- Perhaps it will be difficult, it will be not so difficult;) and the title of this article is also lucavit - the ranks that we are talking about, rather, not difficult, and "rare-earth". However, they are not insured by even joined students, and therefore this seemingly an additional lesson should be taken to the maximum seriousness. After all, after his study, you can deal with almost any "beast"!
Let's start with the classics of the genre:
Example 1.
First, we will pay attention that this is not a power row (I remind you that it has the kind). And, secondly, it is immediately striking the value, which can not know how to enter the region of the row convergence. And this is already a small success of the study!
But still, how to come to success big? I hurry to delight you - such rows can be solved in the same way as power - Relying on the sign of the Dalamber or the Kauchi radical sign!
Decision: The value is not included in the row convergence area. This fact is essential, and it must be mentioned!
The basis of the algorithm works standard. Using the sign of Dalamber, we will find the interval of the convergence of the series: 
A number converges at. We raise the module upstairs:
Immediately monitor the "bad" point: the value is not included in the region of the convergence of the series.
We investigate the convergence of the row on the "internal" ends of the intervals:
If, then 
If, then
Both numeric rows diverge, since it is not fulfilled required sign of convergence.
Answer: Convergence region:
Perform a small analytical check. Let's substitute any value from the right interval, for example,:
- converges in sign of Dalamber.
In the case of substitution of values \u200b\u200bfrom the left interval, convergent rows are also obtained:
If, then.
And finally, if, then a number 
- Really diverge.
A pair of simple example for warming up:
Example 2.
Find the region of the convergence of the functional series
Example 3.
Find the region of the convergence of the functional series
Especially well disperse from "new" module - He will meet today 100500 times!
Brief decisions and answers at the end of the lesson.
The algorithms used seem to be universal and trouble-free, but in fact it is not so - for many functional series they often "slip", and then lead to erroneous conclusions (and such examples I will also consider).
Roughness begins at the level of interpretation of the results: Consider, for example, a number. Here in the limit we get 
(Check yourself)And in theory you need to answer that the series converges in a single point. However, the point is "dried", and therefore our "patient" diverges everywhere!
And for a number "obvious" decision "on Cauchy" does not give anything at all:
- For any meaning "X".
And the question arises, what to do? We use the method to which the main part of the lesson will be devoted! It can be formulated as follows:
Direct analysis of numerical rows at different values
In fact, we have already begun to deal with this in Example 1. First we investigate any particular "X" and the corresponding numerical series. This suggests the value: 
- The resulting numeric row is diverged.
And it immediately pursues to the thought: what if the same thing happens on other points?
Check-ka required sign convergence of a number for arbitrary Values:
Point is taken into account above for all other "X" Standard admission to organize the second wonderful limit:
Output: The row diverges on the whole numerical direct
And this decision is the most worker!
In practice, the functional series often has to be compared with generalized harmonic nearby :
Example 4.
Decision: First of all, we understand with definition area: In this case, the feeding expression should be strictly positive, and, in addition, there must be all members of the series, starting from the 1st. From this it follows that:
. With these values, conditionally convergent rows are obtained: 
etc.
Others, "X" are not suitable, for example, when we get an illegal case where there are no first two members of the series.
It's all good, it's all clear, but another important question remains - how to make a decision correctly? I suggest a scheme that can be zagalno called "Translation of the arrows" into numerical rows:
Consider arbitrary value 
And we investigate the convergence of the numerical series. Routine sign Leibnitsa:
1) This series is alkaline.
2) 
- Members of a number decrease in the module. Each next member of a series of module is less than the previous one: 
So decrease in monotonous.
Conclusion: the series converges on the basis of the leiby. As already noted, convergence here - for the reason that a number 
- Divorces.
So here - neat and correct! For for Alfa, we trickyly hid all the permissible numeric rows.
Answer: The functional series exists and converges conditionally.
Similar example for self solutions:
Example 5.
Explore the convergence of the functional series
An exemplary sample of the definition of the task at the end of the lesson.
So "Working Hypothesis"! - On the interval, the functional series converges!
2) With a symmetric interval, everything is transparent, we consider arbitrary Values \u200b\u200band get: - Absolutely converging numeric rows.
3) And finally, the "middle". Here, too, it is convenient to highlight two gaps.
We are considered arbitrary Value from the interval and get a numerical series: 
! Again - if hard , Substitute any specific number, for example. However, ... you wanted difficulties \u003d)
For all the values \u200b\u200bof "En" performed 
So: 
- Thus, by sign of comparison A number converges with infinitely decreasing progress.
For all the values \u200b\u200bof "X" from the interval we get 
- Absolutely convergent numeric rows.
All "Iks" investigated, "Iks" no more!
Answer: Row convergence region:
I must say an unexpected result! And it should also be added that the use of signs of Dalamber or Cauchy here will definitely mislead!
A direct score is the "highest pilot" of mathematical analysis, but for this, of course, experience is required, and somewhere even intuition.
Or maybe someone will find the way easier? Write! Precedents, by the way, is - several times the readers offered more rational solutions, and I gladly published them.
Successful landing to you :)
Example 11.
Find the region of the convergence of the functional series
My version of the solution is very close.
Additional hardcore can be found in Section VI (Rows)collection of Kuznetsov (Tasks 11-13).On the Internet there are ready-made solutions, but here I have to warmer - Many of them are incomplete, incorrect, and even erroneous. And, by the way, it was one of the reasons for which this article was born.
Let's bring the results of three lessons and systematize our toolkit. So:
To find the interval (s) of the convergence of the functional series, you can use:
1) a sign of Dalamber or a sign of Cauchy. And if a number are not power - We exhibit increased caution, analyzing the resulting result of direct substitution of various values.
2) a sign of uniform convergence of Weierstrass. Do not forget!
3) comparison with typical numeric rows - taxis in the general case.
Then explore the ends of the found intervals (if needed) And we get the region of the convergence of the row.
Now at your disposal is quite a serious arsenal that will cope with almost any thematic task.
I wish you success!
Solutions and answers:
Example 2: Decision: The value is not included in the row convergence area.
We use a sign of Dalamber:
The series converges at:
Thus, the convergence intervals of the functional series: 
.
We investigate the convergence of a number at endpoints:
if, then 
;
if, then 
.
Both numeric rows diverge, because The necessary sign of convergence is not performed.
Answer
: Convergence region: 
Let the function define in the region
Definition. Expression
Called functional nearby.
Example.
With some values, the row can converge for other values \u200b\u200b- to disperse.
Example.
Find a region of the convergence of the row. This series is defined for values.
If then, a row diverges, since the necessary sign of the convergence of the series is not performed; If a series diverges; If - infinitely decreasing geometric progression.
A comparison of this series with a convergent next to give the region of convergence of the series under study.
With the values \u200b\u200bof the functional series, a numerical series is obtained
If for a numerical series converges, the point is called convergence pointfunctional series.
The combination of all points of the convergence of the row forms the area of \u200b\u200bits convergence. The convergence area is usually some axis interval.
If at each point numeric rows converge, then the function range is called convergent in area .
The amount of functional series is some function from a variable defined in the region of the row convergence
What properties have functions if the properties of a number of row are known, that is.
Continuity of functions is not sufficient to make conclusion about continuity.
The convergence of a number of continuous functions to the continuous function is provided with an additional condition expressing one important feature of the convergence of the functional series.
Definition. The functional range is called converging in the area if there is a limit of partial sums of this series, that is.
Definition. The functional series is called evenly converging in the area, if for any positive, there is such a number that inequality is performed for all.
Geometric meaning of uniform convergence
If you surround the graph of the function - strip "determined by the ratio of the graphics all functions starting with quite large value full Lying in this "- strip" surrounding the graph of the limit function.
Properties evenly converging rows .
1. The sum of a uniformly convergent row in some region composed of continuous functions is a function continuous in this area.
2. Such a number can be differentiated
3. A number can be integrated
In order to determine whether the functional series is evenly converging, it is necessary to take advantage of a sufficient sign of Weierstrass convergence.
Definition. Functional range called majorized In some area of \u200b\u200bchange, if there is a similar number of a numeric number with positive members, which inequality are performed for all of this area.
Sign of Weierstrass (uniform convergence of the functional series).
Functional series it converges uniform In the field of convergence, if it is majorized in this area.
In other words, if the functions in a certain area are not exceeded by the absolute value of the corresponding positive numbers and if the numerical series converges, the functional series in this area converges uniformly.
Example. Prove the uniform convergence of the functional series.
Decision. . We replace the general member of this series by a general member of the numerical series, but superior to each member of a number of absolute value. To do this, it is necessary to determine in which the common member of the series will be maximal.
The resulting numerical series converges, it means that the functional series converges evenly according to the sign of Weierstrass.
Example. Find the sum of the row.
To find the sum of a number, we use the well-known formula for the amount of geometric progression
Differentiating the left and right of formula (1), we get sequentially
We highlight in the amount to be calculated, the components proportional to the first and second derivative:
Calculate derivatives:
Power rows.
Among the functional series there is a class of power and trigonometric series.
Definition. Functional series of species
called power in degrees. Expressions are constant numbers.
If a series is powerless in degrees.
The region of the convergence of the power series. Abel theorem.
Theorem. If the power series converges at the point, it converges and moreover, absolutely for any value, in the absolute value of smaller, that is, or in the interval.
Evidence.
Due to the convergence of Rada, his overall member should strive for zero, so all members of this series are uniformly limited: there is such a constant positive number that inequality takes place., That for everyone with the center at the point
The convergence region is functional next to a number of members of which are functions / defined at a numerical axis. For example, the members of the series are determined on the interval, and the members of the series are defined on the segment of the functional series (1) is called the functional rows of the convergence region of the uniform convergence of the Weierstrass feature of the uniform convergent functional series of a numerical series if a number (1) converges At each point of the set D C E and diverged at each point, the set D is not belonging, they say that the series converges on the set D, and called D region of the series of rows. A number (1) is called absolutely converging on the set D if a number converges in this set in the case of the convergence of the row (1) on the set D, its sum s will be a function defined on D, the region of convergence of some functional series can be found using known sufficient signs mounted for rows with positive members, for example, a sign of Dapamber, a sign of Cauchy. Example 1. Find the region of the convergence of the series M Since the numerical series converges when p\u003e 1 and dispel when P ^ 1, then, believing P - IGX, we get this series. which will be converged at igx\u003e c i If x\u003e 10, and diverge with Igx ^ 1, i.e. at 0.< х ^ 10. Таким образом, областью сходимости ряда является луч Пример 2. Найти область сходимости ряда 4 Рассмотрим ряд Члены этого ряда положительны при всех значениях х. Применим к нему признак Даламбера. Имеем пе При ех < 1. т.е. при, этот ряд будет сходиться. Следовательно, заданный ряд сходится абсолютно на интервале При х > 0 row diverges, as l \u003d. The divergence of the series at x \u003d 0 is obvious. Example 3. Neji Area of \u200b\u200bthe convergence of a number of this series are defined and continuous on the set. Applying a sign of Kos and, we will find for anyone. Consequently, the row diverges at all values \u200b\u200bof x. Denote by Sn (x) N-Mu partial sum of the functional series (1). If this series converges on the set D and its sum is 5 (g), it can be represented in the form where there is a sum of the set on the set D of a row called the PN residue of the functional series (1). For all values \u200b\u200bx € D, the ratio is therefore. i.e., the residue Rn (x) of the converging row tends to zero with P oo, which would be x 6 D. Uniform convergence among all converging functional series, the so-called uniformly convergent rows play an important role. Let it be given on the set D functional number of which is equal to S (x). Take it N-Mu partial amount definition. Functional series Functional series of convergence Region Uniform convergence Sign of Weierstrass Properties of uniformly converging functional series is called evenly converging on a set of PS1), if there is a number of LH for any number of E\u003e about such that the inequality will be performed for all numbers n\u003e n and for all x from a set of fi. Comment. Here the number n is the same for all x €, i.e. It does not depend on Z, but it depends on the choice of the number E, so that N \u003d n (E) is written. The uniform convergence of the functional series £ / n (®) to the function 5 (x) on the set Ft is often indicated as follows: determination of the uniform convergence of the row / n (g) on \u200b\u200bthe set of ft can be written in short using logical symbols: we will explain the geometrically meaning of uniform convergence Functional series. Take as a set ft segment [A, 6] and construct graphs of functions. Inequality | performed for numbers n\u003e n and for all a; G [a, b], it can be written in the following form. The obtained inequalities show that the graphs of all functions y \u003d 5 "(g) with the numbers P\u003e n will be fully concluded inside £ -polates, bounded by curves y \u003d s (x) - e and y \u003d 5 (g) + e (Fig. 1). Example 1 Evenly converges on the segment This series is a certain one, satisfies the conditions of the recognition of the leibence at any X € [-1,1] and, therefore, converges on the segment (-1,1]. Let S (x) be its sum, and SN. (x) - His pn is a partial amount. The residue of a number in absolute value does not exceed the absolute value of its first member: And since we will take any e. Then inequality | will be performed if. From here we find that P\u003e \\. If you take a number (here through [a] marked the greatest integer, not exceeding a), then inequality | E will be performed for all numbers P\u003e N and for all x € [-1.1). This means that this series is evenly converged on the segment [-1.1). I. Not any functional series that goes on the set is evenly converging for example 2. We will show that the series converges on the segment, but not evenly. 4 Calculate the PM partial sum of £ "(*) of a number. We have from where this series converges on the segment and its amount if the absolute value of the difference S (x) - 5 "(x) (row residue) is equal to. Take the number e such that. Let allowing the inequality relative to paragraph. We have, from where (as, and when dividing on Inx, the sign of inequality changes to the opposite). The inequality will be performed at. Therefore, such an independent number of N (E) so that the inequality is performed for each) immediately for all x from the segment. , does not exist. If you replace the segment 0 by a smaller segment, where, on the last one, this series will be converged to the function S0 evenly. In fact, when, and therefore, at once for all x §3. The sign of Weierstrass is a sufficient sign of uniform convergence of the functional series is given by Weierstrass theorem. Theorem 1 (sign of Weierstrass). Let for all x from the set q, the members of the functional series in absolute value do not exceed the corresponding members of the concerned numerical series n \u003d 1 with positive members, i.e. for all x € Q. Then the functional series (1) on the set n converges absolutely and evenly . And the tech as, by the condition of the theorem, the members of the series (1) satisfy the condition (3) on the entire set Q, then on the basis of comparison, the series 2 \\ Fn (x) \\ converges at any x € and, and, therefore, a number (1) converges on P Absolutely. Let us prove the uniform convergence of the row (1). Let be denoted by Sn (x) and An partial sums of the series (1) and (2), respectively. We take anyone (how small) the number E\u003e 0. Then, from the convergence of the numerical series (2), the existence of the number n \u003d n (e) is followed, which is therefore for all numbers P\u003e N (E) and for all HBS . A number (1) converges uniformly on the set P. Remark. The numeric number (2) is often referred to as a major, or a major, for a functional series (1). Example 1. Investigate into uniform convergence, a number of inequality is performed for all. And for all. Numeric row converges. By virtue of Weierstrass, the functional series considered is absolutely and evenly on the entire axis. Example 2. To investigate on a uniform convergence, a number of row members are determined and continuous on the segment [-2.2 |. Since on the segment [-2.2) for any natural n, thus, inequality is performed for. Since the numeric series converges, then, on the basis of Weierstrass, the original functional series converges absolutely and evenly on the segment. Comment. The functional series (1) may be evenly on a set of beer when there is no numerical major series (2), i.e., the Weierstrass sign is only a sufficient sign for uniform convergence, but is not necessary. Example. As shown above (example), the series is evenly converged on the segment of 1-1,1]. However, for him a major converging numerical series (2) does not exist. In fact, for all natural P and for all x € [-1.1), inequality is carried out with equality achieved at. Therefore, members of the articulated major series (2) must certainly be satisfied with the condition but a numerical series of functional series of convergence of uniform convergence of the Weierstrass characteristics of uniformly converging functional series diverges. So, a number of £ op will diverge. The properties of uniformly converging functional rows are evenly converging functional series have a number of important properties. Theorem 2. If all members of a series evenly converging on the segment [a, b], multiply by the same function d (x), limited to [A, 6], then the obtained functional row will be evenly converged on. Let on the segment [a, b \\ rod £ Fn (x) evenly converges to the function 5 (g), and the function d (x) is limited, i.e. there is a constant C\u003e 0 such that, by determining the uniform convergence of the series For any number E\u003e 0, there is a number n such that for all P\u003e n and for all x € [A, b] will be carried out inequality where 5N (AR) will be partial amount of the series under consideration. Therefore, we will have for anyone. A row is evenly converged on [A, b | To the function of Theorem 3. Let all members of the Fn (x) of the functional series are continuous and the series converges uniformly on the segment [a, b \\. Then the sum of s (x) of a number is continuous on this segment. M We take on the segment [o, b] two arbitrary gig + ah points. Since this series converges on the segment [A, b] evenly, then for any number E\u003e there is no number N \u003d n (E) such that for all I\u003e N will be carried out inequalities where5 "(g) - partial sums of the FN series (x). These partial sums 5 "(g) are continuous on the segment [A, 6] as the amount of the finite number of continuous on [A, 6) functions Fn (x). Therefore, for a fixed number NO\u003e n (E) and the number of e) there is a number 6 \u003d 6 (E)\u003e 0 such that for the increment of AH, satisfying the condition |, there will be an inequality increment AS sums S (x) can be submitted in the following The form: From where. Given the inequalities (1) and (2), for increments ah, satisfying the condition |, we obtain this means that the sum Six) is continuous at the point x. Since X is an arbitrary point of the segment [A, 6], then 5 (g) continuous on | a, 6 |. Comment. The functional number of whose members are continuous on the segment [A, 6), but which converges on (A, 6] uneven, may have a sum of the discontinuous function. Example 1. Consider the functional series on the segment | 0,1). I calculate its N-Mu partial amount so it breaks on the segment, although the members of the series are continuous on it. By virtue of the proven theorem, this series is not evenly converging on the segment. Example 2. Consider a number as shown above, this series converges at, a number will converge evenly on the basis of Weierstrass, since 1 and numerical series converges. Consequently, for any x\u003e 1, the amount of this series is continuous. Comment. The function is called the Rome function on (this function plays a large role in the theory of numbers). Theorem 4 (about the killed integration of the functional series). Let all members of the Fn (x) of the series are continuous, and the series converges uniformly on the segment [a, b] to the function s (x). Then equality is true due to the continuity of functions f "(x) and the uniform convergence of this series on the segment [A, 6], its sum 5 (g) is continuous and, therefore, integrated on. Consider the difference from the uniform convergence of the series on [O, B] it follows that for any E\u003e 0 there is a number N (E)\u003e 0 such that for all numbers P\u003e N (E) and for all x € [A, 6] Inequality will be performed if the FN range (0 is not evenly converging, it is generally speaking, it is impossible to integrate reactive, i.e. Theorem 5 (about the kinded differentiation of the functional series). Let all members of the converging series 00 have continuous derivatives and a number of compiled from These derivatives are evenly converged on the segment [a, b]. Then, at any point, the equality is true. This series can be differentiated. m We put two any points. Then, by virtue of Theorem 4, we will have the function O- (x) continuous As a sum of a uniformly converging series of continuous functions. Therefore, differentiating equality we obtain exercises. Find areas of the convergence of data of functional series: Using the sign of Weierstrass, prove the uniform convergence of these functional series at the specified intervals:
Functional near called a formally recorded expression
u.1 (x.) + u.2 (x.) + u.3 (x.) + ... + u.n ( x.) + ... , (1)
where u.1 (x.), u.2 (x.), u.3 (x.), ..., u.n ( x.), ... - Sequence of functions from an independent variable x..
Abbreviated recording of a functional series with a sigma :.
Examples of functional series can serve :
(2)
(3)
Giving an independent variable x. Some meaning x.0 and substituting it into the functional series (1), we get a numerical series
u.1 (x.0 ) + u.2 (x.0 ) + u.3 (x.0 ) + ... + u.n ( x.0 ) + ...
If the numeric number obtained converges, they say that the functional series (1) converges when x. = x.0 ; If he diverges, what they say that the row (1) diverges when x. = x.0 .
Example 1. Explore the convergence of the functional series (2) at values x. \u003d 1 I. x. = - 1
.
Decision. For x. \u003d 1 get a numeric row
which converges on the basis of the leiby. For x. \u003d - 1 get a numeric row
,
which is diverged as a product of a diverging harmonic series on - 1. So, a number (2) converges when x. \u003d 1 and diverge x. = - 1 .
If such an inspection on the convergence of the functional series (1) is carried out with respect to all values \u200b\u200bof an independent variable from the field of determining its members, then the points of this region are broken into two sets: with values x.taken in one of them, the row (1) converges, and in the other - diverge.
Many values \u200b\u200bof an independent variable in which the functional series converges is called it region convergence .
Example 2. Find the region of the convergence of the functional series
Decision. Members of a number are defined on the entire numerical direct and form a geometric progression with the denominator q. \u003d SIN x. . Therefore, the series converges if
and diverges if
(values \u200b\u200bare impossible). But at the values \u200b\u200band with other values x.. Consequently, the series converges at all values. x., Besides . The region of its convergence is the whole number straight, with the exception of these points.
Example 3. Find the region of the convergence of the functional series
Decision. The members of the row form a geometric progression with the denominator q.\u003d LN. x. . Therefore, the series converges if, or, from where. This is the region of convergence of this series.
Example 4. Investigate the convergence of the functional series
Decision. Take an arbitrary value. In this case, we get a numerical series
(*)
Find the limit of his common member
Consequently, the series (*) dispels with an arbitrarily selected, i.e. With any meaning x.. The region of its convergence is an empty set.
Uniform convergence of the functional series and its properties
Let us turn to the concept uniform convergence of functional series . Let be s.(x.) - the amount of this series, and s.n ( x.) - Amount n. first members of this series. Functional series u.1 (x.) + u.2 (x.) + u.3 (x.) + ... + u.n ( x.) + ... called evenly converging on the segment [ a., b.] if for any small number ε \u003e 0 There is such a number. N. that at all n. ≥ N. Inequality will be performed
|s.(x.) − s.n ( x.)| < ε
for anyone x. From the segment [ a., b.] .
The above property can be geometrically illustrated as follows.
Consider a graph of the function y. = s.(x.) . Build near this curve band wide 2 ε n., that is, we build curves y. = s.(x.) + ε n. and y. = s.(x.) − ε n. (The figure below they are green).
Then at any ε n. Schedule function s.n ( x.) It will lie entirely in the band under consideration. In the same band, graphics of all subsequent partial sums will be lying.
Any similar functional series, which does not have the sign described above - unevenly moving.
Consider another property of uniformly convergent function series:
the sum of a number of continuous functions, evenly converging on some segment [ a., b.], there is a function continuous on this segment.
Example 5. Determine whether the amount of functional series is continuous
Decision. Find the amount n. The first members of this series:
If a x. \u003e 0, then
,
if a x. < 0 , то
if a x. \u003d 0, then
And therefore .
Our study showed that the sum of this series is a discontinuous function. Its graph is shown in the figure below.
Sign of Weierstrass Uniform Convergence of Functional Rows
To the sign of Weierstrass, come through the concept the majority of functional rows . Functional series
u.1 (x.) + u.2 (x.) + u.3 (x.) + ... + u.n ( x.) + ...
