Basic definitions.
Definition. The sum of the terms of an infinite number sequence is called number series.
At the same time, the numbers 
we will call them members of the series, and u n– a common member of the series.
Definition.
Amounts 
,n
= 1, 2, …
are called private (partial) amounts row.
Thus, it is possible to consider sequences of partial sums of the series S 1 , S 2 , …, S n , …
Definition.
Row 
called convergent, if the sequence of its partial sums converges. Sum of convergent series is the limit of the sequence of its partial sums.
Definition. If the sequence of partial sums of a series diverges, i.e. has no limit, or has an infinite limit, then the series is called divergent and no amount is assigned to it.
Properties of rows.
1) The convergence or divergence of the series will not be violated if you change, discard or add a finite number of terms of the series.
2) Consider two rows 
And 
, where C is a constant number.
Theorem.
If the row 
converges and its sum is equalS, then the series 
also converges, and its sum is equal to CS.
(C
0)
3) Consider two rows 
And 
.Amount or difference of these series will be called a series 
, where the elements are obtained by adding (subtracting) the original elements with the same numbers.
Theorem.
If the rows 
And 
converge and their sums are equal respectivelySAnd
, then the series 
also converges and its sum is equalS
+
.
The difference of two convergent series will also be a convergent series.
The sum of a convergent and a divergent series is a divergent series.
It is impossible to make a general statement about the sum of two divergent series.
When studying series, they mainly solve two problems: studying convergence and finding the sum of the series.
Cauchy criterion.
(necessary and sufficient conditions for the convergence of the series)
In order for the sequence 
was convergent, it is necessary and sufficient that for any 
there was such a numberN, that atn
>
Nand anyp> 0, where p is an integer, the following inequality would hold:
.
Proof. (necessity)
Let 
, then for any number
there is a number N such that the inequality
is fulfilled when n>N. For n>N and any integer p>0 the inequality also holds 
. Taking into account both inequalities, we obtain:
The need has been proven. We will not consider the proof of sufficiency.
Let us formulate the Cauchy criterion for the series.
In order for the series 
was convergent, it is necessary and sufficient that for any 
there was a numberNsuch that atn>
Nand anyp>0 the inequality would hold
.
However, in practice, using the Cauchy criterion directly is not very convenient. Therefore, as a rule, simpler convergence tests are used:
1) If the row
converges, then it is necessary that the common term u n tended to zero. However, this condition is not sufficient. We can only say that if the common term does not tend to zero, then the series definitely diverges. For example, the so-called harmonic series 
is divergent, although its common term tends to zero.
Example. Investigate the convergence of the series 
We'll find 
- the necessary criterion for convergence is not satisfied, which means the series diverges.
2) If a series converges, then the sequence of its partial sums is bounded.
However, this sign is also not sufficient.
For example, the series 1-1+1-1+1-1+ … +(-1) n +1 +… diverges, because the sequence of its partial sums diverges due to the fact that
However, the sequence of partial sums is limited, because 
at any n.
Series with non-negative terms.
When studying series of constant sign, we will limit ourselves to considering series with non-negative terms, because simply multiplying by –1 from these series can yield series with negative terms.
Theorem.
For the convergence of the series 
with non-negative terms it is necessary and sufficient for the partial sums of the series to be bounded.
A sign for comparing series with non-negative terms.
Let two rows be given
And 
at u n ,
v n
0
.
Theorem.
If u n
v n at any n, then from the convergence of the series 
the series converges
, and from the divergence of the series
the series diverges
.
Proof.
Let us denote by S n
And
n partial sums of series
And 
. Because according to the conditions of the theorem, the series 
converges, then its partial sums are bounded, i.e. in front of everyone n n M, where M is a certain number. But because u n
v n, That S n
n then the partial sums of the series
are also limited, and this is sufficient for convergence.
Example. Examine the series for convergence 
Because 
, and the harmonic series 
diverges, then the series diverges 
.
Example.
Because 
, and the series 
converges (like a decreasing geometric progression), then the series 
also converges.
The following convergence sign is also used:
Theorem.
If 
and there is a limit 
, Whereh– a number other than zero, then the series 
And 
behave identically in terms of convergence.
D'Alembert's sign.
(Jean Leron d'Alembert (1717 - 1783) - French mathematician)
If for a series 
with positive terms there is such a numberq<1,
что для всех достаточно больших
ninequality holds
then a series 
converges, if for all there are sufficiently largencondition is met
then a series 
diverges.
D'Alembert's limiting sign.
D'Alembert's limiting criterion is a consequence of the above D'Alembert criterion.
If there is a limit 
, then when
< 1 ряд сходится, а при
> 1 – diverges. If
= 1, then the question of convergence cannot be answered.
Example. Determine the convergence of the series 
.
Conclusion: the series converges.
Example. Determine the convergence of the series 
Conclusion: the series converges.
Cauchy's sign. (radical sign)
If for a series 
with non-negative terms there is such a numberq<1,
что для всех достаточно больших
ninequality holds
,
then a series 
converges, if for all there are sufficiently largeninequality holds
then a series 
diverges.
Consequence.
If there is a limit 
, then when<1
ряд сходится, а при >Row 1 diverges.
Example. Determine the convergence of the series 
.
Conclusion: the series converges.
Example. Determine the convergence of the series 
.
Those. The Cauchy test does not answer the question of the convergence of the series. Let us check that the necessary convergence conditions are satisfied. As mentioned above, if a series converges, then the common term of the series tends to zero.
,
Thus, the necessary condition for convergence is not satisfied, which means the series diverges.
Integral Cauchy test.
If
(x) is a continuous positive function decreasing over the interval And 
then the integrals 
And 
behave identically in terms of convergence.
Alternating series.
Alternating rows.
An alternating series can be written as:
Where 
Leibniz's sign.
If the sign of the alternating row
absolute valuesu i are decreasing 
and the common term tends to zero 
, then the series converges.
Absolute and conditional convergence of series.
Let's consider some alternating series (with terms of arbitrary signs).
(1)
and a series composed of the absolute values of the members of the series (1):
(2)
Theorem. From the convergence of series (2) follows the convergence of series (1).
Proof. Series (2) is a series with non-negative terms. If series (2) converges, then by the Cauchy criterion for any >0 there is a number N such that for n>N and any integer p>0 the following inequality is true:
According to the property of absolute values:
That is, according to the Cauchy criterion, from the convergence of series (2) the convergence of series (1) follows.
Definition.
Row 
called absolutely convergent, if the series converges 
.
It is obvious that for series of constant sign the concepts of convergence and absolute convergence coincide.
Definition.
Row 
called conditionally convergent, if it converges and the series 
diverges.
D'Alembert's and Cauchy's tests for alternating series.
Let 
- alternating series.
D'Alembert's sign.
If there is a limit 
, then when<1
ряд
will be absolutely convergent, and when>
Cauchy's sign.
If there is a limit 
, then when<1
ряд
will be absolutely convergent, and if >1 the series will be divergent. When =1, the sign does not give an answer about the convergence of the series.
Properties of absolutely convergent series.
1)
Theorem.
For absolute convergence of the series 
it is necessary and sufficient that it can be represented as the difference of two convergent series with non-negative terms.
Consequence. A conditionally convergent series is the difference of two divergent series with non-negative terms tending to zero.
2) In a convergent series, any grouping of the terms of the series that does not change their order preserves the convergence and magnitude of the series.
3) If a series converges absolutely, then the series obtained from it by any permutation of terms also converges absolutely and has the same sum.
By rearranging the terms of a conditionally convergent series, one can obtain a conditionally convergent series having any predetermined sum, and even a divergent series.
4) Theorem. For any grouping of members of an absolutely convergent series (in this case, the number of groups can be either finite or infinite, and the number of members in a group can be either finite or infinite), a convergent series is obtained, the sum of which is equal to the sum of the original series.
5) If the rows 
And 
converge absolutely and their sums are equal respectively S
and , then a series composed of all products of the form 
taken in any order, also converges absolutely and its sum is equal to S
- the product of the sums of the multiplied series.
If you multiply conditionally convergent series, you can get a divergent series as a result.
Functional sequences.
Definition. If the members of the series are not numbers, but functions of X, then the series is called functional.
The study of the convergence of functional series is more complicated than the study of numerical series. The same functional series can, with the same variable values X converge, and with others - diverge. Therefore, the question of convergence of functional series comes down to determining those values of the variable X, at which the series converges.
The set of such values is called area of convergence.
Since the limit of each function included in the convergence region of the series is a certain number, the limit of the functional sequence will be a certain function:
Definition. Subsequence ( f n (x) } converges to function f(x) on the segment if for any number >0 and any point X from the segment under consideration there is a number N = N(, x), such that the inequality
is fulfilled when n>N.
With the selected value >0, each point of the segment has its own number and, therefore, there will be an infinite number of numbers corresponding to all points of the segment. If you choose the largest of all these numbers, then this number will be suitable for all points of the segment, i.e. will be common to all points.
Definition. Subsequence ( f n (x) } converges uniformly to function f(x) on the segment , if for any number >0 there is a number N = N() such that the inequality
is fulfilled for n>N for all points of the segment.
Example. Consider the sequence 
This sequence converges on the entire number line to the function f(x)=0 , because
Let's build graphs of this sequence:
sinx 
As can be seen, with increasing number n the sequence graph approaches the axis X.
Functional series.
Definition.
Private (partial) amounts functional range 
functions are called 
Definition.
Functional range 
called convergent at point ( x=x 0
), if the sequence of its partial sums converges at this point. Sequence limit 
called amount row 
at the point X 0
.
Definition.
Set of all values X, for which the series converges 
called area of convergence row.
Definition.
Row 
called uniformly convergent on the interval if the sequence of partial sums of this series converges uniformly on this interval.
Theorem. (Cauchy criterion for uniform convergence of series)
For uniform convergence of the series 
it is necessary and sufficient that for any number
>0 such a number existedN(
), which atn>
Nand any wholep>0 inequality
would hold for all x on the interval [a, b].
Theorem. (Weierstrass test for uniform convergence)
(Karl Theodor Wilhelm Weierstrass (1815 – 1897) – German mathematician)
Row 
converges uniformly and absolutely on the interval [a,
b], if the moduli of its terms on the same segment do not exceed the corresponding terms of a convergent number series with positive terms:
those. there is an inequality:
.
They also say that in this case the functional series 
is majorized number series 
.
Example. Examine the series for convergence 
.
Because 
always, it is obvious that 
.
Moreover, it is known that the general harmonic series 
when=3>1 converges, then, in accordance with the Weierstrass test, the series under study converges uniformly and, moreover, in any interval.
Example. Examine the series for convergence 
.
On the interval [-1,1] the inequality holds 
those. according to the Weierstrass criterion, the series under study converges on this segment, but diverges on the intervals (-, -1) (1, ).
Properties of uniformly convergent series.
1) Theorem on the continuity of the sum of a series.
If the members of the series 
- continuous on the segment [a,
b] function and the series converges uniformly, then its sumS(x) is a continuous function on the interval [a,
b].
2) Theorem on term-by-term integration of a series.
Uniformly converging on the segment [a, b] a series with continuous terms can be integrated term by term on this interval, i.e. a series composed of integrals of its terms over the segment [a, b] , converges to the integral of the sum of the series over this segment.
3) Theorem on term-by-term differentiation of a series.
If the members of the series 
converging on the segment [a,
b] represent continuous functions having continuous derivatives, and a series composed of these derivatives 
converges uniformly on this segment, then this series converges uniformly and can be differentiated term by term.
Based on the fact that the sum of the series is some function of the variable X, you can perform the operation of representing a function in the form of a series (expansion of a function into a series), which is widely used in integration, differentiation and other operations with functions.
In practice, power series expansion of functions is often used.
Power series.
Definition. Power series is called a series of the form
.
To study the convergence of power series, it is convenient to use D'Alembert's test.
Example. Examine the series for convergence 
We apply d'Alembert's sign:
.
We find that this series converges at 
and diverges at 
.
Now we determine the convergence at the boundary points 1 and –1.
For x = 1: 
The series converges according to Leibniz's criterion (see Leibniz's sign.).
At x = -1: 
the series diverges (harmonic series).
Abel's theorems.
(Nils Henrik Abel (1802 – 1829) – Norwegian mathematician)
Theorem.
If the power series 
converges atx
=
x 1
, then it converges and, moreover, for absolutely everyone 
.
Proof. According to the conditions of the theorem, since the terms of the series are limited, then
Where k- some constant number. The following inequality is true:
From this inequality it is clear that when x<
x 1
the numerical values of the terms of our series will be less (at least not more) than the corresponding terms of the series on the right side of the inequality written above, which form a geometric progression. The denominator of this progression 
according to the conditions of the theorem, it is less than one, therefore, this progression is a convergent series.
Therefore, based on the comparison criterion, we conclude that the series 
converges, which means the series 
converges absolutely.
Thus, if the power series 
converges at a point X 1
, then it converges absolutely at any point in the interval of length 2 
centered at a point X
= 0.
Consequence.
If at x = x 1
the series diverges, then it diverges for everyone 
.
Thus, for each power series there is a positive number R such that for all X such that 
the series is absolutely convergent, and for all 
the row diverges. In this case, the number R is called radius of convergence. The interval (-R, R) is called convergence interval.
Note that this interval can be closed on one or both sides, or not closed.
The radius of convergence can be found using the formula:
Example. Find the area of convergence of the series 
Finding the radius of convergence 
.
Therefore, this series converges for any value X. The common term of this series tends to zero.
Theorem.
If the power series 
converges for a positive value x=x 1
, then it converges uniformly in any interval inside 
.
Actions with power series.
This article provides structured and detailed information that may be useful when analyzing exercises and tasks. We will look at the topic of number series.
This article begins with basic definitions and concepts. Next, we will use standard options and study the basic formulas. In order to consolidate the material, the article provides basic examples and tasks.
Yandex.RTB R-A-339285-1
Basic theses
First, let's imagine the system: a 1 , a 2 . . . , a n , . . . , where a k ∈ R, k = 1, 2. . . .
For example, let's take numbers such as: 6, 3, - 3 2, 3 4, 3 8, - 3 16, . . . .
Definition 1
A number series is the sum of terms ∑ a k k = 1 ∞ = a 1 + a 2 + . . . + a n + . . . .
To better understand the definition, consider the given case in which q = - 0. 5: 8 - 4 + 2 - 1 + 1 2 - 1 4 + . . . = ∑ k = 1 ∞ (- 16) · - 1 2 k .
Definition 2
a k is general or k –th member of the series.
It looks something like this - 16 · - 1 2 k.
Definition 3
Partial sum of series looks something like this S n = a 1 + a 2 + . . . + a n , in which n– any number. S n is nth the sum of the series.
For example, ∑ k = 1 ∞ (- 16) · - 1 2 k is S 4 = 8 - 4 + 2 - 1 = 5.
S 1 , S 2 , . . . , S n , . . . form an infinite sequence of numbers.
For a row nth the sum is found by the formula S n = a 1 · (1 - q n) 1 - q = 8 · 1 - - 1 2 n 1 - - 1 2 = 16 3 · 1 - - 1 2 n. We use the following sequence of partial sums: 8, 4, 6, 5, . . . , 16 3 · 1 - - 1 2 n , . . . .
Definition 4
The series ∑ k = 1 ∞ a k is convergent when the sequence has a finite limit S = lim S n n → + ∞ . If there is no limit or the sequence is infinite, then the series ∑ k = 1 ∞ a k is called divergent.
Definition 5
The sum of a convergent series∑ k = 1 ∞ a k is the limit of the sequence ∑ k = 1 ∞ a k = lim S n n → + ∞ = S .
In this example, lim S n n → + ∞ = lim 16 3 t → + ∞ · 1 - 1 2 n = 16 3 · lim n → + ∞ 1 - - 1 2 n = 16 3 , row ∑ k = 1 ∞ (- 16) · - 1 2 k converges. The sum is 16 3: ∑ k = 1 ∞ (- 16) · - 1 2 k = 16 3 .
Example 1
An example of a divergent series is the sum of a geometric progression with a denominator greater than one: 1 + 2 + 4 + 8 +. . . + 2 n - 1 + . . . = ∑ k = 1 ∞ 2 k - 1 .
The nth partial sum is given by S n = a 1 (1 - q n) 1 - q = 1 (1 - 2 n) 1 - 2 = 2 n - 1, and the limit of partial sums is infinite: lim n → + ∞ S n = lim n → + ∞ (2 n - 1) = + ∞ .
Another example of a divergent number series is a sum of the form ∑ k = 1 ∞ 5 = 5 + 5 + . . . . In this case, the nth partial sum can be calculated as Sn = 5n. The limit of partial sums is infinite lim n → + ∞ S n = lim n → + ∞ 5 n = + ∞ .
Definition 6
A sum of the same form as ∑ k = 1 ∞ = 1 + 1 2 + 1 3 + . . . + 1 n + . . . - This harmonic number series.
Definition 7
Sum ∑ k = 1 ∞ 1 k s = 1 + 1 2 s + 1 3 s + . . . + 1 n s + . . . , Where s– a real number, is a generalized harmonic number series.
The definitions discussed above will help you solve most examples and problems.
In order to complete the definitions, it is necessary to prove certain equations.
- ∑ k = 1 ∞ 1 k – divergent.
We use the reverse method. If it converges, then the limit is finite. We can write the equation as lim n → + ∞ S n = S and lim n → + ∞ S 2 n = S . After certain actions we get the equality l i m n → + ∞ (S 2 n - S n) = 0.
Against,
S 2 n - S n = 1 + 1 2 + 1 3 + . . . + 1 n + 1 n + 1 + 1 n + 2 + . . . + 1 2 n - - 1 + 1 2 + 1 3 + . . . + 1 n = 1 n + 1 + 1 n + 2 + . . . + 1 2 n
The following inequalities are valid: 1 n + 1 > 1 2 n, 1 n + 1 > 1 2 n, . . . , 1 2 n - 1 > 1 2 n . We get that S 2 n - S n = 1 n + 1 + 1 n + 2 + . . . + 1 2 n > 1 2 n + 1 2 n + . . . + 1 2 n = n 2 n = 1 2 . The expression S 2 n - S n > 1 2 indicates that lim n → + ∞ (S 2 n - S n) = 0 is not achieved. The series is divergent.
- b 1 + b 1 q + b 1 q 2 + . . . + b 1 q n + . . . = ∑ k = 1 ∞ b 1 q k - 1
It is necessary to confirm that the sum of a sequence of numbers converges at q< 1 , и расходится при q ≥ 1 .
According to the above definitions, the amount n terms is determined according to the formula S n = b 1 · (q n - 1) q - 1 .
If q< 1 верно
lim n → + ∞ S n = lim n → + ∞ b 1 · q n - 1 q - 1 = b 1 · lim n → + ∞ q n q - 1 - lim n → + ∞ 1 q - 1 = = b 1 · 0 - 1 q - 1 = b 1 q - 1
We have proven that the number series converges.
For q = 1 b 1 + b 1 + b 1 + . . . ∑ k = 1 ∞ b 1 . The sums can be found using the formula S n = b 1 · n, the limit is infinite lim n → + ∞ S n = lim n → + ∞ b 1 · n = ∞. In the presented version, the series diverges.
If q = - 1, then the series looks like b 1 - b 1 + b 1 - . . . = ∑ k = 1 ∞ b 1 (- 1) k + 1 . Partial sums look like S n = b 1 for odd n, and S n = 0 for even n. Having considered this case, we will make sure that there is no limit and the series is divergent.
For q > 1, lim n → + ∞ S n = lim n → + ∞ b 1 · (q n - 1) q - 1 = b 1 · lim n → + ∞ q n q - 1 - lim n → + ∞ 1 q - 1 = = b 1 · ∞ - 1 q - 1 = ∞
We have proven that the number series diverges.
- The series ∑ k = 1 ∞ 1 k s converges if s > 1 and diverges if s ≤ 1.
For s = 1 we obtain ∑ k = 1 ∞ 1 k , the series diverges.
When s< 1 получаем 1 k s ≥ 1 k для k, natural number. Since the series is divergent ∑ k = 1 ∞ 1 k , there is no limit. Following this, the sequence ∑ k = 1 ∞ 1 k s is unbounded. We conclude that the selected series diverges when s< 1 .
It is necessary to provide evidence that the series ∑ k = 1 ∞ 1 k s converges for s > 1.
Let's imagine S 2 n - 1 - S n - 1:
S 2 n - 1 - S n - 1 = 1 + 1 2 s + 1 3 s + . . . + 1 (n - 1) s + 1 n s + 1 (n + 1) s + . . . + 1 (2 n - 1) s - - 1 + 1 2 s + 1 3 s + . . . + 1 (n - 1) s = 1 n s + 1 (n + 1) s + . . . + 1 (2 n - 1) s
Let us assume that 1 (n + 1) s< 1 n s , 1 (n + 2) s < 1 n s , . . . , 1 (2 n - 1) s < 1 n s , тогда S 2 n - 1 - S n - 1 = 1 n s + 1 (n + 1) s + . . . + 1 (2 n - 1) s < < 1 n s + 1 n s + . . . + 1 n s = n n s = 1 n s - 1
Let's imagine the equation for numbers that are natural and even n = 2: S 2 n - 1 - S n - 1 = S 3 - S 1 = 1 2 s + 1 3 s< 1 2 s - 1 n = 4: S 2 n - 1 - S n - 1 = S 7 - S 3 = 1 4 s + 1 5 s + 1 6 s + 1 7 s < 1 4 s - 1 = 1 2 s - 1 2 n = 8: S 2 n - 1 - S n - 1 = S 15 - S 7 = 1 8 s + 1 9 s + . . . + 1 15 s < 1 8 s - 1 = 1 2 s - 1 3 . . .
We get:
∑ k = 1 ∞ 1 k s = 1 + 1 2 s + 1 3 s + 1 4 s + . . . + 1 7 s + 1 8 s + . . . + 1 15 s + . . . = = 1 + S 3 - S 1 + S 7 - S 3 + S 15 + S 7 + . . .< < 1 + 1 2 s - 1 + 1 2 s - 1 2 + 1 2 s - 1 3 + . . .
The expression is 1 + 1 2 s - 1 + 1 2 s - 1 2 + 1 2 s - 1 3 + . . . is the sum of the geometric progression q = 1 2 s - 1. According to the initial data at s > 1, then 0< q < 1 . Получаем, ∑ k = 1 ∞ < 1 + 1 2 s - 1 + 1 2 s - 1 2 + 1 2 s - 1 3 + . . . = 1 1 - q = 1 1 - 1 2 s - 1 . Последовательность ряда при s > 1 increases and is limited from above 1 1 - 1 2 s - 1 . Let's imagine that there is a limit and the series is convergent ∑ k = 1 ∞ 1 k s .
Definition 8
Series ∑ k = 1 ∞ a k is positive in that case, if its members > 0 a k > 0 , k = 1 , 2 , . . . .
Series ∑ k = 1 ∞ b k signalternating, if the signs of the numbers are different. This example is presented as ∑ k = 1 ∞ b k = ∑ k = 1 ∞ (- 1) k · a k or ∑ k = 1 ∞ b k = ∑ k = 1 ∞ (- 1) k + 1 · a k , where a k > 0 , k = 1 , 2 , . . . .
Series ∑ k = 1 ∞ b k alternating, since it contains many numbers, negative and positive.
The second option series is a special case of the third option.
Here are examples for each case, respectively:
6 + 3 + 3 2 + 3 4 + 3 8 + 3 16 + . . . 6 - 3 + 3 2 - 3 4 + 3 8 - 3 16 + . . . 6 + 3 - 3 2 + 3 4 + 3 8 - 3 16 + . . .
For the third option, you can also determine absolute and conditional convergence.
Definition 9
The alternating series ∑ k = 1 ∞ b k is absolutely convergent in the case when ∑ k = 1 ∞ b k is also considered convergent.
Let's look at several typical options in detail.
Example 2
If the rows are 6 - 3 + 3 2 - 3 4 + 3 8 - 3 16 +. . . and 6 + 3 - 3 2 + 3 4 + 3 8 - 3 16 + . . . are defined as convergent, then it is correct to assume that 6 + 3 + 3 2 + 3 4 + 3 8 + 3 16 + . . .
Definition 10
An alternating series ∑ k = 1 ∞ b k is considered conditionally convergent if ∑ k = 1 ∞ b k is divergent, and the series ∑ k = 1 ∞ b k is considered convergent.
Example 3
Let us examine in detail the option ∑ k = 1 ∞ (- 1) k + 1 k = 1 - 1 2 + 1 3 - 1 4 + . . . . The series ∑ k = 1 ∞ (- 1) k + 1 k = ∑ k = 1 ∞ 1 k, which consists of absolute values, is defined as divergent. This option is considered convergent since it is easy to determine. From this example we learn that the series ∑ k = 1 ∞ (- 1) k + 1 k = 1 - 1 2 + 1 3 - 1 4 + . . . will be considered conditionally convergent.
Features of convergent series
Let's analyze the properties for certain cases
- If ∑ k = 1 ∞ a k converges, then the series ∑ k = m + 1 ∞ a k is also considered convergent. It can be noted that the row without m terms is also considered convergent. If we add several numbers to ∑ k = m + 1 ∞ a k, then the resulting result will also be convergent.
- If ∑ k = 1 ∞ a k converges and the sum = S, then the series ∑ k = 1 ∞ A · a k , ∑ k = 1 ∞ A · a k = A · S also converges, where A-constant.
- If ∑ k = 1 ∞ a k and ∑ k = 1 ∞ b k are convergent, the sums A And B too, then the series ∑ k = 1 ∞ a k + b k and ∑ k = 1 ∞ a k - b k also converge. The amounts will be equal A+B And A - B respectively.
Determine that the series converges ∑ k = 1 ∞ 2 3 k · k 3 .
Let's change the expression ∑ k = 1 ∞ 2 3 k · k 3 = ∑ k = 1 ∞ 2 3 · 1 k 4 3 . The series ∑ k = 1 ∞ 1 k 4 3 is considered convergent, since the series ∑ k = 1 ∞ 1 k s converges when s > 1. According to the second property, ∑ k = 1 ∞ 2 3 · 1 k 4 3 .
Example 5
Determine whether the series ∑ n = 1 ∞ 3 + n n 5 2 converges.
Let's transform the original version ∑ n = 1 ∞ 3 + n n 5 2 = ∑ n = 1 ∞ 3 n 5 2 + n n 2 = ∑ n = 1 ∞ 3 n 5 2 + ∑ n = 1 ∞ 1 n 2 .
We get the sum ∑ n = 1 ∞ 3 n 5 2 and ∑ n = 1 ∞ 1 n 2 . Each series is considered convergent according to the property. So, as the series converge, so does the original version.
Example 6
Calculate whether the series 1 - 6 + 1 2 - 2 + 1 4 - 2 3 + 1 8 - 2 9 + converges. . . and calculate the amount.
Let's expand the original version:
1 - 6 + 1 2 - 2 + 1 4 - 2 3 + 1 8 - 2 9 + . . . = = 1 + 1 2 + 1 4 + 1 8 + . . . - 2 · 3 + 1 + 1 3 + 1 9 + . . . = = ∑ k = 1 ∞ 1 2 k - 1 - 2 · ∑ k = 1 ∞ 1 3 k - 2
Each series converges because it is one of the members of a number sequence. According to the third property, we can calculate that the original version is also convergent. We calculate the sum: The first term of the series ∑ k = 1 ∞ 1 2 k - 1 = 1, and the denominator = 0. 5, this is followed by, ∑ k = 1 ∞ 1 2 k - 1 = 1 1 - 0 . 5 = 2. The first term is ∑ k = 1 ∞ 1 3 k - 2 = 3 , and the denominator of the descending number sequence = 1 3 . We get: ∑ k = 1 ∞ 1 3 k - 2 = 3 1 - 1 3 = 9 2 .
We use the expressions obtained above to determine the sum 1 - 6 + 1 2 - 2 + 1 4 - 2 3 + 1 8 - 2 9 +. . . = ∑ k = 1 ∞ 1 2 k - 1 - 2 ∑ k = 1 ∞ 1 3 k - 2 = 2 - 2 9 2 = - 7
A necessary condition for determining whether a series is convergent
Definition 11If the series ∑ k = 1 ∞ a k is convergent, then its limit kth term = 0: lim k → + ∞ a k = 0 .
If we check any option, we must not forget about the indispensable condition. If it is not fulfilled, then the series diverges. If lim k → + ∞ a k ≠ 0, then the series is divergent.
It should be clarified that the condition is important, but not sufficient. If the equality lim k → + ∞ a k = 0 holds, then this does not guarantee that ∑ k = 1 ∞ a k is convergent.
Let's give an example. For the harmonic series ∑ k = 1 ∞ 1 k the condition is satisfied lim k → + ∞ 1 k = 0 , but the series still diverges.
Example 7
Determine the convergence ∑ n = 1 ∞ n 2 1 + n .
Let's check the original expression for the fulfillment of the condition lim n → + ∞ n 2 1 + n = lim n → + ∞ n 2 n 2 1 n 2 + 1 n = lim n → + ∞ 1 1 n 2 + 1 n = 1 + 0 + 0 = + ∞ ≠ 0
Limit nth member is not equal to 0. We have proven that this series diverges.
How to determine the convergence of a positive series.
If you constantly use these characteristics, you will have to constantly calculate the limits. This section will help you avoid difficulties when solving examples and problems. In order to determine the convergence of a positive series, there is a certain condition.
For convergence of positive sign ∑ k = 1 ∞ a k , a k > 0 ∀ k = 1 , 2 , 3 , . . . it is necessary to determine a limited sequence of sums.
How to compare series
There are several signs of comparing series. We compare the series whose convergence is proposed to be determined with the series whose convergence is known.
First sign
∑ k = 1 ∞ a k and ∑ k = 1 ∞ b k are positive sign series. The inequality a k ≤ b k is valid for k = 1, 2, 3, ... It follows from this that from the series ∑ k = 1 ∞ b k we can obtain ∑ k = 1 ∞ a k . Since ∑ k = 1 ∞ a k is divergent, the series ∑ k = 1 ∞ b k can be defined as divergent.
This rule is constantly used to solve equations and is a serious argument that will help determine convergence. The difficulty may lie in the fact that it is not possible to find a suitable example for comparison in every case. Quite often, a series is selected according to the principle that the indicator kth term will be equal to the result of subtracting the exponents of the numerator and denominator kth member of the series. Let's assume that a k = k 2 + 3 4 k 2 + 5 , the difference will be equal to 2 – 3 = - 1 . In this case, we can determine that for comparison a series with k-th term b k = k - 1 = 1 k , which is harmonic.
In order to consolidate the obtained material, we will consider in detail a couple of typical options.
Example 8
Determine what the series ∑ k = 1 ∞ 1 k - 1 2 is.
Since the limit = 0 lim k → + ∞ 1 k - 1 2 = 0, we have fulfilled the necessary condition. The inequality will be fair 1 k< 1 k - 1 2 для k, which are natural. From the previous paragraphs we learned that the harmonic series ∑ k = 1 ∞ 1 k is divergent. According to the first criterion, it can be proven that the original version is divergent.
Example 9
Determine whether the series is convergent or divergent ∑ k = 1 ∞ 1 k 3 + 3 k - 1 .
In this example, the necessary condition is satisfied, since lim k → + ∞ 1 k 3 + 3 k - 1 = 0. We represent it as the inequality 1 k 3 + 3 k - 1< 1 k 3 для любого значения k. The series ∑ k = 1 ∞ 1 k 3 is convergent, since the harmonic series ∑ k = 1 ∞ 1 k s converges for s > 1. According to the first criterion, we can conclude that the number series is convergent.
Example 10
Determine what the series ∑ k = 3 ∞ 1 k ln (ln k) is. lim k → + ∞ 1 k ln (ln k) = 1 + ∞ + ∞ = 0 .
In this option, you can mark the fulfillment of the desired condition. Let's define a series for comparison. For example, ∑ k = 1 ∞ 1 k s . To determine what the degree is, consider the sequence (ln (ln k)), k = 3, 4, 5. . . . Members of the sequence ln (ln 3) , ln (ln 4) , ln (ln 5) , . . . increases to infinity. Having analyzed the equation, we can note that, taking N = 1619 as the value, then the terms of the sequence > 2. For this sequence the inequality 1 k ln (ln k) will be true< 1 k 2 . Ряд ∑ k = N ∞ 1 k 2 сходится согласно первому признаку, так как ряд ∑ k = 1 ∞ 1 k 2 тоже сходящийся. Отметим, что согласно первому признаку ряд ∑ k = N ∞ 1 k ln (ln k) сходящийся. Можно сделать вывод, что ряд ∑ k = 3 ∞ 1 k ln (ln k) также сходящийся.
Second sign
Let us assume that ∑ k = 1 ∞ a k and ∑ k = 1 ∞ b k are positive number series.
If lim k → + ∞ a k b k ≠ ∞ , then the series ∑ k = 1 ∞ b k converges, and ∑ k = 1 ∞ a k also converges.
If lim k → + ∞ a k b k ≠ 0, then since the series ∑ k = 1 ∞ b k diverges, then ∑ k = 1 ∞ a k also diverges.
If lim k → + ∞ a k b k ≠ ∞ and lim k → + ∞ a k b k ≠ 0, then the convergence or divergence of a series means the convergence or divergence of another.
Consider ∑ k = 1 ∞ 1 k 3 + 3 k - 1 using the second sign. For comparison ∑ k = 1 ∞ b k we take the convergent series ∑ k = 1 ∞ 1 k 3 . Let's define the limit: lim k → + ∞ a k b k = lim k → + ∞ 1 k 3 + 3 k - 1 1 k 3 = lim k → + ∞ k 3 k 3 + 3 k - 1 = 1
According to the second criterion, it can be determined that the convergent series ∑ k = 1 ∞ 1 k 3 means that the original version also converges.
Example 11
Determine what the series ∑ n = 1 ∞ k 2 + 3 4 k 3 + 5 is.
Let us analyze the necessary condition lim k → ∞ k 2 + 3 4 k 3 + 5 = 0, which is satisfied in this version. According to the second criterion, take the series ∑ k = 1 ∞ 1 k . We are looking for the limit: lim k → + ∞ k 2 + 3 4 k 3 + 5 1 k = lim k → + ∞ k 3 + 3 k 4 k 3 + 5 = 1 4
According to the above theses, a divergent series entails the divergence of the original series.
Third sign
Let's consider the third sign of comparison.
Let us assume that ∑ k = 1 ∞ a k and _ ∑ k = 1 ∞ b k are positive number series. If the condition is satisfied for a certain number a k + 1 a k ≤ b k + 1 b k , then the convergence of this series ∑ k = 1 ∞ b k means that the series ∑ k = 1 ∞ a k is also convergent. The divergent series ∑ k = 1 ∞ a k entails the divergence ∑ k = 1 ∞ b k .
D'Alembert's sign
Let's imagine that ∑ k = 1 ∞ a k is a positive number series. If lim k → + ∞ a k + 1 a k< 1 , то ряд является сходящимся, если lim k → + ∞ a k + 1 a k >1, then divergent.
Note 1
D'Alembert's test is valid if the limit is infinite.
If lim k → + ∞ a k + 1 a k = - ∞ , then the series is convergent, if lim k → ∞ a k + 1 a k = + ∞ , then it is divergent.
If lim k → + ∞ a k + 1 a k = 1, then d’Alembert’s sign will not help and some more research will be required.
Example 12
Determine whether the series is convergent or divergent ∑ k = 1 ∞ 2 k + 1 2 k using d’Alembert’s criterion.
It is necessary to check whether the necessary convergence condition is satisfied. Let's calculate the limit using L'Hopital's rule: lim k → + ∞ 2 k + 1 2 k = ∞ ∞ = lim k → + ∞ 2 k + 1 " 2 k " = lim k → + ∞ 2 2 k ln 2 = 2 + ∞ ln 2 = 0
We can see that the condition is met. Let's use d'Alembert's test: lim k → + ∞ = lim k → + ∞ 2 (k + 1) + 1 2 k + 1 2 k + 1 2 k = 1 2 lim k → + ∞ 2 k + 3 2 k + 1 = 12< 1
The series is convergent.
Example 13
Determine whether the series is divergent ∑ k = 1 ∞ k k k ! .
Let's use d'Alembert's test to determine the divergence of the series: lim k → + ∞ a k + 1 a k = lim k → + ∞ (k + 1) k + 1 (k + 1) ! k k k! = lim k → + ∞ (k + 1) k + 1 · k ! k k · (k + 1) ! = lim k → + ∞ (k + 1) k + 1 k k · (k + 1) = = lim k → + ∞ (k + 1) k k k = lim k → + ∞ k + 1 k k = lim k → + ∞ 1 + 1 k k = e > 1
Therefore, the series is divergent.
Radical Cauchy's sign
Let us assume that ∑ k = 1 ∞ a k is a series with a positive sign. If lim k → + ∞ a k k< 1 , то ряд является сходящимся, если lim k → + ∞ a k k >1, then divergent.
Note 2
If lim k → + ∞ a k k = 1, then this sign does not provide any information - additional analysis is required.
This feature can be used in examples that are easy to identify. The case will be typical when a member of a number series is an exponential power expression.
In order to consolidate the information received, let's consider several typical examples.
Example 14
Determine whether the positive sign series ∑ k = 1 ∞ 1 (2 k + 1) k is convergent.
The necessary condition is considered satisfied, since lim k → + ∞ 1 (2 k + 1) k = 1 + ∞ + ∞ = 0 .
According to the criterion discussed above, we obtain lim k → + ∞ a k k = lim k → + ∞ 1 (2 k + 1) k k = lim k → + ∞ 1 2 k + 1 = 0< 1 . Данный ряд является сходимым.
Example 15
Does the number series ∑ k = 1 ∞ 1 3 k · 1 + 1 k k 2 converge?
We use the feature described in the previous paragraph lim k → + ∞ 1 3 k 1 + 1 k k 2 k = 1 3 lim k → + ∞ 1 + 1 k k = e 3< 1 , следовательно, числовой ряд сходится.
Integral Cauchy test
Let us assume that ∑ k = 1 ∞ a k is a series with positive sign. It is necessary to denote the function of a continuous argument y = f(x), which coincides with a n = f (n) . If y = f(x) greater than zero, is not interrupted and decreases by [ a ; + ∞) , where a ≥ 1
Then, if the improper integral ∫ a + ∞ f (x) d x is convergent, then the series under consideration also converges. If it diverges, then in the example under consideration the series also diverges.
When checking whether a function is decreasing, you can use the material covered in previous lessons.
Example 16
Consider the example ∑ k = 2 ∞ 1 k · ln k for convergence.
The condition for the convergence of the series is considered to be satisfied, since lim k → + ∞ 1 k · ln k = 1 + ∞ = 0 . Consider y = 1 x ln x. It is greater than zero, is not interrupted and decreases by [ 2 ; + ∞) . The first two points are known for certain, but the third should be discussed in more detail. Find the derivative: y " = 1 x · ln x " = x · ln x " x · ln x 2 = ln x + x · 1 x x · ln x 2 = - ln x + 1 x · ln x 2 . It is less than zero on [ 2 ; + ∞).This proves the thesis that the function is decreasing.
Actually, the function y = 1 x ln x corresponds to the characteristics of the principle that we considered above. Let's use it: ∫ 2 + ∞ d x x · ln x = lim A → + ∞ ∫ 2 A d (ln x) ln x = lim A → + ∞ ln (ln x) 2 A = = lim A → + ∞ (ln ( ln A) - ln (ln 2)) = ln (ln (+ ∞)) - ln (ln 2) = + ∞
According to the results obtained, the original example diverges, since the improper integral is divergent.
Example 17
Prove the convergence of the series ∑ k = 1 ∞ 1 (10 k - 9) (ln (5 k + 8)) 3 .
Since lim k → + ∞ 1 (10 k - 9) (ln (5 k + 8)) 3 = 1 + ∞ = 0, then the condition is considered satisfied.
Starting with k = 4, the correct expression is 1 (10 k - 9) (ln (5 k + 8)) 3< 1 (5 k + 8) (ln (5 k + 8)) 3 .
If the series ∑ k = 4 ∞ 1 (5 k + 8) (ln (5 k + 8)) 3 is considered convergent, then, according to one of the principles of comparison, the series ∑ k = 4 ∞ 1 (10 k - 9) ( ln (5 k + 8)) 3 will also be considered convergent. This way we can determine that the original expression is also convergent.
Let's move on to the proof: ∑ k = 4 ∞ 1 (5 k + 8) (ln (5 k + 8)) 3 .
Since the function y = 1 5 x + 8 (ln (5 x + 8)) 3 is greater than zero, it is not interrupted and decreases by [ 4 ; + ∞) . We use the feature described in the previous paragraph:
∫ 4 + ∞ d x (5 x + 8) (l n (5 x + 8)) 3 = lim A → + ∞ ∫ 4 A d x (5 x + 8) (ln (5 x + 8)) 3 = = 1 5 lim A → + ∞ ∫ 4 A d (ln (5 x + 8) (ln (5 x + 8)) 3 = - 1 10 lim A → + ∞ 1 (ln (5 x + 8)) 2 | 4 A = = - 1 10 lim A → + ∞ 1 (ln (5 A + 8)) 2 - 1 (ln (5 4 + 8)) 2 = = - 1 10 1 + ∞ - 1 (ln 28) 2 = 1 10 · ln 28 2
In the resulting convergent series, ∫ 4 + ∞ d x (5 x + 8) (ln (5 x + 8)) 3, we can determine that ∑ k = 4 ∞ 1 (5 k + 8) (ln (5 k + 8 )) 3 also converges.
Raabe's sign
Let us assume that ∑ k = 1 ∞ a k is a positive number series.
If lim k → + ∞ k · a k a k + 1< 1 , то ряд расходится, если lim k → + ∞ k · a k a k + 1 - 1 >1, then it converges.
This determination method can be used if the techniques described above do not give visible results.
Absolute Convergence Study
For the study we take ∑ k = 1 ∞ b k . We use positive sign ∑ k = 1 ∞ b k . We can use any of the suitable features that we described above. If the series ∑ k = 1 ∞ b k converges, then the original series is absolutely convergent.
Example 18
Investigate the series ∑ k = 1 ∞ (- 1) k 3 k 3 + 2 k - 1 for convergence ∑ k = 1 ∞ (- 1) k 3 k 3 + 2 k - 1 = ∑ k = 1 ∞ 1 3 k 3 + 2 k - 1 .
The condition is satisfied lim k → + ∞ 1 3 k 3 + 2 k - 1 = 1 + ∞ = 0 . Let's use ∑ k = 1 ∞ 1 k 3 2 and use the second sign: lim k → + ∞ 1 3 k 3 + 2 k - 1 1 k 3 2 = 1 3 .
The series ∑ k = 1 ∞ (- 1) k 3 k 3 + 2 k - 1 converges. The original series is also absolutely convergent.
Divergence of alternating series
If the series ∑ k = 1 ∞ b k is divergent, then the corresponding alternating series ∑ k = 1 ∞ b k is either divergent or conditionally convergent.
Only d'Alembert's test and Cauchy's radical test will help to draw conclusions about ∑ k = 1 ∞ b k from the divergence from the moduli ∑ k = 1 ∞ b k . The series ∑ k = 1 ∞ b k also diverges if the necessary convergence condition is not satisfied, that is, if lim k → ∞ + b k ≠ 0.
Example 19
Check divergence 1 7, 2 7 2, - 6 7 3, 24 7 4, 120 7 5 - 720 7 6, . . . .
Module kth term is represented as b k = k ! 7 k.
Let us examine the series ∑ k = 1 ∞ b k = ∑ k = 1 ∞ k ! 7 k for convergence using d'Alembert's criterion: lim k → + ∞ b k + 1 b k = lim k → + ∞ (k + 1) ! 7 k + 1 k ! 7 k = 1 7 · lim k → + ∞ (k + 1) = + ∞ .
∑ k = 1 ∞ b k = ∑ k = 1 ∞ k ! 7 k diverges in the same way as the original version.
Example 20
Is ∑ k = 1 ∞ (- 1) k · k 2 + 1 ln (k + 1) convergent.
Let's consider the necessary condition lim k → + ∞ b k = lim k → + ∞ k 2 + 1 ln (k + 1) = ∞ ∞ = lim k → + ∞ = k 2 + 1 " (ln (k + 1)) " = = lim k → + ∞ 2 k 1 k + 1 = lim k → + ∞ 2 k (k + 1) = + ∞ . The condition is not met, therefore ∑ k = 1 ∞ (- 1) k · k 2 + 1 ln (k + 1) the series is divergent. The limit was calculated using L'Hopital's rule.
Conditional convergence criteria
Leibniz's test
Definition 12If the values of the terms of the alternating series decrease b 1 > b 2 > b 3 > . . . > . . . and the modulus limit = 0 as k → + ∞, then the series ∑ k = 1 ∞ b k converges.
Example 17
Consider ∑ k = 1 ∞ (- 1) k 2 k + 1 5 k (k + 1) for convergence.
The series is represented as ∑ k = 1 ∞ (- 1) k 2 k + 1 5 k (k + 1) = ∑ k = 1 ∞ 2 k + 1 5 k (k + 1) . The necessary condition is satisfied: lim k → + ∞ = 2 k + 1 5 k (k + 1) = 0 . Consider ∑ k = 1 ∞ 1 k by the second comparison criterion lim k → + ∞ 2 k + 1 5 k (k + 1) 1 k = lim k → + ∞ 2 k + 1 5 (k + 1) = 2 5
We find that ∑ k = 1 ∞ (- 1) k 2 k + 1 5 k (k + 1) = ∑ k = 1 ∞ 2 k + 1 5 k (k + 1) diverges. The series ∑ k = 1 ∞ (- 1) k 2 k + 1 5 k (k + 1) converges according to the Leibniz criterion: sequence 2 1 + 1 5 1 1 1 + 1 = 3 10, 2 2 + 1 5 · 2 · (2 + 1) = 5 30 , 2 · 3 + 1 5 · 3 · 3 + 1 , . . . decreases and lim k → + ∞ = 2 k + 1 5 k (k + 1) = 0 .
The series converges conditionally.
Abel-Dirichlet test
Definition 13∑ k = 1 + ∞ u k · v k converges if ( u k ) does not increase and the sequence ∑ k = 1 + ∞ v k is bounded.
Example 17
Explore 1 - 3 2 + 2 3 + 1 4 - 3 5 + 1 3 + 1 7 - 3 8 + 2 9 + . . . for convergence.
Let's imagine
1 - 3 2 + 2 3 + 1 4 - 3 5 + 1 3 + 1 7 - 3 8 + 2 9 + . . . = 1 1 + 1 2 (- 3) + 1 3 2 + 1 4 1 + 1 5 (- 3) + 1 6 = ∑ k = 1 ∞ u k v k
where (u k) = 1, 1 2, 1 3, . . . is non-increasing, and the sequence (v k) = 1, - 3, 2, 1, - 3, 2, . . . limited (S k) = 1, - 2, 0, 1, - 2, 0, . . . . The series converges.
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1. If a 1 + a 2 + a 3 +…+a n +…= converges, then the series a m+1 +a m+2 +a m+3 +…, obtained from this series by discarding the first m terms, also converges. This resulting series is called the mth remainder of the series. And, vice versa: from the convergence of the mth remainder of the series, the convergence of this series follows. Those. The convergence and divergence of a series is not violated if a finite number of its terms are added or discarded.
2 . If the series a 1 + a 2 + a 3 +... converges and its sum is equal to S, then the series Ca 1 + Ca 2 +..., where C = also converges and its sum is equal to CS.
3. If the series a 1 +a 2 +... and b 1 +b 2 +... converge and their sums are equal to S1 and S2, respectively, then the series (a 1 +b 1)+(a 2 +b 2)+(a 3 +b 3)+… and (a 1 -b 1)+(a 2 -b 2)+(a 3 -b 3)+… also converge. Their sums are respectively equal to S1+S2 and S1-S2.
4. A). If a series converges, then its nth term tends to 0 as n increases indefinitely (the converse is not true).
-
necessary
sign (condition)convergence
row.
b). If 
then the series is divergent - sufficient
conditiondivergences
row.
-series of this type are studied only according to property 4. This divergent rows.
Sign-positive series.
Signs of convergence and divergence of positive-sign series.
Positive series are series in which all terms are positive. We will consider these signs of convergence and divergence for series with positive signs.
1. The first sign of comparison.
Let two positive-sign series a 1 + a 2 + a 3 +…+a n +…= be given 
(1) иb 1 +b 2 +b 3 +…+b n +…= 
(2).
If the members of the series (1) not more
b n and series (2) converges, then series (1) also converges.
If the members of the series (1) not less corresponding members of series (2), i.e. and n 
b n and row (2) diverges, then series (1) also diverges.
This comparison criterion is valid if the inequality is not satisfied for all n, but only starting from some.
2. Second sign of comparison.
If there is a finite and non-zero limit 
, then both series converge or diverge simultaneously.
- rows of this type diverge according to the second criterion of comparison. They must be compared with the harmonic series.
3. D'Alembert's sign.
If for a positive series (a 1 + a 2 + a 3 +…+a n +…= 
) exists 
(1), then the series converges if q<1,
расходится, если
q>
4. Cauchy's sign is radical.
If there is a limit for a positive series 
(2), then the series converges ifq<1,
расходится, если
q>1. If q=1 then the question remains open.
5. Cauchy's test is integral.
Let us recall improper integrals.
If there is a limit 
. This is an improper integral and is denoted 
.
If this limit is finite, then the improper integral is said to converge. The series, respectively, converges or diverges.
Let the series a 1 + a 2 + a 3 +…+a n +…= 
- positive series.
Let us denote a n =f(x) and consider the function f(x). If f(x) is a positive, monotonically decreasing and continuous function, then if the improper integral converges, then the given series converges. And vice versa: if the improper integral diverges, then the series diverges.
If the series is finite, then it converges.
Rows are very common 
-Derichlet series. It converges if p>1, diverges p<1.
Гармонический ряд является рядом Дерихле
при р=1.
Сходимость
и расходимость данного ряда легко
доказать с помощью интегрального
признака Коши.
1. Number series: basic concepts, necessary conditions for the convergence of the series. The rest of the row.
2. Series with positive terms and tests of their convergence: tests of comparison, D'Alembert, Cauchy.
3. Alternating series, Leibniz’s test.
1. Definition of a number series. Convergence
In mathematical applications, as well as in solving some problems in economics, statistics and other fields, sums with an infinite number of terms are considered. Here we will give a definition of what is meant by such amounts.
Let an infinite number sequence be given
Definition 1.1. Number series or simply near is called an expression (sum) of the form
.
(1.1)
Numbers 
are called members of a number,
–general or n–m member of the series.
To define series (1.1), it is enough to specify the function of the natural argument of calculating the th term of the series by its number 
Example 1.1. Let . Row
(1.2)
called harmonic series.
Example 1.2. Let ,Row
(1.3)
called generalized harmonic series. In a particular case, a harmonic series is obtained.
Example 1.3. Let =. Row
called near geometric progression.
From the terms of series (1.1) we form a numerical sequence of partials
amounts
Where 
– the sum of the first terms of the series, which is called n-th partial amount, i.e.
…………………………….
…………………………….
Number sequence 
with an unlimited increase in number, it can:
1) have a finite limit;
2) have no finite limit (the limit does not exist or is equal to infinity).
Definition 1.2. Series (1.1) is called convergent, if the sequence of its partial sums (1.5) has a finite limit, i.e.
In this case the number is called amount series (1.1) and is written
Definition 1.3. Series (1.1) is called divergent, if the sequence of its partial sums does not have a finite limit.
No sum is assigned to the divergent series.
Thus, the problem of finding the sum of a convergent series (1.1) is equivalent to calculating the limit of the sequence of its partial sums.
Let's look at a few examples.
Example 1.4. Prove that the series
converges and find its sum.
Let's find the nth partial sum of this series.
General member 
represent the series in the form 
.
From here we have: 
. Therefore, this series converges and its sum is equal to 1:
Example 1.5. Examine the series for convergence
For this row 
. Therefore, this series diverges.
Comment. For series (1.6) is the sum of an infinite number of zeros and is obviously convergent.
2. Basic properties of number series
The properties of a sum of a finite number of terms differ from the properties of a series, i.e., the sum of an infinite number of terms. So, in the case of a finite number of terms, they can be grouped in any order, this will not change the sum. There are convergent series (conditionally convergent, which will be considered in Section 5), for which, as Riemann showed * , by appropriately changing the order of their terms, you can make the sum of the series equal to any number, and even a divergent series.
Example 2.1. Consider a divergent series of the form (1.7)
By grouping its members in pairs, we obtain a convergent number series with a sum equal to zero:
On the other hand, by grouping its terms in pairs, starting with the second term, we also obtain a convergent series, but with a sum equal to one:
Convergent series have certain properties that make it possible to treat them as if they were finite sums. So they can be multiplied by numbers, added and subtracted term by term. They can combine any adjacent terms into groups.
Theorem 2.1.(A necessary sign of convergence of a series).
If series (1.1) converges, then its common term tends to zero as n increases indefinitely, i.e.
The proof of the theorem follows from the fact that 
, and if
S is the sum of series (1.1), then
Condition (2.1) is a necessary but not sufficient condition for the convergence of the series. That is, if the common term of the series tends to zero at , this does not mean that the series converges. For example, for the harmonic series (1.2) 
however, as will be shown below, it diverges.
Answer: the series diverges.
Example No. 3
Find the sum of the series $\sum\limits_(n=1)^(\infty)\frac(2)((2n+1)(2n+3))$.
Since the lower limit of summation is 1, the common term of the series is written under the sum sign: $u_n=\frac(2)((2n+1)(2n+3))$. Let's make the nth partial sum of the series, i.e. Let's sum the first $n$ terms of a given number series:
$$ S_n=u_1+u_2+u_3+u_4+\ldots+u_n=\frac(2)(3\cdot 5)+\frac(2)(5\cdot 7)+\frac(2)(7\cdot 9 )+\frac(2)(9\cdot 11)+\ldots+\frac(2)((2n+1)(2n+3)). $$
Why I write exactly $\frac(2)(3\cdot 5)$, and not $\frac(2)(15)$, will be clear from the further narration. However, writing down a partial amount did not bring us one iota closer to our goal. We need to find $\lim_(n\to\infty)S_n$, but if we just write:
$$ \lim_(n\to\infty)S_n=\lim_(n\to\infty)\left(\frac(2)(3\cdot 5)+\frac(2)(5\cdot 7)+\ frac(2)(7\cdot 9)+\frac(2)(9\cdot 11)+\ldots+\frac(2)((2n+1)(2n+3))\right), $$
then this record, completely correct in form, will give us nothing in essence. To find the limit, the expression for the partial sum must first be simplified.
There is a standard transformation for this, which consists in decomposing the fraction $\frac(2)((2n+1)(2n+3))$, which represents the general term of the series, into elementary fractions. A separate topic is devoted to the issue of decomposing rational fractions into elementary ones (see, for example, example No. 3 on this page). Expanding the fraction $\frac(2)((2n+1)(2n+3))$ into elementary fractions, we will have:
$$ \frac(2)((2n+1)(2n+3))=\frac(A)(2n+1)+\frac(B)(2n+3)=\frac(A\cdot(2n +3)+B\cdot(2n+1))((2n+1)(2n+3)). $$
We equate the numerators of the fractions on the left and right sides of the resulting equality:
$$ 2=A\cdot(2n+3)+B\cdot(2n+1). $$
There are two ways to find the values of $A$ and $B$. You can open the brackets and rearrange the terms, or you can simply substitute some suitable values instead of $n$. Just for variety, in this example we will go the first way, and in the next one we will substitute private values $n$. Opening the brackets and rearranging the terms, we get:
$$ 2=2An+3A+2Bn+B;\\ 2=(2A+2B)n+3A+B. $$
On the left side of the equality, $n$ is preceded by a zero. If you like, for clarity, the left side of the equality can be represented as $0\cdot n+ 2$. Since on the left side of the equality $n$ is preceded by zero, and on the right side of the equality $n$ is preceded by $2A+2B$, we have the first equation: $2A+2B=0$. Let's immediately divide both sides of this equation by 2, after which we get $A+B=0$.
Since on the left side of the equality the free term is equal to 2, and on the right side of the equality the free term is equal to $3A+B$, then $3A+B=2$. So, we have a system:
$$ \left\(\begin(aligned) & A+B=0;\\ & 3A+B=2. \end(aligned)\right. $$
We will carry out the proof using the method of mathematical induction. At the first step, you need to check whether the equality being proved is true $S_n=\frac(1)(3)-\frac(1)(2n+3)$ for $n=1$. We know that $S_1=u_1=\frac(2)(15)$, but will the expression $\frac(1)(3)-\frac(1)(2n+3)$ give the value $\frac(2 )(15)$, if we substitute $n=1$ into it? Let's check:
$$ \frac(1)(3)-\frac(1)(2n+3)=\frac(1)(3)-\frac(1)(2\cdot 1+3)=\frac(1) (3)-\frac(1)(5)=\frac(5-3)(15)=\frac(2)(15). $$
So, for $n=1$ the equality $S_n=\frac(1)(3)-\frac(1)(2n+3)$ is satisfied. This completes the first step of the method of mathematical induction.
Let us assume that for $n=k$ the equality is satisfied, i.e. $S_k=\frac(1)(3)-\frac(1)(2k+3)$. Let us prove that the same equality will be satisfied for $n=k+1$. To do this, consider $S_(k+1)$:
$$ S_(k+1)=S_k+u_(k+1). $$
Since $u_n=\frac(1)(2n+1)-\frac(1)(2n+3)$, then $u_(k+1)=\frac(1)(2(k+1)+ 1)-\frac(1)(2(k+1)+3)=\frac(1)(2k+3)-\frac(1)(2(k+1)+3)$. According to the assumption made above $S_k=\frac(1)(3)-\frac(1)(2k+3)$, therefore the formula $S_(k+1)=S_k+u_(k+1)$ will take the form:
$$ S_(k+1)=S_k+u_(k+1)=\frac(1)(3)-\frac(1)(2k+3)+\frac(1)(2k+3)-\ frac(1)(2(k+1)+3)=\frac(1)(3)-\frac(1)(2(k+1)+3). $$
Conclusion: the formula $S_n=\frac(1)(3)-\frac(1)(2n+3)$ is correct for $n=k+1$. Therefore, according to the method of mathematical induction, the formula $S_n=\frac(1)(3)-\frac(1)(2n+3)$ is true for any $n\in N$. Equality has been proven.
In a standard course of higher mathematics, they are usually content with “crossing out” canceling terms, without requiring any proof. So, we got the expression for the nth partial sum: $S_n=\frac(1)(3)-\frac(1)(2n+3)$. Let's find the value of $\lim_(n\to\infty)S_n$:
Conclusion: the given series converges and its sum is $S=\frac(1)(3)$.
The second way to simplify the formula for a partial sum.
Honestly, I prefer this method myself :) Let's write down the partial amount in an abbreviated version:
$$ S_n=\sum\limits_(k=1)^(n)u_k=\sum\limits_(k=1)^(n)\frac(2)((2k+1)(2k+3)). $$
We obtained earlier that $u_k=\frac(1)(2k+1)-\frac(1)(2k+3)$, therefore:
$$ S_n=\sum\limits_(k=1)^(n)\frac(2)((2k+1)(2k+3))=\sum\limits_(k=1)^(n)\left (\frac(1)(2k+1)-\frac(1)(2k+3)\right). $$
The sum $S_n$ contains a finite number of terms, so we can rearrange them as we please. I want to first add all terms of the form $\frac(1)(2k+1)$, and only then move on to terms of the form $\frac(1)(2k+3)$. This means that we will present the partial amount as follows:
$$ S_n =\frac(1)(3)-\frac(1)(5)+\frac(1)(5)-\frac(1)(7)+\frac(1)(7)-\ frac(1)(9)+\frac(1)(9)-\frac(1)(11)+\ldots+\frac(1)(2n+1)-\frac(1)(2n+3)= \\ =\frac(1)(3)+\frac(1)(5)+\frac(1)(7)+\frac(1)(9)+\ldots+\frac(1)(2n+1 )-\left(\frac(1)(5)+\frac(1)(7)+\frac(1)(9)+\ldots+\frac(1)(2n+3)\right). $$
Of course, the expanded notation is extremely inconvenient, so the above equality can be written more compactly:
$$ S_n=\sum\limits_(k=1)^(n)\left(\frac(1)(2k+1)-\frac(1)(2k+3)\right)=\sum\limits_( k=1)^(n)\frac(1)(2k+1)-\sum\limits_(k=1)^(n)\frac(1)(2k+3). $$
Now let's transform the expressions $\frac(1)(2k+1)$ and $\frac(1)(2k+3)$ into one form. I think it is convenient to reduce it to the form of a larger fraction (although it is possible to use a smaller one, this is a matter of taste). Since $\frac(1)(2k+1)>\frac(1)(2k+3)$ (the larger the denominator, the smaller the fraction), we will give the fraction $\frac(1)(2k+3) $ to the form $\frac(1)(2k+1)$.
I will present the expression in the denominator of the fraction $\frac(1)(2k+3)$ as follows:
$$ \frac(1)(2k+3)=\frac(1)(2k+2+1)=\frac(1)(2(k+1)+1). $$
And the sum $\sum\limits_(k=1)^(n)\frac(1)(2k+3)$ can now be written as follows:
$$ \sum\limits_(k=1)^(n)\frac(1)(2k+3)=\sum\limits_(k=1)^(n)\frac(1)(2(k+1) )+1)=\sum\limits_(k=2)^(n+1)\frac(1)(2k+1). $$
If the equality $\sum\limits_(k=1)^(n)\frac(1)(2k+3)=\sum\limits_(k=2)^(n+1)\frac(1)(2k+ 1) $ does not raise any questions, then let's move on. If you have any questions, please expand the note.
How did we get the converted amount? show\hide
We had a series $\sum\limits_(k=1)^(n)\frac(1)(2k+3)=\sum\limits_(k=1)^(n)\frac(1)(2( k+1)+1)$. Let's introduce a new variable instead of $k+1$ - for example, $t$. So $t=k+1$.
How did the old variable $k$ change? And it changed from 1 to $n$. Let's find out how the new variable $t$ will change. If $k=1$, then $t=1+1=2$. If $k=n$, then $t=n+1$. So, the expression $\sum\limits_(k=1)^(n)\frac(1)(2(k+1)+1)$ now becomes: $\sum\limits_(t=2)^(n +1)\frac(1)(2t+1)$.
$$ \sum\limits_(k=1)^(n)\frac(1)(2(k+1)+1)=\sum\limits_(t=2)^(n+1)\frac(1 )(2t+1). $$
We have the sum $\sum\limits_(t=2)^(n+1)\frac(1)(2t+1)$. Question: does it matter which letter is used in this amount? :) Simply writing the letter $k$ instead of $t$, we get the following:
$$ \sum\limits_(t=2)^(n+1)\frac(1)(2t+1)=\sum\limits_(k=2)^(n+1)\frac(1)(2k +1). $$
This is how we get the equality $\sum\limits_(k=1)^(n)\frac(1)(2(k+1)+1)=\sum\limits_(k=2)^(n+1) \frac(1)(2k+1)$.
Thus, the partial sum can be represented as follows:
$$ S_n=\sum\limits_(k=1)^(n)\frac(1)(2k+1)-\sum\limits_(k=1)^(n)\frac(1)(2k+3 )=\sum\limits_(k=1)^(n)\frac(1)(2k+1)-\sum\limits_(k=2)^(n+1)\frac(1)(2k+1 ). $$
Note that the sums $\sum\limits_(k=1)^(n)\frac(1)(2k+1)$ and $\sum\limits_(k=2)^(n+1)\frac(1 )(2k+1)$ differ only in the summation limits. Let's make these limits the same. “Taking away” the first element from the sum $\sum\limits_(k=1)^(n)\frac(1)(2k+1)$ we will have:
$$ \sum\limits_(k=1)^(n)\frac(1)(2k+1)=\frac(1)(2\cdot 1+1)+\sum\limits_(k=2)^ (n)\frac(1)(2k+1)=\frac(1)(3)+\sum\limits_(k=2)^(n)\frac(1)(2k+1). $$
“Taking away” the last element from the sum $\sum\limits_(k=2)^(n+1)\frac(1)(2k+1)$, we get:
$$\sum\limits_(k=2)^(n+1)\frac(1)(2k+1)=\sum\limits_(k=2)^(n)\frac(1)(2k+1 )+\frac(1)(2(n+1)+1)=\sum\limits_(k=2)^(n)\frac(1)(2k+1)+\frac(1)(2n+ 3).$$
Then the expression for the partial sum will take the form:
$$ S_n=\sum\limits_(k=1)^(n)\frac(1)(2k+1)-\sum\limits_(k=2)^(n+1)\frac(1)(2k +1)=\frac(1)(3)+\sum\limits_(k=2)^(n)\frac(1)(2k+1)-\left(\sum\limits_(k=2)^ (n)\frac(1)(2k+1)+\frac(1)(2n+3)\right)=\\ =\frac(1)(3)+\sum\limits_(k=2)^ (n)\frac(1)(2k+1)-\sum\limits_(k=2)^(n)\frac(1)(2k+1)-\frac(1)(2n+3)=\ frac(1)(3)-\frac(1)(2n+3). $$
If you skip all the explanations, then the process of finding a shortened formula for the nth partial sum will take the following form:
$$ S_n=\sum\limits_(k=1)^(n)u_k =\sum\limits_(k=1)^(n)\frac(2)((2k+1)(2k+3)) = \sum\limits_(k=1)^(n)\left(\frac(1)(2k+1)-\frac(1)(2k+3)\right)=\\ =\sum\limits_(k =1)^(n)\frac(1)(2k+1)-\sum\limits_(k=1)^(n)\frac(1)(2k+3) =\frac(1)(3) +\sum\limits_(k=2)^(n)\frac(1)(2k+1)-\left(\sum\limits_(k=2)^(n)\frac(1)(2k+1 )+\frac(1)(2n+3)\right)=\frac(1)(3)-\frac(1)(2n+3). $$
Let me remind you that we reduced the fraction $\frac(1)(2k+3)$ to the form $\frac(1)(2k+1)$. Of course, you can do the opposite, i.e. represent the fraction $\frac(1)(2k+1)$ as $\frac(1)(2k+3)$. The final expression for the partial sum will not change. In this case, I will hide the process of finding the partial amount under a note.
How to find $S_n$ if converted to another fraction? show\hide
$$ S_n =\sum\limits_(k=1)^(n)\frac(1)(2k+1)-\sum\limits_(k=1)^(n)\frac(1)(2k+3 ) =\sum\limits_(k=0)^(n-1)\frac(1)(2k+3)-\sum\limits_(k=1)^(n)\frac(1)(2k+3 )=\\ =\frac(1)(3)+\sum\limits_(k=1)^(n-1)\frac(1)(2k+3)-\left(\sum\limits_(k= 1)^(n-1)\frac(1)(2k+3)+\frac(1)(2n+3)\right) =\frac(1)(3)-\frac(1)(2n+ 3). $$
So, $S_n=\frac(1)(3)-\frac(1)(2n+3)$. Find the limit $\lim_(n\to\infty)S_n$:
$$ \lim_(n\to\infty)S_n=\lim_(n\to\infty)\left(\frac(1)(3)-\frac(1)(2n+3)\right)=\frac (1)(3)-0=\frac(1)(3). $$
The given series converges and its sum $S=\frac(1)(3)$.
Answer: $S=\frac(1)(3)$.
The continuation of the topic of finding the sum of a series will be discussed in the second and third parts.