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 discovered a study of an identical series. And then the solution went according to the model:

Insofar as numerical sequence is considered a special case of the function, then in the limit we will make the substitution: . If , then .

As a result:

Now I have function limit and applicable L'Hopital's rule. In the process of differentiation, one has to take derivative of exponential function, which is technically convenient to find separately from the main solution:

Be patient, since we got here - Barmaley warned at the beginning of the article =) =)

I use L'Hopital's rule twice:

diverges.

A lot of time wasted, but my gate held!

For the sake of interest, I calculated 142 terms of the series in Excel (there was not enough computing power for more) and it seems (but not strictly theoretically guaranteed!), That even the necessary convergence criterion is not satisfied for this series. You can see the epic result here >>> After such misadventures, I could not resist the temptation to test the limit in the same amateur way.

Use on health, the solution is legal!

And this is your baby elephant:

Example 20

Investigate the convergence of a series

If you are well imbued with the ideas of this lesson, then handle this example! It is much simpler than the previous one ;-)

Our trip ended on a bright note, and I hope everyone left an unforgettable impression. Those who wish to continue the banquet can go to the page Ready-made problems in higher mathematics and download the archive with additional tasks on the topic.

Wish you luck!

Solutions and answers:

Example 2: Decision: compare this series with the convergent series . For all natural numbers, the inequality is true, which means that, by comparison, the series under study converges together with next to .

Example 4: Decision: compare this series with the divergent harmonic series. We use the limit comparison criterion:

(the product of an infinitesimal and a bounded one is an infinitesimal sequence)
diverges along with the harmonic series.

Example 5: Decision: we take the multiplier-constant of the common term outside the sum, the convergence or divergence of the series does not depend on it:

Let's compare this series with a converging infinitely decreasing geometric progression. The sequence is bounded: , therefore, the inequality holds for all natural numbers. And, therefore, according to the test of comparison, the series under study converges together with next to .

Example 8: Decision: compare this series with the diverging series (the common term multiplier constant does not affect the convergence or divergence of the series). We use the limit comparison criterion and the remarkable limit :

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

Example 13: Decision

Thus, the series under study converges.

Example 14: Decision: use the d'Alembert test:

Let us replace infinitesimal ones with equivalent ones: for .
We use the second remarkable limit: .

Therefore, the series under study diverges.
Multiply and divide by the adjoint expression:

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

Example 20: Decision: check the necessary condition for the convergence of the series. In the course of calculations, using a typical technique, we organize the 2nd remarkable limit:

Thus, the series under study diverges.

Higher mathematics for correspondence students and not only >>>

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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'Hospital-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.

Dodatkovі mind zbіzhnostі number rows

Take for Kummer's signs as a rozbіzhny row of harmonic rows (3.1). I can't help it. Otrimana of the sign of wealth can be formulated in such a rank. Theorem (sign of Raabe's abbreviation). A number, zbіgaєtsya, as if there is such a thing ...

alternating series

Theorem (Leibniz test). An alternating series converges if: The sequence of absolute values ​​of the terms of the series decreases monotonically, i.e. ; The common term of the series tends to zero:. Moreover, the sum S of the series satisfies the inequalities. Remarks...

Theorem 1 (d'Alembert test). Let a series be given where all > 0. If there is a limit, then at 0<1 ряд сходится, а при >1 row converges.

Alternating and alternating series

Theorem 2 (Cauchy test). Let a series be given. (1) If there is a finite limit, then 1) for , the series converges; 2) for , the series diverges.

Alternating and alternating series

Theorem 3 (integral criterion for convergence). Let the function f(x) be defined, continuous, positive and not increasing on the ray. Then: 1) the number series converges...

Alternating and alternating series

Definition. The number series a1 - a2 + a3 - … + (- 1) n - 1an + … , where all numbers an are positive, is called alternating. Example. A series is sign-alternating, but a series is not sign-alternating...

Integration differential equations using power series

In mathematical applications, as well as in solving some problems in economics, statistics, and other areas, sums with an infinite number of terms are considered. Here we will define what is meant by such amounts...

1.D.P.: Extend AC to AM1=OC and BD to DN1=OB. 2. By the Pythagorean theorem in?M1ON1: M1N1=10. 3. Draw M1KN1D. MK?AK=K. 4. ?BOC=?KAM1 (on the basis: BO=KM1, OC=AM1, by construction, BOC=KM1A=90, lying crosswise at BN1 KM1, M1C - secant) AK=BC. 5. M1KDN1 - parallelogram, DK=M1N1=10; MN=DK/2= (AD+BC)/2=5...

Various methods for solving planimetric problems

1.D.P.: Extend AC to AM1=OC and BD to DN=OB. 2. Consider?OMN, NOM=90°, then by the Pythagorean theorem in?MON MN=10. 3. Let's stand: AEMN, DFMN, OKBC. 4. ?AME = ?KOC and?DFN=?BOK (for trait II) ME=KC, FN=BKMN=BC+AD=a+b=10MN=10/2=5. Answer: MN=5...

Solvability of one boundary value problem

Consider a nonlinear boundary value problem: (1) (2) There is a representation (3) The operator is linear bounded symmetric; has a spectrum in the interval; - is positive, i.e. for any the inequality holds...

Let a positive series be given: , where. (A) Theorem 5. If there is a limit: , (5) then: 1) for the series (A) converges, 2) for the series diverges. Proof. From equality (5), based on the definition of the limit of the numerical sequence, it follows ...

Convergence of positive series

Theorem 6. If there is a limit: (18) then: 1) when the series (A) converges, 2) when - diverges. Proof. Proved using Kummer's scheme. Let be. We consider a series Let's compare it with a series that diverges...

Stability according to Lyapunov

Let be --- decision 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...

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 sign of d'Alembert, taken in his ultimate 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 value of the variable indefinitely. Therefore, each next degree will be an increase in an infinitesimal 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 will have

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

In cases where the signs of d'Alembert and Cauchy do not give a result, sometimes signs based on a comparison with other series that converge or diverge "slower" than a series of geometric progression can give an affirmative answer.

Let us present, without proof, the formulation of four more cumbersome criteria for the convergence of series. The proofs of these criteria are also based on comparison theorems 1–3 (Theorems 2.2 and 2.3) of the series under study with some series whose convergence or divergence has already been established. These proofs can be found, for example, in the fundamental textbook by G. M. Fikhtengol'ts (Vol. 2).

Theorem 2.6. Sign of Raabe. If for members of a positive number series , starting from some number M, the inequality

(Rn £ 1), "n ³ M, (2.10)

then the series converges (diverges).

Raabe's sign in the limiting form. If the terms of the above series satisfy the condition

Remark 6. If we compare the d'Alembert and Raabe tests, we can show that the second one is much stronger than the first one.

If the series has a limit

then the Raabe sequence has a limit

Thus, if the d'Alembert test gives an answer to the question of the convergence or divergence of the series, then the Raabe test also gives it, and these cases are covered by only two of the possible values ​​of R: +¥ and -¥. All other cases of finite R ¹ 1, when the Raabe test gives an affirmative answer to the question of convergence or divergence of the series, correspond to the case D = 1, i.e., the case when the d'Alembert test does not give an affirmative answer to the question of convergence or divergence of the series.

Theorem 2.7. Kummer sign. Let (сn) be an arbitrary sequence of positive numbers. If for members of a positive number series , starting from some number M, the inequality

(Qn £ 0), "n ³ M, (2.11)

then the series converges .

Kummer's test in the limiting form. If there is a limit for the above series

then the series converges .

From Kummer's test, as a corollary, it is easy to obtain evidence for d'Alembert's, Raabe's, and Bertrand's tests. The latter is obtained if we take as the sequence (сn)

cn=nln n, "n н N,

for which the series

diverges (the divergence of this series will be shown in the examples of this section).

Theorem 2.8. Bertrand's criterion in the limiting form. If for members of a positive number series the Bertrand sequence

(2.12)

(Rn is the Raabe sequence) has a limit

then the series converges (diverges).

Below, we formulate the Gauss test, which is the most powerful in the sequence of the applicability ranges of the convergence criteria of the series arranged in ascending order: d'Alembert, Raabe, and Bertrand. The Gauss test generalizes all the power of the previous tests and allows you to study much more complex series, but, on the other hand, its application requires more subtle studies in order to obtain an asymptotic expansion of the ratio of neighboring terms of the series to the second order of smallness with respect to .

Theorem 2.9. Gauss sign. If for members of a positive number series , starting from some number M, the equality

, "n ³ M, (2.13)

where l and p are constants and tn is a bounded value.

a) for l > 1 or l = 1 and p > 1, the series converges;

b) for l< 1 или l = 1 и р £ 1 ряд расходится.

2.5. Cauchy-Maclaurin integral test,

"Telescopic" sign of Cauchy and sign of Ermakov

The criteria for the convergence of series considered above are based on comparison theorems and are sufficient, i.e., if the conditions of the feature for a given series are met, certain statements about its behavior can be made, but if the conditions of the feature are not met for it, then nothing can be asserted about the convergence of the series, it can both converge and diverge.

The Cauchy-Maclaurin integral test differs from those studied above in content, being necessary and sufficient, and also in form, based on a comparison of an infinite sum (series) with an infinite (improper) integral, and demonstrates a natural relationship between the theory of series and the theory of integrals. This interrelation is also easily traced on the example of comparison criteria, the analogues of which take place for improper integrals and their formulations almost verbatim coincide with the formulations for series. A complete analogy is also observed in the formulations of sufficient criteria for the convergence of arbitrary numerical series, which will be studied in the next section, and criteria for the convergence of improper integrals, such as the criteria for the convergence of Abel and Dirichlet.

Below we will also give the “telescopic” Cauchy criterion and the original criterion for the convergence of series, obtained by the Russian mathematician V.P. Ermakov; Ermakov's test in its power has approximately the same scope as the Cauchy–Maclaurin integral test, but does not contain the terms and concepts of integral calculus in the formulation.

Theorem 2.10. The Cauchy-Maclaurin sign. Let for members of a positive number series , starting from some number M, the equality

where the function f(x) is non-negative and non-increasing on the half-line (x ³ M). The number series converges if and only if the improper integral converges

That is, the series converges if there is a limit

, (2.15)

and the series diverges if the limit is I = +¥.

Proof. By virtue of Remark 3 (see § 1), it is obvious that, without loss of generality, we can assume M = 1, since, having discarded (M – 1) terms of the series and making the replacement k = (n – M + 1), we come to consider the series , for which

, ,

and, accordingly, to the consideration of the integral .

Further, we note that the non-negative and non-increasing on the half-line (x ³ 1) function f(x) satisfies the conditions of Riemann integrability on any finite interval, and therefore the consideration of the corresponding improper integral makes sense.

Let's move on to the proof. On any segment of unit length m £ x £ m + 1, since f(x) does not increase, the inequality

By integrating it over a segment and using the corresponding property definite integral, we get the inequality

, . (2.16)

Summing these inequalities term by term from m = 1 to m = n, we obtain

Since f (x) is a non-negative function, the integral

is a non-decreasing continuous function of the argument A. Then

, .

From here and from inequality (15) it follows that:

1) if I< +¥ (т. е. несобственный интеграл сходится), то и неубывающая последовательность частичных сумм bounded, i.e., the series converges;

2) if I = +¥ (i.e., the improper integral diverges),

then the nondecreasing sequence of partial sums is also unbounded, i.e., the series diverges.

On the other hand, denoting , from inequality (16) we obtain:

1) if S< +¥ (т. е. ряд сходится), то для неубывающей continuous function I(A), "A ³ 1 there is a number n such that n + 1 ³ A, and I(A) £ I(n + 1) £ Sn £ S, and hence , i.e., the integral converges;

2) if S = +¥ (i.e., the series diverges), then for any sufficiently large A there exists n £ A such that I(A) ³ I(n) ³ Sn – f(1) ® +¥ (n ® ¥), i.e., the integral diverges. Q.E.D.

Let us present two more interesting criteria for convergence without proof.

Theorem 2.11. "Telescopic" sign of Cauchy. A positive numerical series whose terms are monotonically decreasing converges if and only if the series converges.

Theorem 2.12. Ermakov's sign. Let the members of a positive numerical series be such that, starting from some number M0, the equalities

an = ¦(n), "n ³ М0,

where the function ¦(x) is piecewise continuous, positive, and decreases monotonically as x ³ M0.

Then if there exists a number M ³ M0 such that for all x ³ M the inequality

,

then the series converges (diverges).

2.6. Examples of applying convergence criteria

Using Theorem 2, it is easy to investigate the convergence of the following series

(a > 0, b ³ 0; "a, b Î R).

If a £ 1, then the necessary criterion for convergence (property 2) is violated (see § 1).

,

hence the series diverges.

If а > 1, then сn satisfies the estimate , from which, due to the convergence of the series of geometric progression, the convergence of the considered series follows.

converges by comparison test 1 (Theorem 2.2), since we have the inequality

,

and the series converges as a series of geometric progression.

Let us show the divergence of several series, which follows from the comparison criterion 2 (Corollary 1 of Theorem 2.2). Row

diverges because

.

diverges because

.

diverges because

.

(p > 0)

diverges because

.

converges by the d'Alembert test (Theorem 2.4). Really

.

converges according to the d'Alembert test. Really

.

.

converges in the Cauchy test (Theorem 2.5). Really

.

Let us give an example of the application of the Raabe criterion. Consider the series

,

where the notation (k)!! means the product of all even (odd) numbers from 2 to k (from 1 to k) if k is even (odd). Using the d'Alembert test, we get

Thus, the d'Alembert test does not allow making a definite statement about the convergence of the series. We apply the Raabe sign:

hence the series converges.

Let us give examples of the application of the Cauchy–Maclaurin integral test.

Generalized harmonic series

converges or diverges simultaneously with the improper integral

It is obvious that I< +¥ при p >1 (the integral converges) and I = +¥ for p £ 1 (diverges). Thus, the original series also converges for p > 1 and diverges for p £ 1.

diverges simultaneously with the improper integral

so the integral diverges.

§ 3. Sign-alternating number series

3.1. Absolute and conditional convergence of series

In this section, we study the properties of series whose members are real numbers with an arbitrary sign.

Definition 1. Number series

is called absolutely convergent if the series converges

Definition 2. A number series (3.1) is said to be conditionally convergent or not absolutely convergent if the series (3.1) converges and the series (3.2) diverges.

Theorem 3.1. If a series converges absolutely, then it converges.

Proof. In accordance with the Cauchy criterion (Theorem 1.1), the absolute convergence of the series (3.1) is equivalent to the fulfillment of the relations

" e > 0, $ M > 0 such that " n > M, " p ³ 1 Þ

(3.3)

Since it is known that the modulus of the sum of several numbers does not exceed the sum of their moduli (“triangle inequality”), then from (3.3) the inequality follows (valid for the same numbers as in (3.3), numbers e, M, n, p)

The fulfillment of the last inequality means the fulfillment of the conditions of the Cauchy criterion for the series (3.1), therefore, this series converges.

Corollary 1. Let series (3.1) converge absolutely. Let us compose from the positive terms of the series (3.1), renumbering them in order (as they occur in the process of increasing the index), a positive numerical series

, (uk = ). (3.4)

Similarly, from the modules of the negative terms of the series (3.1), renumbering them in order, we compose the following positive numerical series:

, (vm = ). (3.5)

Then series (3.3) and (3.4) converge.

If we denote the sums of the series (3.1), (3.3), (3.4) respectively by the letters A, U, V, then the formula

A = U - V. (3.6)

Proof. Let us denote the sum of the series (3.2) by A*. By Theorem 2.1, we have that all partial sums of the series (3.2) are limited by the number A*, and since the partial sums of the series (3.4) and (3.5) are obtained by summing some of the terms of the partial sums of the series (3.2), it is obvious that they more limited by A*. Then, introducing the appropriate notation, we obtain the inequalities

;

from which, by virtue of Theorem 2.1, the series (3.4) and (3.5) converge.

(3.7)

Since the numbers k and m depend on n, it is obvious that, as n ® ¥, k ® ¥ and m ® ¥ simultaneously. Then, passing in equality (3.7) to the limit (all limits exist due to Theorem 3.1 and as proved above), we obtain

i.e., equality (3.6) is proved.

Corollary 2. Let the series (3.1) converge conditionally. Then series (3.4) and (3.5) diverge and formula (3.6) for conditionally convergent series is not true.

Proof. If we consider the nth partial sum of the series (3.1), then, as in the previous proof, it can be written

(3.8)

On the other hand, for the nth partial sum of the series (3.2) one can similarly write the expression

(3.9)

Assume the opposite, i.e., let at least one of the series (3.3) or (3.4) converge. Then it follows from formula (3.8) in view of the convergence of series (3.1) that the second of the series (respectively, (3.5) or (3.4)) converges as the difference of two convergent series. And then formula (3.9) implies the convergence of the series (3.2), i.e., the absolute convergence of the series (3.1), which contradicts the condition of the theorem on its conditional convergence.

Thus, from (3.8) and (3.9) it follows that since

Q.E.D.

Remark 1. Associative property for series. The sum of an infinite series essentially differs from the sum of a finite number of elements in that it includes the passage to the limit. Therefore, the usual properties of finite sums are often violated for series, or they are preserved only under certain conditions.

So, for finite sums, the combination (associative) law takes place, namely: the sum does not change if the elements of the sum are grouped in any order

Consider an arbitrary grouping (without permutation) of the terms of the numerical series (3.1). Denote the increasing sequence of numbers

and introduce the notation

Then the series obtained by the above method can be written as

The following theorem, without proof, collects several important statements related to the combination property of series.

Theorem 3.2.

1. If the series (3.1) converges and has the sum A (conditional convergence is sufficient), then an arbitrary series of the form (3.10) converges and has the same sum A. That is, the convergent series has the combination property.

2. The convergence of a series of the form (3.10) does not imply the convergence of the series (3.1).

3. If the series (3.10) is obtained by a special grouping, so that inside each of the brackets there are terms of only one sign, then the convergence of this series (3.10) implies the convergence of the series (3.1).

4. If series (3.1) is positive and some series of the form (3.10) converges for it, then series (3.1) converges.

5. If the sequence of terms of series (3.1) is infinitesimal (i.e., an) and the number of terms in each group, a member of series (3.10), is limited by one constant M (i.e., nk –nk–1 £ M, "k = 1, 2,…), then the convergence of the series (3.10) implies the convergence of the series (3.1).

6. If the series (3.1) converges conditionally, then without permutation it is always possible to group the terms of the series so that the resulting series (3.10) is absolutely convergent.

Remark 2. Commutative property for series. For finite numerical sums, a commutative (commutative) law holds, namely: the sum does not change with any permutation of the terms

where (k1, k2, …, kn) is an arbitrary permutation from the set of natural numbers (1, 2,…, n).

It turns out that a similar property holds for absolutely convergent series and does not hold for conditionally convergent series.

Let there be a one-to-one mapping of the set of natural numbers onto itself: N ® N, i.e., each natural number k corresponds to a unique natural number nk, and the set reproduces without gaps the entire natural series of numbers. Let us denote the series obtained from the series (3.1) with the help of an arbitrary permutation corresponding to the above mapping as follows:

The rules for applying the commutative properties of series are reflected in Theorems 3.3 and 3.4 given below without proof.

Theorem 3.3. If series (3.1) converges absolutely, then series (3.11) obtained by an arbitrary permutation of the terms of series (3.1) also converges absolutely and has the same sum as the original series.

Theorem 3.4. Riemann's theorem. If the series (3.1) converges conditionally, then the terms of this series can be rearranged so that its sum will be equal to any preassigned number D (finite or infinite: ±¥) or will be undefined.

Based on Theorems 3.3 and 3.4, it is easy to establish that the conditional convergence of the series results from mutual cancellation nth growth partial sum as n ® ¥ by adding to the sum either positive or negative terms, and therefore the conditional convergence of the series essentially depends on the order of the terms of the series. The absolute convergence of the series is the result of a rapid decrease in the absolute values ​​of the terms of the series

and does not depend on their order.

3.2. Alternating row. Leibniz sign

Among alternating series, an important particular class of series stands out - alternating series.

Definition 3. Let be a sequence of positive numbers bп > 0, "n н N. Then a series of the form

is called an alternating row. For series of the form (3.12) the following assertion holds.

Theorem 5. Leibniz test. If the sequence composed of the absolute values ​​of the terms of the alternating series (3.8) decreases monotonically to zero

bn > bn+1, "n н N; (3.13)

then such an alternating series (3.12) is called a Leibniz series. The Leibniz series always converges. For the remainder of the Leibniz series

there is an estimate

rn = (–1) nqnbn+1, (0 £ qn £ 1) "nнN. (3.14)

Proof. Let us write an arbitrary partial sum of the series (3.12) with an even number of terms in the form

By condition (3.13), each of the brackets on the right side of this expression is a positive number; therefore, as k increases, the sequence increases monotonically. On the other hand, any member of the B2k sequence can be written as

B2k = b1 – (b2 – b3) – (b4 – b5) –… – (b2k–2 – b2k–1) – b2k,

and since by condition (3.13) there is a positive number in each of the brackets of the last equality, the inequality obviously holds

B2k< b1, "k ³ 1.

Thus, we have a monotonically increasing and bounded from above sequence , and such a sequence, according to the well-known theorem from the theory of limits, has a finite limit

B2k–1 = B2k + b2k,

and taking into account that the common term of the series (according to the hypothesis of the theorem) tends to zero as n ® ¥, we obtain

Thus, we have proved that the series (3.12) converges under condition (3.13) and its sum is equal to B.

Let us prove estimate (3.14). It has been shown above that partial sums of even order B2k, monotonically increasing, tend to the limit B, the sum of the series.

Consider partial sums of odd order

B2k–1 = b1 – (b2 – b3) – (b4 – b5) – … – (b2k–2 – b2k–1).

It is obvious from this expression (since condition (3.13) is satisfied) that the sequence is decreasing and, consequently, by what was proved above, tends to its limit B from above. Thus, we have proved the inequality

0 < B2k < B < B2k–1 < b1. (3.15)

If we now consider the remainder of the series (3.12)

as a new alternating series with the first term bп+1, then for this series, based on inequality (3.15), we can write for even and odd indices, respectively,

r2k = b2k+1 – b2k+2 + …, 0< r2k < b2k+1,

r2k–1 = – b2k + b2k+1 – …, r2k< 0, | r2k–1 | < b2k.

Thus, we have proved that the remainder of the Leibniz series always has the sign of its first term and is less than it in absolute value, i.e., estimate (3.14) is satisfied for it. The theorem has been proven.

3.3. Signs of convergence of arbitrary numerical series

In this subsection, without proof, we present sufficient convergence criteria for numerical series with terms that are arbitrary real numbers (of any sign), moreover, these criteria are also suitable for series with complex terms.

2) the sequence is a sequence converging to zero (bп ® 0 as n ® ¥) with bounded change.

Then series (3.16) converges.

Theorem 3.9. Dirichlet sign. Let the terms of the number series (3.16) satisfy the conditions:

the sequence of partial sums of the series is bounded (inequalities (3.17));

2) the sequence is a monotonic sequence converging to zero (bп ® 0 as n ®¥).

Then series (3.16) converges.

Theorem 3.10. The second generalized sign of Abel. Let the terms of the number series (3.16) satisfy the conditions:

1) the series converges;

2) the sequence is an arbitrary sequence with limited change.

Then series (3.16) converges.

Theorem 3.11. Abel sign. Let the terms of the number series (3.16) satisfy the conditions:

1) the series converges;

2) the sequence is a monotone bounded sequence.

Then series (3.16) converges.

Theorem 3.12. Cauchy's theorem. If the series and converge absolutely and their sums are equal to A and B, respectively, then the series composed of all products of the form aibj (i = 1,2,…, ¥; j = 1,2,…,¥), numbered in any order , also converges absolutely and its sum is equal to AB.

3.4. Examples

Let us first consider several examples of absolute convergence of series. Below we assume that the variable x can be any real number.

2) diverges at |x| > e on the same basis of d'Alembert;

3) diverges for |x| = e by the d'Alembert test in unlimited form, since

due to the fact that the exponential sequence in the denominator tends to its limit, monotonically increasing,

(a ¹ 0 is a real number)

1) converges absolutely for |x/a|< 1, т. е. при |x| < |a|, так как в this case we have a series composed of members of a decreasing geometric progression with the denominator q = x/a, or by the radical Cauchy test (Theorem 2.5);

2) diverges at |x/a| ³ 1, i.e., for |x| ³ |a|, since in this case the necessary criterion for convergence is violated (property 2 (see § 1))

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 overwhelming 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. mathematical analysis. 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 1-2-3 pages of proof, but for now it is enough to show failure necessary condition series convergence, 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

Decision: 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

Decision: 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 smaller 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, for a series, both comparison criteria can be used (the inequality is true), and for a series, only the limiting criterion (the inequality is false).

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

Example 7

Investigate the convergence of a series

Decision: 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. Via 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 technique 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

Decision: 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.

For planning further actions in such examples, the 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
.

Decision: 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

Decision: 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

Decision: 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:

Compare this series with the convergent series . 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

Decision: 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 strong 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.