Series with complex numbers. Series with complex terms. Power complex series

463. What is the geometric meaning of the inequalities: 1) Rez > 0; 2) im z< 0 ?

2. Series with complex terms. Consider the sequence of complex numbers z 1 , z 2 , z 3 , ..., where z p \u003d x p + iy p (n \u003d 1, 2, 3, ...). constant number c = a + bi called limit sequences z 1 , z 2 , z 3 , ..., if for any arbitrarily small number δ>0 there is a number N, what is the meaning z p with numbers n > N satisfy the inequality \z n-With\< δ . In this case, write .

A necessary and sufficient condition for the existence of a limit of a sequence of complex numbers is as follows: the number c=a+bi is the limit of the sequence of complex numbers x 1 + iy 1, x 2 + iy 2, x 3 + iy 3, ... if and only if , .

(1)

whose members are complex numbers is called converging, If nth partial sum of the series S n for n → ∞ tends to a certain end limit. Otherwise, series (1) is called divergent.

Series (1) converges if and only if series with real terms converge

(2) Investigate the convergence of the series This series, whose terms form an infinitely decreasing geometric progression, converges; therefore, the given series with complex terms converges absolutely. ^

474. Find the area of ​​convergence of a series

19.4.1. Numerical series with complex terms. All basic definitions of convergence, properties of convergent series, convergence criteria for complex series do not differ in any way from the real case.

19.4.1.1. Basic definitions. Let an infinite sequence of complex numbers be given z 1 , z 2 , z 3 , …, z n , … .The real part of the number z n we will denote a n , imaginary - b n

(those. z n = a n + i b n , n = 1, 2, 3, …).

Number series- type record.

Partialamountsrow: S 1 = z 1 , S 2 = z 1 + z 2 , S 3 = z 1 + z 2 + z 3 , S 4 = z 1 + z 2 + z 3 + z 4 , …,

S n = z 1 + z 2 + z 3 + … + z n , …

Definition. If there is a limit S sequences of partial sums of the series for
, which is a proper complex number, then the series is said to converge; number S called the sum of the series and write S = z 1 + z 2 + z 3 + … + z n + ... or
.

Find the real and imaginary parts of the partial sums:

S n = z 1 + z 2 + z 3 + … + z n = (a 1 + i b 1) + (a 2 + i b 2) + (a 3 + i b 3) + … + (a n + i b n ) = (a 1 + a 2 + a 3 +…+ a n ) +

Where symbols And the real and imaginary parts of the partial sum are indicated. A numerical sequence converges if and only if the sequences composed of its real and imaginary parts converge. Thus, a series with complex terms converges if and only if the series formed by its real and imaginary parts converge. One of the methods for studying the convergence of series with complex terms is based on this assertion.

Example. Investigate for convergence series .

Let's write out several values ​​of the expression : further values ​​are periodically repeated. A number of real parts: ; series of imaginary parts ; both series converge (conditionally), so the original series converges.

19.4.1.2. Absolute convergence.

Definition. Row called absolutely convergent if the series converges
, composed of the absolute values ​​of its members.

Just as for numerical real series with arbitrary terms, it is easy to prove that if the series converges
, then the series necessarily converges (
, so the series formed by the real and imaginary parts of the series , converge absolutely). If the row converges, and the series
diverges, then the series is called conditionally convergent.

Row
is a series with non-negative terms, therefore, to study its convergence, all known signs (from comparison theorems to the Cauchy integral criterion) can be used.

Example. Investigate for convergence series
.

Let's make a series of modules ():
. This series converges (the Cauchy test
), so the original series converges absolutely.

19.4. 1 . 3 . Properties of convergent series. For convergent series with complex terms, all properties of series with real terms are true:

A necessary criterion for the convergence of a series. The common term of the convergent series tends to zero as
.

If the series converges , then any of its remainder converges. Conversely, if any remainder of the series converges, then the series itself converges.

If the series converges, then the sum of its remainder aftern -th term tends to zero at
.

If all terms of a convergent series are multiplied by the same numberWith , then the convergence of the series is preserved, and the sum is multiplied byWith .

Convergent rows (A ) And (IN ) can be added and subtracted term by term; the resulting series will also converge, and its sum is equal to
.

If the terms of the convergent series are grouped arbitrarily and a new series is made up of the sums of the terms in each pair of parentheses, then this new series will also converge, and its sum will be equal to the sum of the original series.

If a series converges absolutely, then for any permutation of its terms, the convergence is preserved and the sum does not change.

If the rows (A ) And (IN ) converge absolutely to their sums
And
, then their product for an arbitrary order of terms also converges absolutely, and its sum is equal to
.

View Symbol W 1 + W 2 +…+ W n +…= (1), Where W n = u n + i· v n (n = 1, 2, …) complex numbers (sequences of complex numbers) are called near complex numbers.

Numbers W n (n = 1, 2, …) called members of a number, member W n called common member of the series.

Type numbers S n = W 1 + W 2 +…+ W n (2) (n = 1, 2, …) , are called partial sums of the series (1).

Finite or infinite limit S sequences S n called the sum of this series.

If limit S is finite, then the series is called converging, if the limit is infinite, or does not exist at all, then the series divergent.

If S the sum of series (1), then we write
.

Let
, A
. Obviously σ n = u 1 + u 2 +…+ u n , τ n = v 1 + v 2 +…+ v n. How do we know equality
(S of course) is equivalent to two equalities
And
. Therefore, the convergence of series (1) is equivalent to the convergence of two real series: And . Therefore, the basic properties of convergent numerical series extend to convergent complex series.

For example, for complex series, the Cauchy criterion is valid: series (1) converges if and only if for any

, that for alln > Nand anyp= 1, 2, … the inequality.

This criterion directly implies the necessary criterion for the convergence of the series: for series (1) to converge it is necessary and sufficient that its common termW n → 0 .

The following properties of convergent series are valid: if the rows And converge to their sumsSAndd, then the rows
And
converge respectively to the sumsS ± dand λS .

Absolutely convergent series of complex numbers.

Series of complex numbers (1) called absolutely convergent if the series converges
(2).

Theorem.

Every absolutely convergent series (1) of complex numbers converges.

Proof.

Obviously, it suffices for us to establish that for the series (1) the conditions of the Cauchy criterion for the convergence of the series are satisfied. Take any
. Due to the absolute convergence of series (1), series (2) converges. Therefore, for the chosen

, that for any n > N And p=1.2,… the inequality will be satisfied
, But

, and even more so, the inequality
for any n > N And p=1,2,… Consequently, for the series (1), the conditions of the Cauchy criterion for the convergence of the complex series are satisfied. Therefore, series (1) converges. The theorem is correct.

Theorem.

In order for a series of complex numbers (1) is absolutely convergent, it is necessary and sufficient that the real series converge absolutely (3) and (4) , whereW n = u n + i· v n (n = 1, 2,…).

Proof,

relies on the following obvious inequalities

(5)

Necessity. Let the series (1) converge absolutely, let us show that the series (3) and (4) absolutely converge, i.e., the series converge
And
(6). From the absolute convergence of series (1) it follows that series (2)
converges, then, by virtue of the left side of inequality (5), series (6) will converge, i.e., series (3) and (4) converge absolutely.

Adequacy. Let series (3) and (4) converge absolutely, let us show that series (1) also converges absolutely, i.e., that series (2) converges. It follows from the absolute convergence of series (3) and (4) that series (6) converge, so the series also converges
. Therefore, due to the right side of inequality (5), series (2) converges, i.e., series (1) converges absolutely.

So, the absolute convergence of the complex series (1) is equivalent to the absolute convergence of the real number series (3) and (4). Therefore, absolutely convergent complex series are subject to all the basic properties of real absolutely convergent numerical series. In particular, for an absolutely convergent complex series, the theorem on the permutation of its terms is valid, i.e., swapping terms in an absolutely convergent series does not affect the sum of the series. To establish the absolute convergence of a complex series, any criterion for the convergence of a positive series can be used.

Cauchy sign.

Let series (1) have a limit
, then ifq < 1 , то ряд (1) абсолютно сходится, если q>1, then series (1) diverges.

Sign of d'Alembert.

If for the series (1) of complex numbers there is a limit
, then atq < 1 этот ряд абсолютно сходится, а если q> 1, then the series diverges.

Example.

Investigate for absolute convergence series
, Here
.

Let's find
. Obviously
=
=
. Therefore, the series is absolutely convergent.

Absolutely convergent series can be multiplied. The product of an absolutely convergent and a convergent series converges. The product of two convergents can diverge.

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1 8 Complex number series S of the sequence (S) is called the sum of the series (46) The series a k is called the -th remainder of the series (46) For a convergent k series S S r and lm r, those ε > N, N: r< ε Для сходящегося ряда (46) необходимым и достаточным признаком его сходимости является критерий Коши: ряд (46) сходится тогда и только тогда, если ε >, N, N: a< ε p k k Необходимым условием сходимости ряда (46) является требование lm a Действительно, из сходимости ряда (46) следует, согласно критерию Коши, что ε >, N > that for p, it follows that S S< ε Если сходится ряд ak k a (47) с действительными положительными членами, то очевидно, сходится и ряд (46), который в этом случае называется абсолютно сходящимся А для ряда (47) уже можно применить признаки Даламбера и Коши: ряд (47) сходится, если, начиная с a некоторого номера N соотношение l < a, N значит, сходится абсолютно ряд (46)), если a q <, N k ; и ряд (47) сходится (а,

2 9 Function series and their properties Uniform convergence Weierstrass theorem Let an infinite sequence of single-valued functions ((Z)) be defined in a domain G of the complex plane Z ((Z)) An expression of the form U U (48) will be called a functional series Series (48) is called convergent in the domain G if Z G the number series corresponding to it converges If the series (48) converges in the region G, then in this region it is possible to define a single-valued function, the value of which at each point of the region G is equal to the sum of the corresponding number series (48) in the region G Then G, > k () U k()< ε Заметим, что в общем случае N зависит и от ε и от Определение Если ε >, N(ε), N(ε): ε, N (ε,), N(ε,) : the domain G k U k< ε G, то ряд (48) называется равномерно сходящимся в k k Если остаток ряда обозначить r U, то тогда условие равномерной сходимости ряда (48) можем записать в виде: r < ε, N(ε), G Достаточным признаком равномерной сходимости ряда (48) является признак Вейерштрасса: Если всюду в области G члены функционального ряда (48) могут быть мажорированы членами некоторого абсолютно сходящегося числового ряда a, те

3 a U, G, (49) then the series (48) converges uniformly N Indeed, since the series a converges, then >< ε, U U a < ε при N, что и доказывает равномерную k k k k k k сходимость ряда (48) в области G Приведем некоторые теоремы о равномерно сходящихся рядах Они доказываются совершенно также, как соответствующие теоремы вещественного анализа и поэтому приведем их без доказательства Теорема 5 Если функции U непрерывны в области G, а ряд U сходится в этой области равномерно к функции, то также непрерывна в G Теорема 6 Если ряд (48) непрерывных функций U сходится равномерно в области G к функции, то интеграл от этой функции по любой кусочногладкой кривой, целиком лежащей в области G, можно вычислить путем почленного интегрирования ряда (48), те Теорема 7 Если члены d U d U сходящегося в области G ряда U имеют непрерывные производные в этой области и ряд U равномерно сходится в G, то данный ряд U можно почленно дифференцировать в области G, причем U U, где U - сумма ряда

4 For functional series in complex analysis, there is the Weierstrass theorem, which allows us to significantly strengthen the theorem on the possibility of term-by-term differentiation of a functional series, known from real analysis. Before stating and proving it, we note that the series U, which converges uniformly along the line l, remains uniformly after multiplying all its terms by the function ϕ bounded on l Indeed, let the inequality ϕ () be satisfied on the line l< M Тогда для остатков ρ и r рядов U и U ϕ справедливо соотношение ϕ U U r < M r ρ ϕ ε и, тк N, >N:r< и одновременно с ним ρ < ε, то этим доказано M высказанное утверждение Если сумма данного ряда есть S, то сумма ряда, полученного после умножения на ϕ, очевидно будет ϕ S Теорема 8 (Вейерштрасса) Если члены ряда - аналитические в некоторой области G функции и этот ряд сходится в области G равномерно, то его сумма также является функцией аналитической в G, ряд можно почленно дифференцировать и полученный ряд F равномерно сходится к () F Выберем любую внутреннюю точку области G и построим круг столь малого радиуса с центром в этой точке, чтобы он целиком лежал внутри G (рис) В силу равномерной сходимости данного ряда в G, G ρ Рис он, в частности, равномерно сходится на окружности этого круга Пусть - любая точка на Умножим ряд () () () () () (5) на величину Полученный ряд

5 also converges uniformly to its sum () () () () (), since function (5) is limited to, because for points of this circle ρ is the radius of the circle (recall: - here is a constant) Then, according to the above, series (5) can be integrated term by term: () d () d () d d π π π π Due to the analyticity of the functions, the Cauchy formula can be applied to them, on the basis of which we obtain () d π, (5) and the sum of the series on the right in (5) is and, therefore, we obtain the equality π () d at the point Tk - any point of the domain G, then the first part of the theorem is proved. we obtain that the series converges uniformly, and its sum is equal to (k) (k)

6 series of the form where Power series Abel's theorem A very important case of general functional series are power series (), (53) - some complex numbers, and - a fixed point of the complex plane. series, the general theorems of the previous sections can be applied. As it was established in them, many properties are a consequence of uniform convergence. To determine the region of convergence of the power series (53), the following theorem turns out to be essential. Theorem 9 (Abel) If the power series (53) converges at some point, then it converges absolutely and at any point that satisfies the condition, moreover, in the circle< ρ, радиусом ρ, меньшим < сходится равномерно, ряд Δ Выберем произвольную точку, удовлетворяющую условию < Обозначим q сходимости ряда следовательно M >, that M, q< В силу необходимого признака его члены стремятся к нулю при, отсюда () M M q M, Тогда, где q < (54) Ряд справа в (54) бесконечно убывающая геометрическая прогрессия со знаменателем q < Тогда из (54) следует сходимость и рассматриваемого ряда

7p< достаточно в силу признака Вейерштрасса (53) В круге построить сходящийся числовой ряд, можорирующий данный ряд в рассматриваемой области Очевидно, таковым является ряд ρ M, также представляющий собой сумму бесконечной геометрической прогрессии со знаменателем, меньшим единицы Из теоремы Абеля можно вывести ряд следствий, в известной мере аналогичным следствиям из теоремы Абеля в теории степенных рядов вещественного анализа Если степенной ряд (53) расходится в некоторой точке, то он расходится и во всех точках, удовлетворяющих неравенству >The exact upper limit of the distances from the point to the point at which the series (53) converges is called the radius of convergence of the power series, and the region<, называется кругом сходимости степенного ряда В точках границы ряд может как сходиться так и расходиться Пример Найти область сходимости ряда Δ Находим радиус сходимости по признаку Даламбера lm () и наш ряд сходится в круге < При <, те, исследуется особо В этом случае и, значит, областью абсолютной сходимости является

8p< В круге любого радиуса ρ, меньшего чем радиус сходимости, степенной ряд (53) сходится равномерно 3 Внутри круга сходимости степенной ряд сходится к аналитической функции В самом деле, члены ряда u есть функции, аналитические на всей плоскости Z, ряд сходится в любой замкнутой подобласти круга сходимости Тогда по теореме Вейерштрасса сумма ряда есть аналитическая функция 4 Степенной ряд внутри круга сходимости можно почленно интегрировать и дифференцировать любое число раз, причем радиус сходимости полученных рядов равен радиусу сходимости исходного ряда 5 Коэффициенты степенного ряда (53) находятся по формулам! () () (55) Доказательство этого факта приводится методами, аналогичными методам вещественного анализа Ряд Тейлора Теорема Тейлора Нули аналитических функций Итак степенной ряд внутри круга сходимости определяет некоторую аналитическую функцию Возникает вопрос: можно ли функции, аналитической внутри некоторого круга, сопоставить степенной ряд, сходящийся в этом круге к данной функции? < Теорема 9 (Тейлора) Функция, аналитическая внутри круга, может быть представлена в этом круге сходящимся степенным рядом, причем этот ряд определен однозначно

9 Let us choose an arbitrary point inside the circle ρ ρ< и построим окружность ρ точке радиусом < с центром в ρ (рис), содержащую точку внутри Такое построение возможно для любой точки внутри этого круга Так как < ρ, а внутри круга < Рис аналитична, то по формуле Коши имеем π ρ () d (56) Преобразуем подынтегральное выражение: (57) <, то < Так как Поэтому второй сомножитель справа в (57) можно представить как сумму степенного ряда (прогрессии), ту которая первый член есть, а знаменатель прогрессии есть Так как, те () () (58) ρ, то ряд (58) сходится равномерно по, так как он мажорируется сходящимся числовым рядом Подставляя (58) в (56) и интегрируя почленно, получаем ρ (< ρ)

10 Let us introduce the notation () d () ρ π () d () π ρ () and rewrite (59) as a power series converging at the chosen point: (59) (6) () (6) In formula (6), the neighborhood ρ can be replaced, by virtue of the Cauchy theorem, by any closed contour lying in the region< и содержащим точку внутри Так как - произвольная точка данной области, то отсюда следует, что ряд (6) сходится к круге ρ < этот ряд сходится равномерно Итак, функция всюду внутри круга < аналитическая внутри круга <, причем в разлагается в этом круге в сходящийся степенной ряд Коэффициенты разложения (6) на основании формулы Коши для производных аналитической функции имеет вид () d () π ρ () ()! (6) Для доказательства единственности разложения (6) допустим, что имеет еще место формула разложения (), (6)

11 where there would also be one coefficient<, поэтому на основании формулы (55) Ряд (6) сходящимся в круге () () (6) Тем самым единственность определения коэффициентов доказана Разложение функции, аналитической в круге! <, что совпадает с, в сходящийся степенной ряд (6), часто называется разложением Тейлора, а сам ряд (6) Рядом Тейлора Доказанная теорема устанавливает взаимнооднозначное соответствие между функцией, аналитической в окрестности некоторой точки и степенным рядом с центром в этой точке, это означает эквивалентность конкретной аналитической функции, как функции бесконечное число раз дифференцируемой и функцией, представимой в виде суммы степенного ряда G и Заметим, наконец, что, если функция является аналитической в области G - внутренняя точка, то радиус сходимости ряда Тейлора () () () этой функции не меньше расстояния от точки до! границы области G (имеется в виду ближайшее расстояние) Пример Разложить в ряд Тейлора по степеням Δ Эта функция является аналитической на всей комплексной плоскости за исключением точек, Поэтому в круге < функция может быть ± разложена в ряд Тейлора При условии < выражение рассматриваться как сумма бесконечно убывающей прогрессии может q, q < Поэтому

12 , < Пример 3 Найти разложение в ряд Тейлора в круге < Определение по формуле (6) здесь довольно затруднительно Поэтому, представим π Так как < и <, то, используя геометрическую, получаем q q, Используя показательную форму чисел и находим окончательно 4 s π (63) Тк расстояние от центра разложения до ближайших особых точек (те до границы аналитичности) есть, то радиус сходимости ряда (63) есть Рис X Y

13 4 4 3 Example<, 4 3 < Ближайшей к центру разложения особой точкой является точка, до которой расстояние равно, поэтому В заключение приведем основные разложения: e (<)!! 3! cos! 4 3 4! ; (<)! ; s () m 3 3! 5 5! m m m!! (<) ()! ; m(m)(m)! ; l 3 3 () 4 (<) Если для аналитической функции (), то точка называется нулем аналитической функции В этом случае разложение функции в ряд Тейлора в окрестности точки имеет вид () () тк () Если в разложении функции окрестности точки и, следовательно, разложение имеет вид, в ряд Тейлора в,

14 then the point () (), (64) is called the zero of the function If, then the zero is called a prime of order or multiplicity From the formulas for the coefficients of the Taylor series, we see that if the point is a zero of the order, then where () () can be rewritten in the form, but () () () [ () ] () ϕ, ϕ () (), () ϕ, and the circle of convergence of this series is obviously the same as that of the series (64) converse statement where Any function of the form is an integer, ϕ () and zero of order are zeros, and (±) Example 6 Find the order of zero for the function 8 s Expand the denominator in powers: 3 3! 8 5 5! ! 5! 3! 5 5! ϕ

15 5 ϕ, where ϕ, and ϕ and a point of the function 3!, so that the point 5! ϕ is analytic in is the zero of the 5th order for the original Laurent series and its convergence region Decomposition of an analytic function in a Laurent series Consider a series of the form () where is a fixed point of the complex plane, (65) are some complex numbers. Let us establish its area of ​​convergence To do this, we represent (65) in the form centered at a point of some radius, and in particular, it can be equal to zero or infinity Inside the circle of convergence, this series converges to some analytic function of a complex variable, those (),< (67)

16 To determine the region of convergence of a series of variables, setting () () Then this series will take the form let's make a replacement - an ordinary power series converging inside its circle of convergence to some analytical function ϕ () with a complex variable Let the radius of convergence of the resulting power series be r Then ϕ,< r Возвращаясь к старой переменной и полагая ϕ () () (68), >r It follows that the region of convergence of the series, the region external to the circle r, we obtain (69) () is<, то существует общая область сходимости этих рядов круговое кольцо r < <, в которой ряд (65) сходится к аналитической функции (), r < < (7) Так как ряды (67) и (68) являются обычными степенными рядами, то в указанной области функция обладает всеми свойствами суммы степенного ряда Это означает, что ряд Лорана сходится внутри своего кольца сходимости к некоторой функции, аналитической в данном кольце

17 If r >, then the series (67) and (68) do not have a common region of convergence, thus in this case the series (65) does not converge anywhere to any function Note that the series is a regular part of the series (7), and Example 7 Expand - the main part of the row (65) () a)< < ; б) >; V)< < называется правильной частью или в ряд Лорана в кольцах: Во всех кольцах функция регулярна (аналитична) и поэтому может быть представлена рядом Лорана (доказательство этого факта в следующем пункте) Перепишем функцию в виде а) Так как <, то второе слагаемое есть сумма убывающей геометрической прогрессии Поэтому () Здесь главная часть состоит из одного слагаемого < б) в этом случае, поэтому () 3

18 This expansion lacks a regular part< в) Для случая < функцию также надо привести к сходящейся геометрической прогрессии, но со знаменателем Это даст: 3 Заметим, что в главной части этого разложения присутствует одно слагаемое Возникает вопрос: можно ли функции аналитической в некотором круговом кольце, сопоставить ряд Лорана, сходящийся к этой функции в данном кольце? На этот вопрос отвечает Теорема Функция, аналитическая в круговом кольце < <, однозначно представляется в этом кольце сходящимся рядом Лорана дробь На Рис 3 Δ Зафиксируем произвольную точку внутри данного кольца и контурами окружности и с центром в, радиусы которых удовлетворяют условиям < < < < < (рис 3) Согласно формуле Коши для многосвязной области имеем π () d () выполняется неравенство q, можно представить в виде d (7) Поэтому

19 Let us carry out term-by-term integration in (7), which is possible due to the uniform convergence of the series in, we obtain d π, (7) where d π, (73) (7) we will have π π d d, (for d), (74) where d π (75) Changing the direction of integration in (75), we obtain

20 π () () d ()() d π, > (76) Due to the analyticity of the integrands in (73) and (76) in the circular ring< < в соответствии с теоремой Коши значения интегралов не изменятся при произвольной деформации контуров интегрирования в области аналитичности Это позволяет объединить формулы (73) и (76): π () d (), ±, ±, (77) где - произвольный замкнутый контур, лежащий в указанном кольце и содержащий точку внутри Возвратимся теперь к формуле (7), получим где коэффициенты () (), (78) () для всех определяются однообразной формулой (77) Так как - люба точка кольца < <, то отсюда следует, что ряд (78) сходится к внутри данного кольца причем в замкнутом кольце < < ряд сходится к равномерно Доказательство единственности разложения (78) опускаем Из полученных результатов следует, что областью сходимости ряда (78) Лорана является круговое кольцо < <, на границах которого имеется хотя бы по одной особой точке аналитической функции ряд (78), к которой сходится Замечание Формула (77) для определения коэффициентов разложения в ряд Лорана (78) не всегда практически удобна Поэтому часто прибегают к разложению рациональной дроби на простейшие с использованием геометрической прогрессии, а также используют разложение в ряд Тейлора элементарные функции Приведем примеры

21 Example 8 Expand the Laurent series (those in powers) Y in the vicinity of the point ()() in Δ In this case, we will construct two circular rings centered at the point (Fig. 4): a) a circle “without a center”< < ; Рис 4 X б) внешность круга >It is analytic in each of these rings, and has singular points on the boundaries. Let us expand the function in powers in each of these regions)< < ; ; [ () () () ] () < Этот ряд сходится, так как Так что ()() () () () (), ; >) Here we have 3, () () () () () is a convergent series, since<

22 s As a result ()() () () those, 3, 3 Example 9 Expand the function Δ in a Laurent series in the vicinity of the point We have:, s s s cos cos s s! cos 4 () () 3 4! 3! () 5! () (scos)!! 5

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Standard methods, but reached a dead end with another example.

What is the difficulty and where can there be a snag? Let's put aside the soapy rope, calmly analyze the reasons and get acquainted with the practical methods of solution.

First and most important: in the vast majority of cases, to study the convergence of a series, it is necessary to apply some familiar method, but the common term of the series is filled with such tricky stuffing that it is not at all obvious what to do with it. And you go around in circles: the first sign does not work, the second does not work, the third, fourth, fifth method does not work, then the drafts are thrown aside and everything starts anew. This is usually due to a lack of experience or gaps in other sections of calculus. In particular, if running sequence limits and superficially disassembled function limits, then it will be difficult.

In other words, a person simply does not see the necessary solution due to a lack of knowledge or experience.

Sometimes “eclipse” is also to blame, when, for example, the necessary criterion for the convergence of the series is simply not fulfilled, but due to ignorance, inattention or negligence, this falls out of sight. And it turns out like in that bike where the professor of mathematics solved a children's problem with the help of wild recurrent sequences and number series =)

In the best traditions, immediately living examples: rows and their relatives - diverge, since in theory it is proved sequence limits. Most likely, in the first semester, you will be beaten out of your soul for a proof of 1-2-3 pages, but now it is quite enough to show that the necessary condition for the convergence of the series is not met, referring to known facts. Famous? If the student does not know that the root of the nth degree is an extremely powerful thing, then, say, the series put him in a rut. Although the solution is like two and two: , i.e. for obvious reasons, both series diverge. A modest comment “these limits have been proven in theory” (or even its absence at all) is quite enough for offset, after all, the calculations are quite heavy and they definitely do not belong to the section of numerical series.

And after studying the next examples, you will only be surprised at the brevity and transparency of many solutions:

Example 1

Investigate the convergence of a series

Solution: first of all, check the execution necessary criterion for convergence. This is not a formality, but a great chance to deal with the example of "little bloodshed".

"Inspection of the scene" suggests a divergent series (the case of a generalized harmonic series), but again the question arises, how to take into account the logarithm in the numerator?

Approximate examples of tasks at the end of the lesson.

It is not uncommon when you have to carry out a two-way (or even three-way) reasoning:

Example 6

Investigate the convergence of a series

Solution: first, carefully deal with the gibberish of the numerator. The sequence is limited: . Then:

Let's compare our series with the series . By virtue of the double inequality just obtained, for all "en" it will be true:

Now let's compare the series with the divergent harmonic series.

Fraction denominator less the denominator of the fraction, so the fraction itself – more fractions (write down the first few terms, if not clear). Thus, for any "en":

So, by comparison, the series diverges along with the harmonic series.

If we change the denominator a little: , then the first part of the reasoning will be similar: . But to prove the divergence of the series, only the limit test of comparison is already applicable, since the inequality is false.

The situation with converging series is “mirror”, that is, for example, for a series, both comparison criteria can be used (the inequality is true), and for a series, only the limiting criterion (the inequality is false).

We continue our safari through the wild, where a herd of graceful and succulent antelopes loomed on the horizon:

Example 7

Investigate the convergence of a series

Solution: the necessary convergence criterion is satisfied, and we again ask the classic question: what to do? Before us is something resembling a convergent series, however, there is no clear rule here - such associations are often deceptive.

Often, but not this time. By using Limit comparison criterion Let's compare our series with the convergent series . When calculating the limit, we use wonderful limit , where as infinitesimal stands:

converges together with next to .

Instead of using the standard artificial technique of multiplication and division by a "three", it was possible to initially compare with a convergent series.
But here a caveat is desirable that the constant-multiplier of the general term does not affect the convergence of the series. And just in this style the solution of the following example is designed:

Example 8

Investigate the convergence of a series

Sample at the end of the lesson.

Example 9

Investigate the convergence of a series

Solution: in the previous examples, we used the boundedness of the sine, but now this property is out of play. The denominator of a fraction of a higher order of growth than the numerator, so when the sine argument and the entire common term infinitely small. The necessary condition for convergence, as you understand, is satisfied, which does not allow us to shirk from work.

We will conduct reconnaissance: in accordance with remarkable equivalence , mentally discard the sine and get a series. Well, something like that….

Making a decision:

Let us compare the series under study with the divergent series . We use the limit comparison criterion:

Let us replace the infinitesimal with the equivalent one: for .

A finite number other than zero is obtained, which means that the series under study diverges along with the harmonic series.

Example 10

Investigate the convergence of a series

This is a do-it-yourself example.

For planning further actions in such examples, the mental rejection of the sine, arcsine, tangent, arctangent helps a lot. But remember, this possibility exists only when infinitesimal argument, not so long ago I came across a provocative series:

Example 11

Investigate the convergence of a series
.

Solution: it is useless to use the limitedness of the arc tangent here, and the equivalence does not work either. The output is surprisingly simple:


Study Series diverges, since the necessary criterion for the convergence of the series is not satisfied.

The second reason"Gag on the job" consists in a decent sophistication of the common member, which causes difficulties of a technical nature. Roughly speaking, if the series discussed above belong to the category of “figures you guess”, then these ones belong to the category of “you decide”. Actually, this is called complexity in the "usual" sense. Not everyone will correctly resolve several factorials, degrees, roots and other inhabitants of the savannah. Of course, factorials cause the most problems:

Example 12

Investigate the convergence of a series

How to raise a factorial to a power? Easily. According to the rule of operations with powers, it is necessary to raise each factor of the product to a power:

And, of course, attention and attention again, the d'Alembert sign itself works traditionally:

Thus, the series under study converges.

I remind you of a rational technique for eliminating uncertainty: when it is clear order of growth numerator and denominator - it is not at all necessary to suffer and open the brackets.

Example 13

Investigate the convergence of a series

The beast is very rare, but it is found, and it would be unfair to bypass it with a camera lens.

What is double exclamation point factorial? The factorial "winds" the product of positive even numbers:

Similarly, the factorial “winds up” the product of positive odd numbers:

Analyze what is the difference between

Example 14

Investigate the convergence of a series

And in this task, try not to get confused with the degrees, wonderful equivalences And wonderful limits.

Sample solutions and answers at the end of the lesson.

But the student gets to feed not only tigers - cunning leopards also track down their prey:

Example 15

Investigate the convergence of a series

Solution: the necessary criterion of convergence, the limiting criterion, the d'Alembert and Cauchy criteria disappear almost instantly. But worst of all, the feature with inequalities, which has repeatedly rescued us, is powerless. Indeed, comparison with a divergent series is impossible, since the inequality incorrect - the multiplier-logarithm only increases the denominator, reducing the fraction itself in relation to the fraction. And another global question: why are we initially sure that our series is bound to diverge and must be compared with some divergent series? Does he fit in at all?

Integral feature? Improper integral evokes a mournful mood. Now, if we had a row … then yes. Stop! This is how ideas are born. We make a decision in two steps:

1) First, we study the convergence of the series . We use integral feature:

Integrand continuous on

Thus, a number diverges together with the corresponding improper integral.

2) Compare our series with the divergent series . We use the limit comparison criterion:

A finite number other than zero is obtained, which means that the series under study diverges along with side by side .

And there is nothing unusual or creative in such a decision - that's how it should be decided!

I propose to independently draw up the following two-move:

Example 16

Investigate the convergence of a series

A student with some experience in most cases immediately sees whether the series converges or diverges, but it happens that a predator deftly disguises itself in the bushes:

Example 17

Investigate the convergence of a series

Solution: at first glance, it is not at all clear how this series behaves. And if we have fog in front of us, then it is logical to start with a rough check of the necessary condition for the convergence of the series. In order to eliminate uncertainty, we use an unsinkable multiplication and division method by adjoint expression:

The necessary sign of convergence did not work, but brought our Tambov comrade to light. As a result of the performed transformations, an equivalent series was obtained , which in turn strongly resembles a convergent series .

We write a clean solution:

Compare this series with the convergent series . We use the limit comparison criterion:

Multiply and divide by the adjoint expression:

A finite number other than zero is obtained, which means that the series under study converges together with next to .

Perhaps some have a question, where did the wolves come from on our African safari? Don't know. They probably brought it. You will get the following trophy skin:

Example 18

Investigate the convergence of a series

An example solution at the end of the lesson

And, finally, one more thought that visits many students in despair: instead of whether to use a rarer criterion for the convergence of the series? Sign of Raabe, sign of Abel, sign of Gauss, sign of Dirichlet and other unknown animals. The idea is working, but in real examples it is implemented very rarely. Personally, in all the years of practice, I have resorted to only 2-3 times sign of Raabe when nothing really helped from the standard arsenal. I reproduce the course of my extreme quest in full:

Example 19

Investigate the convergence of a series

Solution: Without any doubt a sign of d'Alembert. In the course of calculations, I actively use the properties of degrees, as well as second wonderful limit:

Here's one for you. D'Alembert's sign did not give an answer, although nothing foreshadowed such an outcome.

After going through the manual, I found a little-known limit proven in theory and applied a stronger radical Cauchy criterion:

Here's two for you. And, most importantly, it is not at all clear whether the series converges or diverges (an extremely rare situation for me). Necessary sign of comparison? Without much hope - even if in an unthinkable way I figure out the order of growth of the numerator and denominator, this still does not guarantee a reward.

A complete d'Alembert, but the worst thing is that the series needs to be solved. Need to. After all, this will be the first time that I give up. And then I remembered that there seemed to be some more powerful signs. Before me was no longer a wolf, not a leopard and not a tiger. It was a huge elephant waving a big trunk. I had to pick up a grenade launcher:

Sign of Raabe

Consider a positive number series.
If there is a limit , That:
a) At a row diverges. Moreover, the resulting value can be zero or negative.
b) At a row converges. In particular, the series converges for .
c) When Raabe's sign does not give an answer.

We compose the limit and carefully simplify the fraction:


Yes, the picture is, to put it mildly, unpleasant, but I was no longer surprised. lopital rules, and the first thought, as it turned out later, turned out to be correct. But first, for about an hour, I twisted and turned the limit using “usual” methods, but the uncertainty did not want to be eliminated. And walking in circles, as experience suggests, is a typical sign that the wrong way of solving has been chosen.

I had to turn to Russian folk wisdom: "If nothing helps, read the instructions." And when I opened the 2nd volume of Fichtenholtz, to my great joy I found a study of an identical series. And then the solution went according to the model.