Number series: definitions, properties, convergence criteria, examples, solutions. Numerical series of increased complexity Sufficient criteria for the convergence of a positive sign series
This article has collected and structured the information necessary to solve almost any example on the topic of number series, from finding the sum of a series to examining its convergence.
Article review.
Let's start with the definitions of a positive-sign, alternating-sign series and the concept of convergence. Next, consider standard series, such as a harmonic series, a generalized harmonic series, and recall the formula for finding the sum of an infinitely decreasing geometric progression. After that, we turn to the properties of convergent series, dwell on the necessary condition for the convergence of the series, and state sufficient conditions for the convergence of the series. We will dilute the theory by solving typical examples with detailed explanations.
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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 number 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: .
In the study of any numerical series for convergence, first of all, it is necessary to check the fulfillment of the necessary condition for convergence. Failure to comply with this condition indicates the divergence of the numerical series, that is, if , then the series diverges.
On the other hand, it must be understood that this condition is not sufficient. That is, the fulfillment of equality does not indicate the convergence of the numerical series. For example, for the harmonic series necessary condition convergence 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 feature is used very often and is very powerful tool research of 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 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.
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 the Russian folk wisdom: "If all else fails, 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:
Because the 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 good at ideas this lesson, then deal with 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 tasks in higher mathematics and download the archive with additional tasks on the topic.
Wish you success!
Solutions and answers:
Example 2: Solution: 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: Solution: 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: Solution: 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: Solution: 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: Solution
Thus, the series under study converges.
Example 14: Solution: 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: Solution: 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 part-time students and not only >>>
(Go to main page)
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 непрерывной функции I(А), "А ³ 1 существует номер n такой, что n + 1 ³ А, и I(А) £ I(n + 1) £ Sn £ S, а следовательно, , 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 a > 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 from formula (3.8), in view of the convergence of series (3.1), it follows that the second of the series (respectively, (3.5) or (3.4)) converges as the difference of two convergent series. But 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 associative (associative) law holds, 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 theorem below, without proof, collects several important statements related to associative property rows.
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 given 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 positive number, therefore, as k increases, the sequence monotonically increases. 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 variation.
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 данном случае имеем ряд, составленный из членов убывающей геометрической прогрессии со знаменателем q = x/a, либо по радикальному признаку Коши (теорема 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))
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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.
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