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Functional series of the theory. Functional series and convergence: uniform and uneven. Power Rows: Basic Concepts, Abel Theorem

13.07.2020

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.) + ...

- 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, even joined students are insured against them, and therefore this seemingly seemingly additional lesson You should consider 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.

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:

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 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. Prove uniform convergence 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 converged, 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, evenly converges on the segment [a, b]. Then, at any point, equality that is, this series can be differentiated. M We take two any points. Then, by virtue of Theorem 4, we will have a function of O- (X) continuously as the sum of a uniformly converging series of continuous functions. Therefore, the 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 rows at the specified intervals:

Functional series. Power rows.
Region of the convergence of the row

Laughter for no reason - a sign of Dalamber

So the hour of functional series struck. For the successful development of the topic, and, in particular, this lesson, you need to understand well in conventional numeric rows. It should be good to understand what a number is to be able to apply signs of comparison for a study of a number of convergence. Thus, if you just started studying the topic or are a kettle in the highest mathematics, necessarythree lessons to work consistently: Rows for teapots, Sign of Dalamber. Signs of Cauchy and Aligning rows. Sign Leibnitsa. Necessarily all three! If there is elementary knowledge and skills to solve problems with numeric rows, it will be quite simple to cope with the functional rows, since the new material is not very much.

In this lesson, we will consider the concept of a functional series (which is generally this), we will get acquainted with power rows, which are found in 90% of practical tasks, and learn to solve a common model task for finding the radius of convergence, the convergence interval and the regions of the convergence of the power series. Next, I recommend to consider the material about decomposition of functions in power rows, and "ambulance" beginner will be provided. Having abandoned a little, go to the next level:

Also in the section of the functional series there are their numerous applications to approximate calculationsAnd some mansion goes from Fourier, which in the educational literature, as a rule, a separate chapter stands out. I have only one article, but it is long and many, many additional examples!

So, the guidelines are arranged, went:

The concept of functional series and power series

If the limit is infinityThe algorithm of the decision also finishes its work, and we give the final answer of the task: "A number converges at" (or with either "). See the case number 3 of the previous paragraph.

If the limit is not zero and not infinity, We have the most common case No. 1 in practice - a series converges at some interval.

In this case, the limit is equal. How to find a row convergence interval? Make up inequality:

IN Any task of this type In the left part of inequality should be the result of calculating the limit, and in the right part of inequality - strictly unit. I will not explain why this is such inequality and why on the right one. Lessons are practical, and it is already very good that some theorems have become clearer at my stories. Some theorems have become clearer.

Technique of working with a module and solutions of double inequalities were considered in detail in the first year in the article Function definition areaBut for convenience I will try to comment in the most detail all the actions. Reveal the inequality with a module for school rules . In this case:

Half the path behind.

At the second stage, it is necessary to investigate the convergence of the row at the ends of the interval found.

First, take the left end of the interval and substitute it in our power row:

For

A numerical series is obtained, and we need to explore it for convergence (already familiar to the task of the previous lessons).

1) A number is alkaline.
2) - Members of a number decrease in the module. At the same time, each next member of a series of module is less than the previous one: So decrease in monotonous.
Conclusion: a series converges.

With the help of a series composed of modules, find out exactly how:
- converges ("reference" row from the family of a generalized harmonic series).

Thus, the resulting numerical series converges absolutely.

for - converges.

! Remind that any convergent positive series is also completely converging.

Thus, the power series converges, and absolutely, at both ends of the interval found.

Answer: The region of convergence of the studied power series:

Has the right to life and other decoration of the answer: the series converges if

Sometimes in the condition of the task, you need to specify the radius of convergence. Obviously, in the considered example.

Example 2.

Find the convergence area of \u200b\u200bthe power row

Decision: The interval of the convergence of the row will find via Sign of Dalamber (but not on the basis of! - for the functional series there is no such trait):

A number converges as

Left We need to leave only, Therefore, we multiply both parts of inequality by 3:

- A number is a sign.
– - 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: a series converges.

We investigate it on the nature of convergence:

Compare this series with divergent.
We use a marking sign of comparison:

A finite number is obtained, different from zero, it means that a row diverges along with nearby.

Thus, the series converges conditionally.

2) for - Divorces (proven).

Answer: The region of the convergence of the studied power series :. With a series converge converge.

In the considered example, the convergence area of \u200b\u200bthe power series is semi-interval, and at all points of the interval of the power row it converges absolutely, and at the point, as it turned out - Conditional.

Example 3.

Find the convergence interval of the power series and explore his convergence at the ends of the interval found

This is an example for an independent solution.

Consider a couple of examples that are rare, but are found.

Example 4.

Find the region of the series of the row:

Decision: Using the sign of the Dalamber, we will find the convergence interval of this series:

(1) Compile the ratio of the next member of the series to the previous one.

(2) Get rid of four-story fractions.

(3) Cubes and on the rule of action with degrees are summarized for a single degree. In the numerator, cunning decay the degree, i.e. Unlock in such a way as to reduce the fraction on the next step. Factorials describe in detail.

(4) Under the cube reinforce the numerator to the denominator, indicating that. In the fraction cut everything that can be reduced. The multiplier we take out the limit for the sign, it can be discussed, because it does not have anything that depends on the "dynamic" variable "En". Please note that the module sign is not drawn - for the reason that it takes non-negative values \u200b\u200bfor any "X".

The limit received zero, and therefore, it is possible to give the final answer:

Answer: A number converges as

And first, it seemed that this series with a "terrible stuffing" would be difficult to solve. Zero or infinity in the limit is almost a gift, because the solution is noticeably reduced!

Example 5.

Find a region of the convergence of the row

This is an example for an independent solution. Be careful;-) Complete solution The answer at the end of the lesson.

Consider some more examples containing novelty elements in terms of using technical techniques.

Example 6.

Find the interval of the convergence of a series and explore its convergence at the ends of the interval found

Decision: The total member of the power series includes a multiplier that provides a smoke. The solution algorithm is fully saved, but when you draw up the limit, we ignore (do not write) this multiplier, since the module destroys all the "minuses".

The interval of the convergence of the row will find with the help of a sign of Dalamber:

We compile standard inequality:
A number converges as
Left We need to leave only module, Therefore, we multiply both parts of inequality on 5:

Now reveal the module already familiar to:

In the middle of double inequality, you need to leave only "X", for these purposes, from each part of inequality, we subtract 2:

- Interval convergence of the studied power series.

We investigate the convergence of the row at the ends of the interval found:

1) substitute the value in our power row :

Be extremely attentive, the multiplier does not provide an alignment, with any natural "en." The resulting minus we endure outside the row and forget about it, since it (like any constant factor) does not affect the convergence or divergence of the numerical series.

Notice againthat in the course of the substitution of the value in the general member of the power series, the multiplier decreased. If this had not happened, it would mean that we either incorrectly calculated the limit, or the module was incorrectly discovered.

So, it is required to explore the convergence of a numerical series. Here it is easiest to use a marginal sign of comparison and compare this series with a divergent harmonic nearby. But, honestly, the marginal sign of comparison to horror I'm tired, so I will make some diversity in the decision.

So, the series converges when

Multiply both parts of inequality by 9:

Remove the root from both parts, while remembering the old school joke:

Reveal the module:

and add to all parts one:

- Interval convergence of the studied power series.

We investigate the convergence of the power row at the ends of the interval found:

1) If, then the following numeric number is obtained:

The multiplier disappeared without a trace, because with any natural meaning "En".

4.1. Functional series: Basic concepts, convergence region

Definition 1.. A number, whose members are functions of one or
several independent variables defined on some set is called functional near.

Consider the functional series, whose members are functions of one independent variable h.. The amount of first n. The members of the series are partial sum of this functional series. General Member There is a function OT h.defined in some area. Consider the functional series at the point. . If the corresponding numeric row converges, i.e. There is a limit of partial sums of this series
(Where - amount of numerical row), then the point is called convergence point Functional series . If numerical series diverged the point is called point of divergence Functional series.

Definition 2.. Region convergence Functional series called many of all such values h., in which the functional series converges. The region of convergence consisting of all points of convergence is indicated . Note that R.

Functional series converges in the area if for any It converges as a numeric row, while its sum will be some feature. . This is the so-called limit function sequences : .

How to find the region of the convergence of the functional series ? You can use a sign similar to Dalamber. For row Make up and consider the limit for fixed h.:
. Then It is a solution to inequality and solving the equation (we take only those solutions of the equation in
which corresponding numeric rows converge).

Example 1.. Find the area of \u200b\u200bthe convergence of the row.

Decision. Denote , . We will comply and calculate the limit, then the region of the convergence of the series is determined by inequality and equation . We investigate further convergence of the initial series at points that are roots of the equation:

what if , then the spending row ;

b) if , , then ranks convergence (by

symptom leibher, example 1, lecture 3, section. 3.1).

Thus, the region of convergence A number of: .

4.2. Power Rows: Basic Concepts, Abel Theorem

Consider a special case of a functional series, the so-called power row where
.

Definition 3.. Power nearby called functional series of species,

where - constant numbers called coefficients of the rank.

The power series is the "endless polynomial", located in increasing degrees . Any numerical series is an
a private case of a power series when .

Consider a private case of a power series when :
. Find out what kind of
Region of the convergence of this series .

Theorem 1 (Abel Theorem). 1) if a power row converges at point then it is absolutely converged at all h.for which the inequality is fair .

2) if the power row is diverged when then he dispersed at all h., for which .

Evidence. 1) under the condition of the power series converges at the point ,

i.e. it converges a numeric row

(1)

and according to the necessary sign of convergence, its general member seeks 0, i.e. . Consequently, there is such a number that all members of a number are limited to this number:
.

Consider anyone now h., for which and make up a number of absolute values \u200b\u200b:.
We write this row in another form: as , then (2).

From inequality
We get, i.e. row

it consists of members who are more relevant members of the series (2). Row represents a series of geometric progression with denominator Moreover , as . Consequently, a number (2) converges when . Thus, the power row Absolutely converges.

2) Let a series Different to , in other words,

different with numerical series . We prove that for any h. () A row diverges. The proof is leading from the opposite. Let with some

fixed ( ) a series converges, then it converges at all (See the first part of this theorem), in particular, with, which contradicts the condition 2) of Theorem 1. The theorem is proved.

Corollary. Abel Theorem allows you to judge the location of the point of convergence of the power row. If point It is a point of convergence of a power row, the interval filled with convergence points; If the divergence point is point T.
Infinite intervals filled with divergence points (Fig. 1).

Fig. 1. Intervals of convergence and divergence of a number

It can be shown that there is such a number that at all
Power row absolutely converges, and when - Divorces. We will assume that if a series converges only at one point 0, then , and if a series converges at all T. .

Definition 4.. Convergence interval Power Row called such an interval that at all This series converges and moreover absolutely, and for all h.lying outside of this interval, a row diverges. Number R. called radius of convergence Power series.

Comment. At the ends of the interval The question of convergence or divergence of the power series is solved separately for each specific series.

Let us show one of the ways to determine the interval and the radius of the convergence of the power series.

Consider a power group And denote .

We will make a number of the absolute values \u200b\u200bof its members:

and we apply to him a sign of Dalamber.

Let exist

On the basis of Dalamber a number converges if , and diverge, if . Hence the series converges at, then the convergence interval: . With a number diverged because .
Using the designation , I get a formula for determining the radius of the convergence of the power series:

where - Ceffs of the power series.

If it turns out that the limit , then believe .

To determine the interval and radius of the convergence of the power series, you can also use the radical sign of Cauchy, the radius of the convergence of the row is determined from the ratio .

Definition 5.. Generalized Power Near called a number of species

. It is also called near the degrees .
For such a number, the convergence interval has the form: where - Radius of convergence.

We will show how the convergence radius is for a generalized power series.

those. where .

If a T. , and convergence R; if a T. and the region of convergence .

Example 2.. Find a region of the convergence of the row .

Decision. Denote . Let's make a limit

We solve the inequality: , , therefore, the interval

convergence has the form: Moreover R. \u003d 5. Additionally, we investigate the ends of the convergence interval:
but) , , get a number who diverges;
b) , , get a number that converges
conditionally. Thus, the region of convergence: , .