more thought that visits many students in despair: instead of whether to use a rarer criterion for the convergence of the series? Sign of Raabe, sign of Abel, sign of Gauss, sign of Dirichlet and other unknown animals. The idea is working, but in real examples it is implemented very rarely. Personally, in all the years of practice, I have only 2-3 times resorted to sign of Raabe when nothing really helped from the standard arsenal. I reproduce the course of my extreme quest in full:

Example 19

Investigate the convergence of a series

Solution: Without any doubt a sign of d'Alembert. In the course of calculations, I actively use the properties of degrees, as well as second wonderful limit:

Here's one for you. D'Alembert's sign did not give an answer, although nothing foreshadowed such an outcome.

After going through the manual, I found a little-known limit proven in theory and applied a stronger radical Cauchy criterion:

Here's two for you. And, most importantly, it is not at all clear whether the series converges or diverges (an extremely rare situation for me). Necessary sign of comparison? Without much hope - even if in an unthinkable way I figure out the order of growth of the numerator and denominator, this still does not guarantee a reward.

A complete d'Alembert, but the worst thing is that the series needs to be solved. Need. After all, this will be the first time that I give up. And then I remembered that there seemed to be some more powerful signs. Before me was no longer a wolf, not a leopard and not a tiger. It was a huge elephant waving a big trunk. I had to pick up a grenade launcher:

Sign of Raabe

Consider a positive number series.
If there is a limit , then:
a) At a row diverges. Moreover, the resulting value can be zero or negative.
b) At a row converges. In particular, the series converges for .
c) When Raabe's sign does not give an answer.

We compose the limit and carefully simplify the fraction:


Yes, the picture is, to put it mildly, unpleasant, but I was no longer surprised. lopital rules, and the first thought, as it turned out later, turned out to be correct. But first, for about an hour, I twisted and turned the limit using “usual” methods, but the uncertainty did not want to be eliminated. And walking in circles, as experience suggests, is a typical sign that the wrong way of solving has been chosen.

I had to turn to Russian folk wisdom: "If nothing helps, read the instructions." And when I opened the 2nd volume of Fichtenholtz, to my great joy I found a study of an identical series. And then the solution went according to the model.

6. Sign of Raabe

Theorem 6. If there is a limit:

then: 1) for the series (A) converges, 2) for the series diverges.

Proof. An auxiliary assertion is proved:

Statement 1. (12)

Proof. The expression is considered:

We took the logarithms of both sides of the equation:

Returned to the limit:

From equality (11), based on the definition of the limit of a numerical sequence, it follows that for any arbitrarily small there exists such that for the inequality:

1) Let, then. We denoted, then, starting from the number, it follows from inequality (13) that the following inequality is true:

take any number. According to (12), for sufficiently large, the following will be true:

From here, according to (14), it follows:

On the right - the ratio of two consecutive members of the Dirichlet series at; after applying Theorem 4, the convergence of the series (A) becomes obvious.

2) Let, then, similarly to paragraph (1), from (13) the following inequality follows:

From here we immediately found:

after applying Theorem 4 to the series (A) and the Dirichlet series, the divergence of the series (A) becomes clear.

Remark 5. Raabe's test is much stronger than d'Alembert's test

Remark 6. Raabe's criterion does not give an answer to the question posed.

11) Explore the series using the signs of d'Alembert and Raabe:

d'Alembert's test does not give an answer to the question of the convergence of this series. The series is investigated using the Raabe test:

This resulted in type uncertainty, so we applied the 1st L'Hospital-Bernoulli rule:

Rad diverges at, converges at, and at , the Raabe sign does not answer the question of convergence.

12) Explore the series using the Raabe sign:

It turned out the type uncertainty, but before applying the 1st L'Hopital-Bernoulli rule, the derivative of the expression is found, for this it is logarithmized and the derivative of the logarithm is sought:

Now you can find the derivative of the expression:

Back to the limit. The 1st L'Hospital-Bernoulli rule applies:

The expression is considered. After applying the 1st L'Hospital-Bernoulli rule to it:

From this it follows that:

Substitute this equality into the expression:

From here, according to the Raabe test, it follows that the given series diverges at, converges at, and when the Raabe test does not answer the question of the convergence of the series.

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Let --- solution system of equations, defined on some interval, and --- solution of the same system of equations, defined on some interval. We say that a solution is an extension of a solution if...

Formulation in limit form

Comment. If a Unable to parse expression (executable file texvc not found; See math/README for setup help.): R=1, then the Raabe criterion does not answer the question about the convergence of the series.

Proof

The proof is based on the use of a generalized comparison criterion when compared with a generalized harmonic series

see also

• The d'Alembert convergence test is a similar test based on the ratio of neighboring terms.

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Literature

• Arkhipov, G. I., Sadovnichiy, V. A., Chubarikov, V. N. Lectures on mathematical analysis: Textbook of Universities and Ped. universities / Ed. V. A. Sadovnichy. - M .: Higher School, 1999. - 695 p. - ISBN 5-06-003596-4..
• - article from the Mathematical Encyclopedia

Links

• Weisstein, Eric W.(English) on the Wolfram MathWorld website.

Let us take in Kummer's criterion as a divergent series (12.1) the harmonic series

In this case we have

The obtained criterion of convergence can be formulated as follows.

Theorem (Raabe convergence test). Row

converges if there is such that

This series diverges if, starting from some

The limit form of the Raabe sign is as follows:

then the series (12.9) converges, and if

then it diverges.

The Raabe convergence test is much more sensitive than the similar d'Alembert convergence test. Indeed, where the d'Alembert criterion, taken in its limiting form, establishes the convergence of the series (12.9):

there Raabe gives a sign.

Similarly, for a series whose divergence is indicated by the d'Alembert test, according to the Raabe test,

1. Consider the series

Here so that for each specific x

and the application of d'Alembert's sign is ineffectual here. The sign of Raabe gives

This shows that the series under consideration converges at and diverges at at . We note in passing that at , the series (12.10) turns into a harmonic one, which, as is known, diverges. The fact that the Raabe criterion in its original (non-limiting) form establishes the divergence of the harmonic series cannot be considered an independent result, since the assertion that constitutes the Raabe criterion is based on this divergence.

Compose the ratio of neighboring members of this series:

We will expand the logarithms on the right and square roots according to Taylor's power formula. In this and in the following examples, we will use limit tests for convergence. This means that we will have to increase the values ​​​​of the variable indefinitely. Therefore, each next degree will be infinitely small when increasing higher order compared to the previous ones. Discarding all degrees, starting with a certain one, we will make an error that will be small not only absolutely, but also in comparison with the last of the retained terms. This relative error will be the smaller, the larger the value and disappears in the limit with an unlimited increase in . Depending on the required accuracy of reasoning, we will retain one or another number of terms in the Taylor formulas for the corresponding functions. Further, we will connect with a sign expressions that differ from each other by values ​​that are small compared to the accuracy given by the retained and written out terms.

First, we confine ourselves to terms of logarithms and roots containing in a degree not higher than the first. We'll have

Consequently, d'Alembert's convergence criterion cannot give us any answer here either.