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).

Alternating series are series whose terms are alternately positive and negative. . Most often, alternating series are considered, in which the terms alternate one after the other: each positive is followed by a negative, each negative is followed by a positive. But there are alternating rows in which members alternate through two, three, and so on.

Consider an example of an alternating series, the beginning of which looks like this:

3 − 4 + 5 − 6 + 7 − 8 + ...

and immediately the general rules for recording alternating rows.

As with any series, to continue a given series, you need to specify a function that determines the common term of the series. In our case it is n + 2 .

How to set the alternation of signs of the members of a series? Multiplying a function by minus one to some degree. In what degree? Let us immediately emphasize that not every degree ensures the alternation of signs for the terms of the series.

Let's say we want the first term of the alternating series to have a positive sign, as is the case in the example above. Then minus one must be to the power n− 1 . Start substituting numbers starting from one into this expression and you will get as an exponent at minus one, then even, then odd number. That's what it is necessary condition alternating signs! We get the same result when n+ 1 . If we want the first term of the alternating series to be with a negative sign, then we can define this series by multiplying the function of the common term by one to the power n. We get an even number, an odd number, and so on. As we can see, the already described condition for alternating signs is fulfilled.

Thus, we can write the above alternating series in general form:

To alternate the signs of a series member, the power minus one can be the sum n and any positive or negative, even or odd number. The same applies to 3 n , 5n, ... That is, alternating the signs of the members of the alternating series provides the degree at minus one in the form of a sum n, multiplied by any odd number and any number.

What powers at minus one do not ensure the alternation of signs of the terms of the series? Those that are present in the form n, multiplied by any even number, to which any number, including zero, even or odd, has been added. Examples of indicators of such degrees: 2 n , 2n + 1 , 2n − 1 , 2n + 3 , 4n+ 3 ... In the case of such powers, depending on what number “en” is added to, multiplied by an even number, either only even or only odd numbers are obtained, which, as we have already found out, does not give alternation of signs of the terms of the series .

Alternating series - a special case alternating series . Alternating series are series with terms of arbitrary signs , that is, those that can be positive and negative in any order. An example of an alternating series:

3 + 4 + 5 + 6 − 7 + 8 − ...

Next, we consider the signs of convergence of alternating and alternating series. The conditional convergence of alternating series of signs can be established using the Leibniz test. And for a wider range of series - alternating series (including alternating series) - the criterion of absolute convergence applies.

Convergence of alternating series of signs. Leibniz's test

For series of alternating signs, the following criterion of convergence holds—the Leibniz criterion.

Theorem (Leibniz test). The series converges, and its sum does not exceed the first term, if the following two conditions are simultaneously satisfied:

• the absolute values ​​of the terms of the alternating series decrease: u1 > u 2 > u 3 > ... > u n>...;
• limit of its common term with unlimited increase n equal to zero.

Consequence. If we take the sum of the alternating series as the sum of its n terms, then the error allowed will not exceed the absolute value of the first discarded term.

Example 1. Investigate the convergence of the series

Solution. This is an alternating series. The absolute values ​​of its members decrease:

and the limit of the common term

equal to zero:

Both conditions of the Leibniz test are satisfied, so the series converges.

Example 2. Investigate the convergence of the series

Solution. This is an alternating series. First we prove that:

, .

If N= 1, then for all n > N inequality 12 holds n − 7 > n. In turn, for everyone n. Therefore, that is, the terms of the series decrease in absolute value. Let us find the limit of the general term of the series (using L'Hopital's rule):

The limit of the common term is zero. Both conditions of the Leibniz test are satisfied, so the answer to the question of convergence is positive.

Example 3. Investigate the convergence of the series

Solution. Given an alternating series. Let us find out whether the first condition of the Leibniz criterion is satisfied, that is, the requirement. For the requirement to be met, it is necessary that

We have made sure that the requirement is met for everyone n > 0 . Leibniz's first criterion is satisfied. Let's find the limit of the general term of the series:

.

The limit is not zero. Thus, the second condition of the Leibniz criterion is not satisfied, so convergence is out of the question.

Example 4. Investigate the convergence of the series

Solution. In this series, two negative terms are followed by two positive ones. This series is also alternating. Let's find out whether the first condition of Leibniz's test is satisfied.

The requirement is met for everyone n > 1 . Leibniz's first criterion is satisfied. Let's find out whether the limit of the general term is equal to zero (applying L'Hopital's rule):

.

We got zero. Thus, both conditions of the Leibniz criterion are satisfied. Convergence is taking place.

Example 5. Investigate the convergence of the series

Solution. This is an alternating series. Let's find out whether the first condition of Leibniz's test is satisfied. Because

,

Because n ≥ 0 , then 3 n+ 2 > 0 . In turn, for everyone n, That's why . Consequently, the terms of the series decrease in absolute value. Leibniz's first criterion is satisfied. Let's find out whether the limit of the general term of the series is equal to zero (applying L'Hopital's rule):

.

We got a zero value. Both conditions of the Leibniz test are satisfied, so this series converges.

Example 6. Investigate the convergence of the series

Solution. Let us find out whether the first condition of the Leibniz test is satisfied for this alternating series:

The terms of the series decrease in absolute value. Leibniz's first criterion is satisfied. Let's find out whether the limit of the common term is equal to zero:

.

The limit of the common term is not zero. The second condition of Leibniz's criterion is not satisfied. Therefore, this series diverges.

Leibniz's test is a sign conditional convergence of the series. This means that the conclusions about the convergence and divergence of the alternating series considered above can be supplemented: these series converge (or diverge) conditionally.

Absolute convergence of alternating series

Let the row

– alternating sign. Let us consider a series composed of the absolute values ​​of its members:

Definition. A series is said to be absolutely convergent if a series composed of the absolute values ​​of its members converges. If an alternating series converges, and a series composed of the absolute values ​​of its members diverges, then such an alternating series is called conditionally or non-absolutely convergent .

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

Example 7. Determine whether a series converges

Solution. Corresponding to this series next to the positive terms is the series This generalized harmonic series, in which , therefore the series diverges. Let's check whether the conditions of the Leibniz test are met.

Let's write the absolute values ​​of the first five terms of the series:

.

As we can see, the terms of the series decrease in absolute value. Leibniz's first criterion is satisfied. Let's find out whether the limit of the common term is equal to zero:

We got a zero value. Both conditions of Leibniz's criterion are satisfied. That is, according to Leibniz’s criterion, convergence takes place. And the corresponding series with positive terms diverges. Therefore, this series converges conditionally.

Example 8. Determine whether a series converges

absolutely, conditionally, or diverges.

Solution. Corresponding to this series next to the positive terms is the series This is a generalized harmonic series, in which, therefore, the series diverges. Let's check whether the conditions of the Leibniz test are met.

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.

with (generally speaking) complex terms, for which the series converges

For the absolute convergence of the series (1) it is necessary and sufficient (Cauchy criterion for the absolute convergence of the series) that for any there exists a number such that for all numbers and all integers the following holds:

If a series is absolutely convergent, then it converges. Row

absolutely converges and a row

converges, but not absolutely. Let

A series composed of the same terms as series (1), but taken, generally speaking, in a different order. From the absolute convergence of series (1) it follows that absolute series(3), and series (3) has the same sum as series (1). If the rows

absolutely converge, then: any linear combination of them

also absolutely converges; a series obtained from all possible pairwise products of terms of these series, arranged in an arbitrary order, is also absolutely convergent and its sum is equal to the product of the sums of these series. The listed properties of absolutely convergent series carry over to multiple rows

absolutely converges, i.e., all series obtained by sequentially summing the members of series (4) by indices converge absolutely, and the sums of multiple series (4) and repeated series (5) are equal and coincide with the sum of any single series formed from all members of series (4 ).

If the terms of series (1) are elements of a certain Banach space with the norm of elements, then series (1) is called. absolutely convergent if the series converges

In case of A. s. R. elements of a Banach space, the above-considered properties of absolutely convergent number series, in particular, algebraic systems, are generalized. R. elements of a Banach space converges in this space. In a similar way, the concept of A. s. R. carries over to multiple series in Banach space.

Mathematical encyclopedia. - M.: Soviet Encyclopedia. I. M. Vinogradov. 1977-1985.

See what “ABSOLUTELY CONVERGENT SERIES” is in other dictionaries:

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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.