Series, in mathematics
1. Definitions. R. is a sequence of elements composed according to some law. If R. is given, then this means that a law is indicated, with the help of which it is possible to compose as many elements of R. According to the property of elements, R. of numbers, R. of functions, and R. of actions are distinguished. Let's give some examples. 1, 2, 3, 4,..., n,... there are R. natural numbers; 1, 4, 9, 16,..., P 2 ... R. squares; a 0 , a 1 x, a 2 a 2 ,..., a n x n ,... R. power functions or power R. 1, x, x 2 /(1.2), x 3 /(1.2.3),... x n /(1.2...n),... 0, x, x 2 /2, x 3 /3, x 4 /4... (-1) n-1 x n /n.. In order to calculate the numerical value of some expression, it is necessary to perform R. actions. Eg. √[(35 - 3)/2] = √ = √16 = 4. With the help of R. actions, the greatest divisor of two given numbers is found. R. u 0 , u 1 , u 2 ,... u n... name endless, if after any element u k there is an element u k+1 ; otherwise R. naming. final. Eg. 1. 2, 3,... 9, 10 is a final R. because there are no elements after element 10. 2. A number defined next to. Of particular importance are the infinite R. of the form (1)... a 1 /10, a 2 /10 2 ,
...
a n/10n,..., where a 1 , a 2 , a 3 ,
... a n,... positive integers, a 0 is arbitrarily large; each of the other numbers a 1 , a 2 , a 3 ,
... less than 10. Such a series can be called a number, since it is possible to compare this series with rational numbers (see), you can establish the concepts of equality, sum, product, difference and quotient of such series. R. (1) will be denoted for brevity by one letter a. They say that but more rational number p/q, if for a sufficiently large n there is an inequality a 0 + a 1 /10 + a 2 /10 2 +... + a n/10n> p/q If, for any n a 0 + a 1 /10 + a 2 /10 2 +... + a n /10 n not > p/q but with a large enough n a 0 + a 1 /10 + a 2 /10 2 +... + a n/10n> r/s where r/s an arbitrary number less than p/q, then they consider and equal to p/q. On this basis, R. 9/10, 9/10 2 , 9/10 3 ,... is equal to one. This equality is denoted as follows: 0, 999 ... = 1. If a a not equal to 9, but all subsequent numbers a k +1 , a k +2 , a k+3 ,... are equal to 9, then the number a, determined by R. (1), is equal to a 0 + a 1 /10 + a 2 /10 2 +... + (a k + 1)/10 k . If not all numbers a k+1 , a k+2 , a k+3 ...are 9, then a = a 0 + a 1 /10 + a 2 /10 2 +... + a k /10k It may happen that all elements of series (1), starting from a k+1 ,
are equal to zero. In this case, according to the stated definition a a 0 + a 1 /10 + a 2 /10 2 +... + (a k+1)/10 k This kind of number is called final decimal. It is known from arithmetic that when an ordinary fraction is converted to a decimal, a finite fraction or an infinite periodic fraction is obtained. Any periodic decimal fraction can be converted to an ordinary fraction. It follows from this that an infinite non-periodic decimal fraction cannot equal a rational number and therefore represents a number of a special kind called irrational(cm.). 3. Convergence and divergence of series. R. numbers (2)... u 0 , u 1 , u 2 ,... u n,... called converging, if there is such a number a(rational or irrational) that with increasing n numerical value of the difference a - (u 0 + u 1 + u 2 +... u n- 1) becomes and remains arbitrarily small. Such a number a called sum R. In this case, they write (3)... a = u 0 + u 1 + u 2 +... and this equality is called decomposition numbers a into infinite R. If such a number a does not exist, then R. (2) is called. divergent. The most important example of a convergent R. is a geometric progression (see). 1, q, q 2 ,..., whose denominator q numerically less than one. In this case, there is a decomposition 1/(1 - q) = 1 + q + q 2 +... An example of a divergent R. is 1/1, 1/2, 1/3,... 1 + 1/2 + 1/3 +... doesn't make any sense. If, however, the terms of the harmonic R. are taken alternately with the signs + and -, then we obtain a convergent R. The expression 1 - 1/2 + 1/3 - 1/4 +... is equal to the logarithm of 2 taken at the base e(cm.). Not being able to state in detail the criteria for convergence, we note only the following theorems. A given R. is convergent if the R. of modules (see) of its members is convergent. R. v 0 ,
-v 1 , v 2 , -v 3 ..., in which the numbers v 0 , v 1 , v 2 , v 3 ... positive, converging if increasing n lim v n = 0. R. with positive members u 0 , u 1 , u 2 ,..., u n,... converging if lim(u n + 1)/u n
lim(u n + 1)/u n > 1 If for R. with positive members but, and 0 , and 1 , u 2 ,
.., and n... attitude lim(u n + 1)/u n = 1 - r/n+θ (n) /nα , where r do not depend on n, α
> 1 and θ ( n) remains constantly less than some positive number in numerical value, then R. convergent at r> 1 and diverging when r is less than or = 1 (Tannery, "Introduction à la theorie des fonctions d"une variable", p. 84). 4. Conditional and absolute convergence. If R. (4) v 0 , v 1 , v 2 ,... v n,... convergent, but the R. of the modules of its members is divergent, then we say that R. (4) conditionally convergent. Eg. 1, -1/2, 1/3, -1/4,... R. naz. absolutely convergent, if the R. modules of its members are convergent. The sum of a conditionally convergent R. changes with a change in the order of its members. Eg. 1 - 1/2 + 1/3 - 1/4 +... = log2, but 1 - 1/2 - 1/4 + 1/3 - 1/6 - 1/8 +... 1/2 - 1/4 + 1/6 - 1/8 +.... = 1/2 log 2. The sum of an absolutely convergent R. does not depend on the order of its members. If numbers a and b decompose into absolutely convergent R. a = a 0 + a 1 + a 2 +....., b = b 0 + b 1 + b 2 +..... ., a 0 b 0 , a 0 b 1 + a 1 b 0 , a 0 b 2 + a 1 b 2 + a 2 b 0 ,... absolutely convergent and, moreover, a 0 b 0 + (a 0 b 1 + a 1 b 0) + (a 0 b 2 + a 1 b 2 + a 2 b 0) +... = ab. 5. Uniform convergence. Suppose given R. (5)... f 0 (x), f 1 (x), f 2 (x),
..., f n(x),
... whose members are functions of one variable x, which can take both real and imaginary (see) values. Set of values X, under which this R. converges, forms the so-called area of convergence. R. 1, X, 1.2x 2 , 1.2.3x 3 ,...... ., convergent only at x = 0. R. 1, X, (1/2 + 1.2x 2), (1/3 + 1.2.3x 3),... diverging for every X. R. 1, X/ 1, (x 2 /1.2), (x 3 /1.2.3),... gathering. for every meaning X. If power R. α 0 , α 1 x,α2 x 2 ,... gathering. at some value X, not equal to zero, then this R. gathering. and at every x, whose modulus is less than some number R. If we use the geometric representation of imaginary quantities (see), then we can say that the region of convergence of this R. is a circle of radius R. An example is a geometric progression 1, x, x 2 , x 3 ,...., whose radius circle of convergence is equal to one. If a X belongs to the area of gathering. R. (5), then for any n, greater than some number t mod[ f n(x) + fn+ 1 (x) + fn+ 2 (x) +...]
Generally t depends on X and from ε, but it is possible, in special cases, that t depends only on ε if the values X belong to some area (S). In this case, R. (5) is called. uniformly converging in the region (S). For example, consider R. (6)... (1 - X), X (1 - X), X 2 (1 - X).... limited to real and positive values X. In order for there to be an inequality (7)... x n(1 -x) + xn+ 1 (1 -x) +...xn need to take n> Log ε /Log x Next, in the case under consideration t= Log ε /Log x. As we see, t depends on X. No matter how big m, there are such values X in the interval (0, 1) that inequality (7) will not be satisfied for any n, more t. If a X= 1, then inequality (7) is satisfied when n is greater than or = 1 Let's pretend that t= Log ε /Log (1 - α) and n is greater or = m Track. R. (6) uniformly coming off. in the interval (0, 1 - α). If in the region of uniform convergence the terms of the series f 0 (x), f 1 (x), f 2 (x)... are continuous functions of x, then the sum of this R. is also a continuous function (see Discontinuity). Evenly converge. R. can be term-by-term integrated or differentiated. Power R. a 0 , a 1 x, a 2 X 2 ... have uniform convergence inside the circle of convergence. 6. Decomposition of functions into series. In what follows, we will assume that the independent variable is real. Using the Maclaurin formula (see), the following expansions are obtained: (these formulas are valid for any x). In order to calculate, for example, cos 2 ° using formula (9), instead of x substitute the ratio to the radius of the length of an arc containing 2 degrees. In forms. (11) the logarithm is taken at the base e. This form. is inconvenient for calculating logarithms, since it is necessary to take a lot of terms of the R. to obtain even insignificant accuracy. More convenient for calculation is formula 13, which is derived from formula (11) assuming (1 + X)/(1 - X) = (a + z)/z in the expansion of the function log(1 + x) - log(l - x). Assuming a = 1, z= 1, find log2; " a = 1, z= 1, "log5; a + z = 3 4 , a= 80, "log3; a + z = 7 4 , a= 2400, "log7; Multiplying the found natural logarithms of these numbers by M \u003d 1 / log10 \u003d 0.43429 44819 03251 82765 ..., we get ordinary logarithms (based on 10) of the same numbers (see). Form. (12) is valid for X= 1 if m> -1, and at x= -1 if m> 0 (Abel, "Oeuvres complètes", 1881, p. 245). Using direct division, rational functions are expanded into power R. functions. You can also use the method of indefinite coefficients for this purpose. Assuming, for example 1/(1 + 2t + 5t 3 + 3t 3) = y 0 + y 1 t + y 2 t 2 + y 3 t 3 +..., y 0 = 1, y 1 + 2y 0 = 0, y 2 + 2y 1 + 5y 0 = 0, y 3 + 2y 2 + 5 at 1 +
3 at 0 = 0, y 4 + 2y 3 + 5 at 2 +
3 at 1 = 0 etc. R. coefficients y 0 , at 1 , y 2 ... has the property that four consecutive coefficients. related by the ratio y n +3 + 2y n +2 + 5 at n +1 +
3 at n = 0. This kind of R. called. returnable. From the written equations, y 0 is successively determined, at 1 , y 2 ... The expansion of a given function in R. can be found with the help of integral calculus, if the expansion in R. of the derivative is known. In this way, a decomposition is obtained (14)... arc tg x = x - (x 3 /3) + (x 5 /5) -... (15)... arc sin X = x/1 + 1/2(x 3/3) + (1.2/2.4)(x5/5) +... valid for values X, satisfying the conditions R. (14) using the formula of Machen (Machin) π /4 = 4 arc tg(1/5) - arc tg(1/239) makes it possible to calculate π very quickly with a large number of decimal places. Thus Shanks calculated π with 707 decimal places. The expansion of functions into trigonometric functions and the expansion of elliptic functions will be presented later. Encyclopedic Dictionary F.A. Brockhaus and I.A. Efron. - St. Petersburg: Brockhaus-Efron.
1890-1907
.
See what "Series, in mathematics" is in other dictionaries:
SERIES, an infinite series, the expression of which members a1, a2,..., an,... are numbers (number series) or functions (functional series). If the sum of the first n terms of the series (partial sum): Sn= a1+ a2+ ... + an with an unlimited increase in n tends to ... ... encyclopedic Dictionary
Content. 1) Definition. 2) The number determined next. 3) Convergence and divergence of series. 4) Conditional and absolute convergence. 5) Uniform convergence. 6) Expansion of functions into series. 1. Definitions. R. is a sequence of elements, ... ... Encyclopedic Dictionary F.A. Brockhaus and I.A. Efron
It has several meanings: A series is a collection of homogeneous, similar objects located in one line. A series is a collection of any phenomena that follow one after another in a certain order. A number of some, a considerable number, for example, "a number of countries" ... Wikipedia
A series, an infinite sum, for example of the form u1 + u2 + u3 +... + un +... or, in short, . (1) One of the simplest examples of R., already found in elementary mathematics, is the sum of an infinitely decreasing geometric progression 1 + q + q 2 +... + q… … Great Soviet Encyclopedia
Taylor series decomposition of a function into an infinite sum of power functions. The series is named after the English mathematician Brooke Taylor, although the Taylor series was known long before Taylor's publications, it was used back in the 17th century by Gregory, and ... ... Wikipedia
Taylor series decomposition of a function into an infinite sum of power functions. The series is named after the English mathematician Taylor, although the Taylor series was known long before Taylor's publications; it was used in the 17th century by Gregory, as well as by Newton. Rows ... ... Wikipedia
Decomposition of a function into an infinite sum of power functions. The series is named after the English mathematician Taylor, although the Taylor series was known long before Taylor's publications; it was used in the 17th century by Gregory, as well as by Newton. Taylor series ... ... Wikipedia
The Möbius series is a functional series of the form This series was investigated by Möbius, who found an inversion formula for this series: where is the Möbius function ... Wikipedia
I m. 1. A set of homogeneous objects located in one line. ott. Build in one line; line. 2. Linear sequence of seats in theater, cinema, etc. ott. Persons occupying such positions. 3. Stalls located in one line ... Modern explanatory dictionary of the Russian language Efremova
Books
- The Mathematics of Observers and Its Applications to Quantum Mechanics, Relativity and Classical Mathematics, B. S. Hots, D. B. Hots. This book presents the results of the authors related to the Mathematics of Observers (author's title Observer s Mathematics). This mathematics was first introduced by the authors, it was studied ...
Rows for teapots. Solution examples
All survivors welcome to the second year! In this lesson, or rather, in a series of lessons, we will learn how to manage rows. The topic is not very difficult, but to master it you will need knowledge from the first course, in particular, you need to understand what is the limit, and be able to find the simplest limits. However, it's okay, in the course of the explanations I will give the appropriate links to the necessary lessons. For some readers, the topic of mathematical series, methods of solving, signs, theorems may seem peculiar, and even pretentious, absurd. In this case, you do not need to “load” much, we accept the facts as they are, and simply learn to solve typical, common tasks.
1) Rows for teapots, and for samovars immediately content :)
For ultra-fast preparation on a topic there is an express course in pdf format, with the help of which it is really possible to "raise" the practice in just a day.
The concept of a number series
In general number series can be written like this:
Here:
- mathematical icon of the sum;
– common term of the series(remember this simple term);
- variable - "counter". The record means that the summation is carried out from 1 to “plus infinity”, that is, first we have , then , then , and so on - to infinity. A variable or is sometimes used instead of a variable. Summation does not necessarily start from one, in some cases it can start from zero, from two, or from any natural number.
In accordance with the “counter” variable, any series can be painted in detail: 
– and so on ad infinitum.
Terms 
- this is NUMBERS, which are called members row. If they are all non-negative (greater than or equal to zero), then such a series is called positive number line.
Example 1
By the way, this is already a “combat” task - in practice, quite often it is required to record several members of the series.
First, then:
Then, then:
Then, then:
The process can be continued indefinitely, but according to the condition, it was required to write the first three terms of the series, so we write down the answer:
Note the fundamental difference from number sequence,
in which the terms are not summed, but are treated as such.
Example 2
Write down the first three terms of the series
This is an example for self-solving, the answer is at the end of the lesson.
Even for a seemingly complex series, it is not difficult to describe it in expanded form:
Example 3
Write down the first three terms of the series
In fact, the task is performed orally: mentally substitute in the common term of the series first , then and . Eventually: 
Leave the answer like this it is better not to simplify the obtained terms of the series, that is do not comply actions: , , . Why? Answer in the form 
much easier and more convenient for the teacher to check.
Sometimes there is a reverse
Example 4
There is no clear solution algorithm here. you just have to see the pattern.
In this case:
For verification, the resulting series can be “painted back” in expanded form.
But the example is a little more difficult for an independent solution:
Example 5
Write the sum in collapsed form with a common term of the series 
Check again by writing the series in expanded form
Convergence of number series
One of the key objectives of the topic is examination of a series for convergence. In this case, two cases are possible:
1) Rowdiverges. This means that an infinite sum is equal to infinity: either sums in general does not exist, as, for example, in the series 
(by the way, here is an example of a series with negative terms). A good example of a divergent number series came across at the beginning of the lesson: 
. Here it is quite obvious that each next term of the series is greater than the previous one, therefore 
and hence the series diverges. An even more trivial example: 
.
2) Rowconverges. This means that an infinite sum is equal to some final number: . Please: 
This series converges and its sum is zero. A more meaningful example is infinitely decreasing geometric progression, known to us since school: 
. The sum of the members of an infinitely decreasing geometric progression is calculated by the formula: , where is the first member of the progression, and is its base, which, as a rule, is written as correct fractions. In this case: , . In this way: 
A finite number is obtained, which means that the series converges, which was required to be proved.
However, in the vast majority of cases find the sum of the series is not so simple, and therefore, in practice, to study the convergence of the series, special signs are used, which have been proven theoretically.
There are several signs of convergence of a series: necessary criterion for the convergence of a series, comparison criteria, d'Alembert's criterion, Cauchy's criteria, sign of Leibniz and some other signs. When to apply what sign? It depends on the common term of the series, figuratively speaking - on the "stuffing" of the series. And very soon we will put everything on the shelves.
! For further learning, you need understand well, what is the limit and it is good to be able to reveal the uncertainty of the form. For repetition or study of the material, refer to the article Limits. Solution examples.
A necessary criterion for the convergence of a series
If the series converges, then its common term tends to zero: .
The converse is not true in the general case, i.e., if , then the series can both converge and diverge. And so this sign is used to justify divergence row:
If the common term of the series does not go to zero, then the series diverges
Or in short: if , then the series diverges. In particular, a situation is possible when the limit does not exist at all, as, for example, limit. Here they immediately substantiated the divergence of one series :)
But much more often the limit of the divergent series is equal to infinity, while instead of "x" it acts as a "dynamic" variable. Let's refresh our knowledge: limits with "x" are called limits of functions, and limits with a variable "en" - limits of numerical sequences. The obvious difference is that the variable "en" takes discrete (discontinuous) natural values: 1, 2, 3, etc. But this fact has little effect on the methods for solving the limits and methods for disclosing uncertainties.
Let us prove that the series from the first example diverges.
Common member of the series:
Conclusion: row diverges
The necessary feature is often used in real practical tasks:
Example 6
We have polynomials in the numerator and denominator. The one who carefully read and comprehended the method of disclosure of uncertainty in the article Limits. Solution examples, certainly caught that when the highest powers of the numerator and denominator equal, then the limit is final number .
Divide the numerator and denominator by 
Study Series diverges, since the necessary criterion for the convergence of the series is not satisfied.
Example 7
Examine the series for convergence
This is a do-it-yourself example. Full solution and answer at the end of the lesson
So, when we are given ANY number series, first of all we check (mentally or on a draft): does its common term tend to zero? If it does not strive, we draw up a solution following the example of examples No. 6, 7 and give the answer that the series diverges.
What types of apparently divergent series have we considered? It is immediately clear that rows like or diverge. The series from examples No. 6, 7 also diverge: when the numerator and denominator contain polynomials, and the highest degree of the numerator is greater than or equal to the highest degree of the denominator. In all these cases, when solving and designing examples, we use the necessary criterion for the convergence of the series.
Why is the sign called necessary? Understand in the most natural way: in order for the series to converge, necessary so that its common term tends to zero. And everything would be fine, but this not enough. In other words, if the common term of the series tends to zero, THIS DOES NOT MEAN that the series converges- it can both converge and diverge!
Meet:
This row is called harmonic series. Please remember! Among the numerical series, he is a prima ballerina. More precisely, a ballerina =)
It is easy to see that 
, BUT. In the theory of mathematical analysis, it is proved that the harmonic series diverges.
You should also remember the concept of a generalized harmonic series:
1) This row diverges at . For example, the series diverge, , .
2) This row converges at . For example, the series , , . I emphasize once again that in almost all practical tasks it does not matter to us at all what the sum of, for example, the series is, the very fact of its convergence is important.
These are elementary facts from the theory of series that have already been proven, and when solving some practical example, one can safely refer, for example, to the divergence of the series or the convergence of the series.
In general, the material under consideration is very similar to study of improper integrals, and those who have studied this topic will find it easier. Well, for those who have not studied, it is doubly easier :)
So, what to do if the common term of the series GOES to zero? In such cases, to solve examples, you need to use others, sufficient signs of convergence / divergence:
Comparison criteria for positive number series
I draw your attention that here we are talking only about positive numerical series (with non-negative members).
There are two signs of comparison, one of them I will simply call sign of comparison, another - limiting sign of comparison.
First consider comparison sign, or rather, the first part of it:
Consider two positive numerical series and . If known, that the row is converges, and, starting from some number , the inequality holds, then the series converges too.
In other words: The convergence of a series with larger terms implies the convergence of a series with smaller terms. In practice, the inequality is often satisfied in general for all values of :
Example 8
Examine the series for convergence
First, we check(mentally or on a draft) execution: 
, which means that it was not possible to “get off with little blood”.
We look into the "package" of the generalized harmonic series and, focusing on the highest degree, we find a similar series: It is known from theory that it converges.
For all natural numbers, the obvious inequality holds:
and larger denominators correspond to smaller fractions: 
, which means that, according to the criterion of comparison, the series under study converges together with next to .
If you have any doubts, then the inequality can always be painted in detail! Let us write down the constructed inequality for several numbers "en":
If , then
If , then
If , then
If , then
….
and now it is quite clear that the inequality 
holds for all natural numbers "en".
Let us analyze the comparison criterion and the solved example from an informal point of view. Still, why does the series converge? Here's why. If the series converges, then it has some final amount : 
. And since all members of the series less corresponding members of the series, then the stump is clear that the sum of the series cannot be greater than the number , and even more so, cannot be equal to infinity!
Similarly, we can prove the convergence of "similar" series: 
, , 
etc.
! note that in all cases we have “pluses” in the denominators. The presence of at least one minus can seriously complicate the use of the considered comparison feature. For example, if the series is compared in the same way with a convergent series (write down several inequalities for the first terms), then the condition will not be fulfilled at all! Here you can dodge and choose for comparison another convergent series, for example, , but this will entail unnecessary reservations and other unnecessary difficulties. Therefore, to prove the convergence of a series, it is much easier to use marginal comparison criterion(see next paragraph).
Example 9
Examine the series for convergence
And in this example, I suggest you consider for yourself the second part of the comparison feature:
If known, that the row is diverges, and starting from some number (often from the very first) inequality holds, then the series also diverges.
In other words: The divergence of the series with smaller terms implies the divergence of the series with larger terms.
What should be done?
It is necessary to compare the series under study with a divergent harmonic series. For a better understanding, construct some specific inequalities and make sure that the inequality is true.
Solution and sample design at the end of the lesson.
As already noted, in practice the comparison feature just considered is rarely used. The real "workhorse" of the number series is marginal comparison criterion, and in terms of frequency of use, only sign of d'Alembert.
Limit sign of comparison of numerical positive series
Consider two positive numerical series and . If the limit of the ratio of the common members of these series is equal to finite non-zero number: , then both series converge or diverge at the same time.
When is the limit comparison criterion used? The limit sign of comparison is used when the “stuffing” of the series is polynomials. Either one polynomial in the denominator, or polynomials in both the numerator and the denominator. Optionally, polynomials can be under roots.
Let's deal with the series for which the previous sign of comparison stalled.
Example 10
Examine the series for convergence
Compare this series with the convergent series . We use the limit test of comparison. It is known that the series converges. If we can show that it is final non-zero number, it will be proved that the series also converges.
A finite, non-zero number is obtained, which means that the series under study converges together with next to .
Why was the series chosen for comparison? If we had chosen any other series from the “clip” of the generalized harmonic series, then we would not have succeeded in the limit final non-zero numbers (you can experiment).
Note: when we use the marginal comparison feature, irrelevant, in what order to compose the relation of common members, in the considered example, the relation could be drawn in reverse: - this would not change the essence of the matter.
Number lines. Convergence and divergence of numerical series. d'Alembert convergence test. Variable rows. Absolute and conditional convergence of series. functional rows. Power series. Expansion of elementary functions in the Maclaurin series.
Guidelines on topic 1.4:
Number rows:
A number series is a sum of the form
where are the numbers u 1 , u 2 , u 3 , n n , called members of the series, form an infinite sequence; the term un is called the common term of the series.
. . . . . . . . .
composed of the first terms of the series (27.1) are called partial sums of this series.
Each row can be associated with a sequence of partial sums S1, S2, S3. If, as the number n increases infinitely, the partial sum of the series S n tends to the limit S, then the series is called convergent, and the number S- the sum of a convergent series, i.e.
This entry is equivalent to the entry
If a partial amount S n series (27.1) with an unlimited increase n has no finite limit (in particular, tends to + ¥ or to - ¥), then such a series is called divergent
If the series converges, then the value S n for sufficiently large n is an approximate expression for the sum of the series S.
Difference r n = S - S n is called the remainder of the series. If the series converges, then its remainder tends to zero, i.e. r n = 0, and vice versa, if the remainder tends to zero, then the series converges.
The series of a species is called geometric line.
called harmonic.
if N®¥, then S n®¥, i.e. the harmonic series diverges.
Example 1. Write a series by its given common term:
1) assuming n = 1, n = 2, n = 3, we have an infinite sequence of numbers: , , , Adding its terms, we get the series 
2) Doing the same, we get the series
3) Giving n the values 1, 2, 3, and taking into account that 1! = 1, 2! = 1 × 2, 3! = 1 × 2 × 3, we get the series
Example 2. Find n-th term of the series by its given first numbers:
1) ; 2) 
; 3) .
Example 3. Find the sum of the terms of the series:
2) 
.
1) Find the partial sums of the terms of the series:
; 
;
… .
Let us write down the sequence of partial sums: …, , … .
The common term of this sequence is . Consequently,
.
The sequence of partial sums has a limit equal to . So the series converges and its sum is .
2) This is an infinitely decreasing geometric progression, in which a 1 = , q= . Using the formula, we get So, the series converges and its sum is equal to 1.
Convergence and divergence of numerical series. Convergence sign d'Alembert :
A necessary criterion for the convergence of a series. A series can only converge if its common term is u n with unlimited number increase n goes to zero:
If , then the series diverges - this is a sufficient sign of the solubility of the series.
Sufficient conditions for the convergence of a series with positive terms.
Sign of comparison of series with positive terms. The series under study converges if its members do not exceed the corresponding members of another, obviously convergent series; the series under study diverges if its terms exceed the corresponding terms of another obviously divergent series.
In the study of series for convergence and solubility on this basis, the geometric series is often used
which converges for |q|
,
being divergent.
In the study of series, the generalized harmonic series is also used
.
If a p= 1, then this series turns into a harmonic series, which is divergent.
If a p< 1, то члены данного ряда больше соответствующих членов гармонического ряда и, значит, он расходится. При p> 1 we have a geometric series in which | q| < 1; он является сходящимся. Итак, обобщенный гармонический ряд сходится при p> 1 and diverges at p£1.
Sign of d'Alembert. If for a series with positive terms
(u n>0)
condition is satisfied, then the series converges at l l > 1.
d'Alembert's sign does not give an answer if l= 1. In this case, other methods are used to study the series.
Variable rows.
Absolute and conditional convergence of series:
Number series
u 1 + u 2 + u 3 + u n
is called alternating if among its members there are both positive and negative numbers.
A number series is called sign-alternating if any two adjacent members have opposite signs. This series is a special case of an alternating series.
Convergence criterion for alternating series. If the terms of the alternating series monotonically decrease in absolute value and the common term u n tends to zero as n® , then the series converges.
A series is called absolutely convergent if the series also converges. If a series converges absolutely, then it is convergent (in the usual sense). The converse is not true. A series is said to be conditionally convergent if it itself converges and the series composed of the modules of its members diverges. Example 4. Examine the series for convergence  .
|
Let us apply the Leibniz sufficient test for alternating series. We get  because the . Therefore, this series converges. Example 5. Examine the series for convergence  .
|
Let's try to apply the Leibniz sign:  It can be seen that the modulus of the general term does not tend to zero when n→∞. Therefore, this series diverges. Example 6. Determine if the series is absolutely convergent, conditionally convergent or divergent. |
Applying the d'Alembert criterion to a series composed of the modules of the corresponding terms, we find  Therefore, this series converges absolutely. |
Example 7. Examine for convergence (absolute or conditional) an alternating series:
1) The terms of this series monotonically decrease in absolute value 
and 
. Therefore, according to the Leibniz test, the series converges. Let us find out whether this series converges absolutely or conditionally.
2) The terms of this series monotonically decrease in absolute value: 
, but
.
Functional series:
The usual number series consists of numbers:
All members of the series 
- this is numbers.
The functional line consists of functions:
In the general term of the series, in addition to polynomials, factorials, etc. certainly includes the letter "x". It looks like this, for example: Like a number series, any functional series can be written in expanded form:
As you can see, all members of the functional series are functions.
The most popular type of functional series is power series.
Power series:
power next is called a series
,
where are the numbers a 0, a 1, a 2, a n are called the coefficients of the series, and the term a n x n is a common member of the series.
The convergence region of a power series is the set of all values x for which the series converges.
Number R is called the radius of convergence of the series if, for | x|
Example 8. Given a series
Investigate its convergence at points x= 1 and X= 3, x= -2.
When x = 1, this series turns into a number series
.
Let us investigate the convergence of this series by the d'Alembert test. We have
those. the series converges.
For x = 3 we get the series
Which diverges, since the necessary criterion for the convergence of the series is not satisfied
For x = -2 we get
This is an alternating series, which, according to the Leibniz test, converges.
So at the points x= 1 and X= -2. the series converges, and at the point x= 3 diverges.
Expansion of elementary functions in the Maclaurin series:
Near Taylor for function f(x) is called a power series of the form
Basic definitions
Definition. The sum of the members of an infinite number sequence is called a number series.
In this case, the numbers will be called members of the series, and un - the common member of the series.
Definition. Sums, n = 1, 2, ... are called partial (partial) sums of the series.
Thus, it is possible to consider sequences of partial sums of the series S1, S2, …, Sn, …
Definition. A series is called convergent if the sequence of its partial sums converges. The sum of a convergent series is the limit of the sequence of its partial sums.
Definition. If the sequence of partial sums of the series diverges, i.e. has no limit, or has an infinite limit, then the series is called divergent and no sum is assigned to it.
Row Properties
1) The convergence or divergence of the series will not be violated if you change, discard or add a finite number of terms in the series.
2) Consider two series and, where C is a constant number.
Theorem. If a series converges and its sum is equal to S, then the series also converges and its sum is equal to CS. (C0)
3) Consider two rows and. The sum or difference of these series will be called a series where the elements are obtained as a result of addition (subtraction) of the original elements with the same numbers.
Theorem. If the series and converge and their sums are equal to S and, respectively, then the series also converges and its sum is equal to S + .
The difference of two convergent series will also be a convergent series.
The sum of a convergent and divergent series will be a divergent series.
It is impossible to make a general statement about the sum of two divergent series.
When studying series, two problems are mainly solved: the study of 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 to be convergent, it is necessary and sufficient that for any there exists a number N such that for n > N and any p > 0, where p is an integer, the inequality would hold:
Proof. (need)
Let, then for any number there is a number N such that the inequality
is performed for n>N. For n>N and any integer p>0, the inequality also holds. Considering both inequalities, we get:
The need has been proven. We will not consider the proof of sufficiency.
Let us formulate the Cauchy criterion for the series.
For a series to be convergent it is necessary and sufficient that for any there exists a number N such that for n>N and any p>0 the inequality
However, in practice, it is not very convenient to use the Cauchy criterion directly. Therefore, as a rule, simpler convergence criteria are used:
1) If the series converges, then it is necessary that the common term un tends 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 exactly diverges. For example, the so-called harmonic series is divergent, although its common term tends to zero.
Answer: the series diverges.
Example #3
Find the sum of the series $\sum\limits_(n=1)^(\infty)\frac(2)((2n+1)(2n+3))$.
Since the lower summation limit is 1, the common term of the series is written under the sum sign: $u_n=\frac(2)((2n+1)(2n+3))$. Compose the nth partial sum of the series, i.e. sum the first $n$ members of the given numerical 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, recording a partial sum did not bring us one iota closer to the goal. After all, 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 not give us anything in essence. To find the limit, the partial sum expression 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 common 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 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 a change, in this example we will go the first way, and the next - we will substitute the private values of $n$. Expanding 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 equation, $n$ is preceded by zero. If you like, the left side of the equality can be represented for clarity 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 $2A+2B$ precedes $n$, we have the first equation: $2A+2B=0$. We immediately divide both parts of this equation by 2, after which we get $A+B=0$.
Since the free term on the left side of the equality 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. $$
The proof will be carried out by the method of mathematical induction. At the first step, we need to check whether the required equality $S_n=\frac(1)(3)-\frac(1)(2n+3)$ holds for $n=1$. We know that $S_1=u_1=\frac(2)(15)$, but will the expression $\frac(1)(3)-\frac(1)(2n+3)$ give the value $\frac(2 )(15)$ if $n=1$ is substituted into it? Let's check:
$$ \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). $$
So, for $n=1$ the equality $S_n=\frac(1)(3)-\frac(1)(2n+3)$ is satisfied. This completes the first step of the method of mathematical induction.
Assume that for $n=k$ the equality holds, i.e. $S_k=\frac(1)(3)-\frac(1)(2k+3)$. Let us prove that the same equality will hold for $n=k+1$. To do this, consider $S_(k+1)$:
$$ S_(k+1)=S_k+u_(k+1). $$
Since $u_n=\frac(1)(2n+1)-\frac(1)(2n+3)$, then $u_(k+1)=\frac(1)(2(k+1)+ 1)-\frac(1)(2(k+1)+3)=\frac(1)(2k+3)-\frac(1)(2(k+1)+3)$. According to the above assumption $S_k=\frac(1)(3)-\frac(1)(2k+3)$, so the formula $S_(k+1)=S_k+u_(k+1)$ takes the form:
$$ 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). $$
Conclusion: the formula $S_n=\frac(1)(3)-\frac(1)(2n+3)$ is true for $n=k+1$. Therefore, according to the method of mathematical induction, the formula $S_n=\frac(1)(3)-\frac(1)(2n+3)$ is true for any $n\in N$. Equality has been proven.
In a standard course in higher mathematics, one is usually content with "deleting" the canceling terms, without requiring any proof. So, we got the expression for the nth partial sum: $S_n=\frac(1)(3)-\frac(1)(2n+3)$. Find the value of $\lim_(n\to\infty)S_n$:
Conclusion: the given series converges and its sum is $S=\frac(1)(3)$.
The second way is to simplify the formula for the partial sum.
To be honest, I prefer this method myself :) Let's write down the partial sum in an abbreviated form:
$$ S_n=\sum\limits_(k=1)^(n)u_k=\sum\limits_(k=1)^(n)\frac(2)((2k+1)(2k+3)). $$
We got earlier that $u_k=\frac(1)(2k+1)-\frac(1)(2k+3)$, so:
$$ 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). $$
The sum $S_n$ contains a finite number of terms, so we can rearrange them however we like. I want to first add all the terms of the form $\frac(1)(2k+1)$, and only then go to the terms of the form $\frac(1)(2k+3)$. This means that we will represent the partial sum in this form:
$$ 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). $$
Now we transform the expressions $\frac(1)(2k+1)$ and $\frac(1)(2k+3)$ to the same form. I think it's convenient to make it look like a larger fraction (although you can use a smaller one, it's a matter of taste). Since $\frac(1)(2k+1)>\frac(1)(2k+3)$ (the larger the denominator, the smaller the fraction), we will reduce the fraction $\frac(1)(2k+3) $ to the form $\frac(1)(2k+1)$.
I will present the expression in the denominator of the fraction $\frac(1)(2k+3)$ as follows:
$$ \frac(1)(2k+3)=\frac(1)(2k+2+1)=\frac(1)(2(k+1)+1). $$
And the sum $\sum\limits_(k=1)^(n)\frac(1)(2k+3)$ can now be written like this:
$$ \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). $$
If the equality $\sum\limits_(k=1)^(n)\frac(1)(2k+3)=\sum\limits_(k=2)^(n+1)\frac(1)(2k+ 1) $ does not raise questions, then let's go further. If there are questions, please expand the note.
How did we get the converted amount? show/hide
We had the series $\sum\limits_(k=1)^(n)\frac(1)(2k+3)=\sum\limits_(k=1)^(n)\frac(1)(2( k+1)+1)$. Let's introduce a new variable instead of $k+1$ - for example, $t$. So $t=k+1$.
How did the old variable $k$ change? And it changed from 1 to $n$. Let's find out how the new variable $t$ will change. If $k=1$, then $t=1+1=2$. If $k=n$, then $t=n+1$. So the expression $\sum\limits_(k=1)^(n)\frac(1)(2(k+1)+1)$ is now: $\sum\limits_(t=2)^(n +1)\frac(1)(2t+1)$.
$$ \sum\limits_(k=1)^(n)\frac(1)(2(k+1)+1)=\sum\limits_(t=2)^(n+1)\frac(1 )(2t+1). $$
We have the sum $\sum\limits_(t=2)^(n+1)\frac(1)(2t+1)$. Question: does it matter which letter to use in this sum? :) Tritely writing the letter $k$ instead of $t$, we get the following:
$$ \sum\limits_(t=2)^(n+1)\frac(1)(2t+1)=\sum\limits_(k=2)^(n+1)\frac(1)(2k +1). $$
This is how the equality $\sum\limits_(k=1)^(n)\frac(1)(2(k+1)+1)=\sum\limits_(k=2)^(n+1) is obtained \frac(1)(2k+1)$.
Thus, the partial sum can be represented in the following form:
$$ 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 ). $$
Note that the sums $\sum\limits_(k=1)^(n)\frac(1)(2k+1)$ and $\sum\limits_(k=2)^(n+1)\frac(1 )(2k+1)$ differ only in the limits of summation. Let's make these limits the same. "Taking" the first element from the sum $\sum\limits_(k=1)^(n)\frac(1)(2k+1)$ we get:
$$ \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). $$
"Taking" the last element from the sum $\sum\limits_(k=2)^(n+1)\frac(1)(2k+1)$, we get:
$$\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). $$
If you skip all the explanations, then the process of finding an abbreviated formula for the n-th partial sum will take the following form:
$$ 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). $$
Let me remind you that we reduced the fraction $\frac(1)(2k+3)$ to the form $\frac(1)(2k+1)$. Of course, you can do the opposite, i.e. represent the fraction $\frac(1)(2k+1)$ as $\frac(1)(2k+3)$. The final expression for the partial sum will not change. In this case, I will hide the process of finding a partial sum under a note.
How to find $S_n$, if you bring to the form of a different fraction? show/hide
$$ 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 is $S=\frac(1)(3)$.
Answer: $S=\frac(1)(3)$.
The continuation of the topic of finding the sum of a series will be considered in the second and third parts.
