Now we will move on to the study of series whose members are real numbers of any sign.

Definition 1. We will call the series

absolutely convergent if the series converges

Note that this definition does not say anything about whether the series (1.49) itself is assumed to converge. It turns out that such an assumption would be unnecessary, since the following theorem is true.

Theorem 1.9. The convergence of series (1.50) implies the convergence of series (1.49).

Proof. Let us use the Cauchy criterion for the series (i.e., Theorem 1.1). It is required to prove that for any number there is a number such that for all numbers satisfying the condition and for any natural number the following inequality is true:

We fix any . Since the series (1.50) converges, then, by Theorem 1.1, there is a number such that for all numbers satisfying the condition and for any natural number the following inequality holds:

Since the modulus of the sum of several terms does not exceed the sum of their moduli, then

Comparing inequalities (1.52) and (1.53), we obtain inequalities (1.51). The theorem has been proven.

Definition 2. The series (1.49) is called conditionally convergent if this series converges, while the corresponding series of moduli (1.50) diverges.

An example of an absolutely convergent series is a series.

This series converges absolutely, because when the series (1.33) converges.

Let us give an example of a conditionally convergent series. Let us prove the conditional convergence of the series

Since the corresponding series of modules (harmonic series), as we already know, diverges, then to prove the conditional convergence of the series (1.54) it is enough to prove that this series converges. Let us prove that series (1.54) converges to the number . In paragraph 2 § 9 ch. 6 part 1 we obtained the decomposition according to the Maclaurin formula of the function

There, for all x from the segment, the following estimate of the remainder term was obtained.

Alternating rows. Leibniz's sign.
Absolute and conditional convergence

In order to understand the examples of this lesson, you need to have a good understanding of positive number series: understand what a series is, know the necessary sign for the convergence of a series, be able to apply comparison tests, d'Alembert's test, Cauchy's test. The topic can be raised almost from scratch by consistently studying the articles Rows for dummies And D'Alembert's sign. Cauchy's signs. Logically, this lesson is the third in a row, and it will allow you not only to understand the alternating rows, but also to consolidate the material already covered! There will be little novelty, and mastering the alternating rows will not be difficult. Everything is simple and accessible.

What is an alternating series? This is clear or almost clear from the name itself. Just a simple example.

Let's look at the series and describe it in more detail:

And now there will be a killer comment. The members of an alternating series have alternating signs: plus, minus, plus, minus, plus, minus, etc. to infinity.

Alignment provides a multiplier: if even, then there will be a plus sign, if odd, there will be a minus sign (as you remember from the lesson about number sequences, this thing is called a “flashing light”). Thus, an alternating series is “identified” by minus one to the degree “en”.

In practical examples, the alternation of the terms of the series can be provided not only by the multiplier, but also by its siblings: , , , …. For example:

The pitfall is “deceptions”: , , etc. - such multipliers do not provide sign change. It is absolutely clear that for any natural: , , . Rows with deceptions are slipped not only to especially gifted students, they arise from time to time “by themselves” during the solution functional series.

How to examine an alternating series for convergence? Use Leibniz's test. I don’t want to say anything about the German giant of thought Gottfried Wilhelm Leibniz, since in addition to mathematical works, he wrote several volumes on philosophy. Dangerous for the brain.

Leibniz's test: If the members of an alternating series monotonously decrease in modulus, then the series converges.

Or in two points:

1) The series is alternating.

2) The terms of the series decrease in modulus: , and decrease monotonically.

If these conditions are met, then the series converges.

Brief information about the module is given in the manual Hot formulas for school mathematics course, but for convenience once again:

What does “modulo” mean? The module, as we remember from school, “eats” the minus sign. Let's go back to the row . Mentally erase all the signs with an eraser and let's look at the numbers. We will see that every next series member less than the previous one. Thus, the following phrases mean the same thing:

– Members of the series regardless of sign are decreasing.
– Members of the series decrease modulo.
– Members of the series decrease By absolute value.
– Module the common term of the series tends to zero:

// End of help

Now let's talk a little about monotony. Monotony is boring consistency.

Members of the series strictly monotonous decrease in modulus if EVERY NEXT member of the series modulo LESS than previous: . For a row The strict monotonicity of decreasing is fulfilled; it can be described in detail:

Or we can say briefly: each next member of the series modulo less than the previous one: .

Members of the series not strictly monotonous decrease in modulo if EACH FOLLOWING member of the series modulo is NOT GREATER than the previous one: . Consider a series with factorial: Here there is a loose monotonicity, since the first two terms of the series are identical in modulus. That is, each next member of the series modulo no more than the previous one: .

Under the conditions of Leibniz's theorem, decreasing monotonicity must be satisfied (it does not matter whether it is strict or non-strict). In addition, members of the series can even increase in modulus for some time, but the “tail” of the series must necessarily be monotonically decreasing.

There is no need to be afraid of what I have piled up; practical examples will put everything in its place:

Example 1

The common term of the series includes the factor , and this prompts a natural idea to check whether the conditions of the Leibniz test are met:

1) Checking the row for alternation. Usually at this point the decision series is described in detail and pronounce the verdict “The series is alternating.”

2) Do the terms of the series decrease in absolute value? Here you need to solve the limit, which is most often very simple.

– the terms of the series do not decrease in modulus, and this automatically implies its divergence – for the reason that the limit does not exist *, that is, the necessary criterion for the convergence of the series is not fulfilled.

Example 9

Examine the series for convergence

Example 10

Examine the series for convergence

After a high-quality study of numerical positive and alternating series, with a clear conscience, you can move on to functional series, which are no less monotonous and monotonously interesting.

Example 2.

Investigate whether the series converges.

Because the

Then the series converges.

Integral convergence test

The integral criterion for convergence is expressed by the following theorem

Theorem 1.8.

Given a series with positive terms

If at the function is continuous, positive and does not increase, and at points takes on the values ​​, then the series(1.23) and improper integral(1.24) converge or diverge at the same time.

Proof.

If , then where

;

If integral (1.24) converges and , That with any natural Hence,

.

Since the sequence is monotonically increasing and bounded, then there exists, i.e. series (1.23) also converges. If series (1.23) converges and , then for any .

From equality (1.26) it follows that at any . The improper integral also converges.

Using the integral test, one can prove that the series

(1.27)

where is any real number, converges at and diverges at .

Indeed, it converges at and diverges at .

Alternating rows. Leibniz's test

Alternating next is a series that has any two terms with numbers and have opposite signs, i.e. series of the form

(1.30)

Proof.

Let us consider the partial sums of series (1.28) with even and odd numbers:

Let's transform the first of these sums:

Due to condition (1.29), the difference in each bracket is positive, so the sum and for everyone. So, the sequence of even partial sums is monotonically increasing and bounded. It has a limit, which we denote by , i.e. . Because the , then, taking into account the previous equality and condition (1.30), we obtain

So, the sequence of partial sums of a given series with even and odd numbers, respectively, have the same limit. It follows that the sequence of all partial sums of a series has a limit; those. the series converges.

Example.

Investigate whether a series converges

(1.31)

This series is alternating. It converges because it satisfies the conditions of the theorem

The estimate for the remainder of an alternating series is determined using the following theorem.

Theorem 1.10.

The sum of the remainder of an alternating series that satisfies the conditions of Leibniz's theorem has the sign of the first remaining term and does not exceed it in absolute value.

Proof.

Let us consider the remainder of series (1.28) after the terms. Let its sum, -i partial sum, then

Since the conditions of Theorem 1.9 are satisfied, then in front of everyone, i.e. , where

or

It is similarly proven that the sum of the remainder of the series after the terms satisfies the conditions , i.e. And .

Therefore, regardless of even or odd

Consider a series composed of modules of members of this series:

(1.34)

Theorem 1.11.

If the row(1.34) converges, then the series converges(1.33).

Proof.

Since the series (1.34) converges, then by virtue of the Cauchy criterion (Theorem 1.1) for any there exists such a number , then for all and any integer the inequality holds

.

That . This means that series (1.33) also converges.

Comment.

The convergence of series (1.33) does not imply the convergence of series (1.34). For example, a series converges (see Section 1.6), and the series of moduli of its members diverges (harmonic series, see Section 1.2).

absolutely convergent, if a series of moduli of its terms converges. For example, a series

is absolutely convergent, since the series of moduli of its terms converges, i.e. series (geometric progression with denominator , ).

An alternating series is called non-absolutely convergent (conditionally convergent), if it converges, but the series of moduli of its members diverges. For example, the series is not absolutely convergent (see remark).

Actions on rows.

Product of the series

Theorem 1.12.

If the row(1.35) converges, then the series(1.36) also converges, and

(1.37)

Proof.

Let us denote by u - e the partial sums of series (1.35) and (1.36), i.e.

Obviously, . If series (1.35) converges and its sum is equal to , i.e. , , That

In addition to series (1.35), consider the series

also converges absolutely and its sum is equal to

Comment.

The rules for operating on series do not always coincide with the rules for operating on finite sums. In particular, in finite sums you can arbitrarily change the order of the terms, group the terms however you like, and the sum will not change. The terms of a final sum can be added in reverse order; this is not possible for a series, because it does not have a last term.

It is not always possible to group members in a series. For example, a series

is divergent because

and there is no limit to its partial amounts. After grouping members

we obtain a convergent series, its sum is zero. With a different grouping of members

we obtain a convergent series whose sum is equal to one.

We present two theorems without proof.

Theorem 1.14.

Rearranging the terms of an absolutely convergent series does not violate its convergence; the sum of the series remains the same.

Theorem 1.15.

If a series does not converge absolutely, then by properly rearranging its terms it is always possible to give the sum of the series an arbitrary value and even make the series divergent.

Row

Let a series be given ∑ a n (\displaystyle \sum a_(n)) And α = lim ¯ n → ∞ ⁡ | a n | n (\displaystyle \alpha =\varlimsup _(n\to \infty )(\sqrt[(n)](|a_(n)|))). Then

The statement about convergence in the Cauchy and d'Alembert tests is derived from a comparison with geometric progression (with denominators lim ¯ n → ∞ ⁡ | a n + 1 a n | (\displaystyle \varlimsup _(n\to \infty )\left|(\frac (a_(n+1))(a_(n)))\right|) And α (\displaystyle \alpha ) respectively), about divergence - from the fact that the common term of the series does not tend to zero.

Cauchy's test is stronger than D'Alembert's test in the sense that if D'Alembert's test indicates convergence, then Cauchy's test indicates convergence; if Cauchy's test does not allow one to draw a conclusion about convergence, then D'Alembert's test also does not allow one to draw any conclusions; There are series for which Cauchy's test indicates convergence, but D'Alembert's test does not indicate convergence.

Integral Cauchy-Maclaurin test

Let a series be given ∑ n = 1 ∞ a n , a n ⩾ 0 (\displaystyle \sum _(n=1)^(\infty )a_(n),a_(n)\geqslant 0) and function f (x) : R → R (\displaystyle f(x):\mathbb (R) \to \mathbb (R) ) such that:

Then the series ∑ n = 1 ∞ a n (\displaystyle \sum _(n=1)^(\infty )a_(n)) and integral ∫ 1 ∞ f (x) d x (\displaystyle \int \limits _(1)^(\infty )f(x)dx) converge or diverge simultaneously, and ∀ k ⩾ 1 ∑ n = k ∞ a n ⩾ ∫ k ∞ f (x) d x ⩾ ∑ n = k + 1 ∞ a n (\displaystyle \forall k\geqslant 1\ \sum _(n=k)^(\infty )a_(n)\geqslant \int \limits _(k)^(\infty )f(x)dx\geqslant \sum _(n=k+1)^(\infty )a_(n))

Raabe's sign

Let a series be given ∑ a n (\displaystyle \sum a_(n)), a n > 0 (\displaystyle a_(n)>0) And R n = n (a n a n + 1 − 1) (\displaystyle R_(n)=n\left((\frac (a_(n))(a_(n+1)))-1\right)).

Raabe's test is based on comparison with the generalized harmonic series

Actions on rows

Examples

Consider the series 1 2 + 1 3 + 1 2 2 + 1 3 2 + 1 2 3 + . . . (\displaystyle (\frac (1)(2))+(\frac (1)(3))+(\frac (1)(2^(2)))+(\frac (1)(3^( 2)))+(\frac (1)(2^(3)))+...). For this row:

Thus, Cauchy's test indicates convergence, while D'Alembert's test does not allow us to draw any conclusions.

Consider the series ∑ n = 1 ∞ 2 n − (− 1) n (\displaystyle \sum _(n=1)^(\infty )2^(n-(-1)^(n)))

Thus, Cauchy's test indicates divergence, while D'Alembert's test does not allow us to draw any conclusions.

Row ∑ n = 1 ∞ 1 n α (\displaystyle \sum _(n=1)^(\infty )(\frac (1)(n^(\alpha )))) converges at α > 1 (\displaystyle \alpha >1) and diverges at α ⩽ 1 (\displaystyle \alpha \leqslant 1), however:

Thus, Cauchy's and d'Alembert's signs do not allow us to draw any conclusions.

Row ∑ n = 1 ∞ (− 1) n n (\displaystyle \sum _(n=1)^(\infty )(\frac ((-1)^(n))(n))) converges conditionally according to the Leibniz criterion, but not absolutely, since the harmonic series ∑ n = 1 ∞ | (− 1) n n | = ∑ n = 1 ∞ 1 n (\displaystyle \sum _(n=1)^(\infty )\left|(\frac ((-1)^(n))(n))\right|=\sum _(n=1)^(\infty )(\frac (1)(n))) diverges.

, is unlimited in the left neighborhood of the point b (\displaystyle b). Improper integral of the second kind ∫ a b f (x) d x (\displaystyle \int \limits _(a)^(b)f(x)dx) called absolutely convergent, if the integral converges ∫ a b | f(x) | d x (\displaystyle \int \limits _(a)^(b)|f(x)|dx).

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

Definition 2.2. A number series whose members after any number have different signs is called alternating sign .

For alternating series the following holds: general sufficient test for convergence.

Theorem 2.2. Let an alternating series be given

If a series composed of moduli of members of this series converges

then the alternating series (2.2) itself converges.

It should be noted that the converse statement is not true: if series (2.2) converges, this does not mean that series (2.3) will converge.

Definition 2.3. absolutely convergent , if a series composed of the moduli of its members converges.

An alternating series is called conditionally convergent , if it itself converges, but the series composed of the moduli of its members diverges.

Among alternating series, absolutely convergent series occupy a special place. Such series have a number of properties, which we will formulate without proof.

The product of two absolutely convergent series with sums is an absolutely convergent series whose sum is equal to .

Thus, absolutely convergent series are summed, subtracted, and multiplied like ordinary series. The sum of such series does not depend on the order in which the terms are written.

In the case of conditionally convergent series, the corresponding statements (properties), generally speaking, do not hold.

Thus, by rearranging the terms of a conditionally convergent series, it is possible to ensure that the sum of the series changes. For example, a series conditionally converges according to Leibniz's criterion. Let the sum of this series be equal to . Let's rewrite its terms so that after one positive term there will be two negative ones. We get a series

The amount has been halved!

Moreover, by rearranging the terms of a conditionally convergent series, one can obtain a convergent series with a predetermined sum or a divergent series (Riemann's theorem).

Therefore, operations on the series cannot be performed without ensuring their absolute convergence. To establish absolute convergence, all the signs of convergence of number series with positive terms are used, replacing the common term with its module everywhere.

Example 2.1. .

Solution. The original series is alternating. Let us consider a series composed of absolute values ​​of the members of a given series, i.e. row . Since , then the terms of a similar series are not greater than the terms of the Dirichlet series , which is known to converge. Therefore, based on the comparison criterion, this series converges absolutely. ,

Example 2.2. Examine the series for convergence.

Solution.

2) Consider a series composed of absolute terms. We examine it for convergence using d'Alembert's test

According to d'Alembert's criterion, a series composed of absolute terms converges. This means that the original alternating series converges absolutely. ,

Example 2.3. Examine the series for convergence .

Solution. 1) This row is alternating. We use Leibniz's criterion. Let's check if the conditions are met.

Therefore, the original series converges.

2) Consider a series composed of absolute terms. We examine it for convergence using the limiting comparison test. Consider a harmonic series that diverges.

Consequently, both series behave identically, i.e. a series composed of absolute terms also diverges. This means that the original alternating series converges conditionally. ,