Section 12.3

Finding Sums of Series by Differentiation; Some Series Expansions

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Problem 5081

By differentiating the equality 1+x+x2+x3+…=11−x1+x+x^{2}+x^{3}+\ldots = \frac{1}{1-x}, obtain the following formula (for ∣x∣<1\left|x\right| < 1): ∑n=0∞(n+1)xn=1(1−x)2\sum_{n=0}^{\infty } (n+1)x^{n} = \frac{1}{(1-x)^{2}}.

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Problem 5082

By differentiating the equality 1+x+x2+x3+…=11−x1+x+x^{2}+x^{3}+\ldots = \frac{1}{1-x}, obtain the following formula (for ∣x∣<1\left|x\right| < 1): ∑n=0∞(n+1)(n+2)xn=2(1−x)3\sum_{n=0}^{\infty } (n+1)(n+2)x^{n} = \frac{2}{(1-x)^{3}}.

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Problem 5083

By differentiating the equality 1+x+x2+x3+…=11−x1+x+x^{2}+x^{3}+\ldots = \frac{1}{1-x}, obtain the following formula (for ∣x∣<1\left|x\right| < 1): ∑n=0∞(n+1)(n+2)(n+3)3!xn=1(1−x)4\sum_{n=0}^{\infty } \frac{(n+1)(n+2)(n+3)}{3!}x^{n} = \frac{1}{(1-x)^{4}}.

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Problem 5084

By differentiating the equality 1+x+x2+x3+…=11−x1+x+x^{2}+x^{3}+\ldots = \frac{1}{1-x}, obtain the following formula (for ∣x∣<1\left|x\right| < 1): ∑n=1∞n2xn=x+x2(1−x)3\sum_{n=1}^{\infty } n^{2}x^{n} = \frac{x+x^{2}}{(1-x)^{3}}.

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Problem 5085

By differentiating the equality 1+x+x2+x3+…=11−x1+x+x^{2}+x^{3}+\ldots = \frac{1}{1-x}, obtain the following formula (for ∣x∣<1\left|x\right| < 1): ∑n=1∞n3xn=x+4x2+x3(1−x)4\sum_{n=1}^{\infty } n^{3}x^{n} = \frac{x+4x^{2}+x^{3}}{(1-x)^{4}}.

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Problem 5086

Prove the equality x+x55+x99+…=12arctg⁡x+14ln⁡1+x1−xx+\frac{x^{5}}{5}+\frac{x^{9}}{9}+\ldots = \frac{1}{2}\operatorname {arctg}x+\frac{1}{4}\ln \frac{1+x}{1-x}.

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Problem 5087

Prove the equality ∑n=0∞x4n(4n)!=14(ex+e−x+2cos⁡x)\sum_{n=0}^{\infty } \frac{x^{4n}}{(4n)!} = \frac{1}{4}(e^{x}+e^{-x}+2\cos x).

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Problem 5088

Prove the equality ∑n=0∞x3n(3n)!=13(ex+2e−x2cos⁡x32)\sum_{n=0}^{\infty } \frac{x^{3n}}{(3n)!} = \frac{1}{3}\left(e^{x}+2e^{-\frac{x}{2}}\cos \frac{x\sqrt{3}}{2}\right).

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Problem 5089

Prove that ∑n=1∞xn1−xn=∑n=1∞τ(n)xn\sum_{n=1}^{\infty } \frac{x^{n}}{1-x^{n}} = \sum_{n=1}^{\infty } \tau (n)x^{n}; ∣x∣<1\left|x\right| < 1, where τ(n)\tau (n) is the number of integer divisors of nn (Lambert's series).

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Problem 5090

Prove that ∑n=1∞nxn1−xn=∑n=1∞T(n)xn\sum_{n=1}^{\infty } \frac{nx^{n}}{1-x^{n}} = \sum_{n=1}^{\infty } T(n)x^{n}; ∣x∣<1\left|x\right| < 1, where T(n)T(n) is the sum of the divisors of nn.

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Problem 5091

Prove that

(∑n=1∞xn2)2=∑n=1∞r(n)xn;∣x∣<1, \left(\sum _{n=1}^{\infty } x^{n^{2}}\right)^{2} = \sum _{n=1}^{\infty } r(n)x^{n}; \quad \left|x\right| < 1,

where r(n)r(n) is the number of distinct ways of writing nn as a sum of two integer squares.

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Problem 5092

The function f(x)=xex+e−xex−e−xf(x) = x\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}} is even, i.e., f(−x)=f(x)f(-x) = f(x). Hence we may set f(x)=A0+A1x2+A2x4+…f(x) = A_{0}+A_{1}x^{2}+A_{2}x^{4}+\ldots, where A0,A1,A2,…A_{0}, A_{1}, A_{2}, \ldots are unknown coefficients. We may set An=(−1)n−1Bn4n(2n)!A_{n} = \frac{(-1)^{n-1}B_{n}4^{n}}{(2n)!}, where BnB_{n} are other coefficients, called the Bernoulli numbers. Expanding exe^{x} and e−xe^{-x} in a series in the equality written above and expressing AnA_{n} in terms of the numbers BnB_{n}, we get:

1+x22!+x44!+…1+x23!+x45!+…=B0+22B12!x2+24B24!x4+… \frac{1+\frac{x^{2}}{2!}+\frac{x^{4}}{4!}+\ldots }{1+\frac{x^{2}}{3!}+\frac{x^{4}}{5!}+\ldots } = B_{0}+\frac{2^{2}B_{1}}{2!}x^{2}+\frac{2^{4}B_{2}}{4!}x^{4}+\ldots

Prove that the Bernoulli numbers BnB_{n} can be obtained one after another by means of the recurrence formula:

(−1)n−14nBn(2n)!+(−1)n−24n−1Bn−13!(2n−2)!+(−1)n−34n−2Bn−25!(2n−4)!+…+4B1(2n−1)!2!+B0(2n+1)!=1(2n)!. (-1)^{n-1}\frac{4^{n}B_{n}}{(2n)!} + (-1)^{n-2}\frac{4^{n-1}B_{n-1}}{3!(2n-2)!} + (-1)^{n-3}\frac{4^{n-2}B_{n-2}}{5!(2n-4)!} + \ldots + \frac{4B_{1}}{(2n-1)!2!} + \frac{B_{0}}{(2n+1)!} = \frac{1}{(2n)!}.

Applying it, one can obtain the following values of the first Bernoulli numbers: B0=1B_{0} = 1; B1=16B_{1} = \frac{1}{6}; B2=130B_{2} = \frac{1}{30}; B3=142B_{3} = \frac{1}{42}; B4=130B_{4} = \frac{1}{30}; B5=566B_{5} = \frac{5}{66}; B6=6912730B_{6} = \frac{691}{2730}; B7=76B_{7} = \frac{7}{6}; B8=3617510B_{8} = \frac{3617}{510}; B9=43867798B_{9} = \frac{43867}{798}; B10=174611300B_{10} = \frac{174611}{300}; B11=854513138B_{11} = \frac{854513}{138}; B12=2363640912730B_{12} = \frac{236364091}{2730}; and so on.

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Problem 5093

Using the recurrence formula for the Bernoulli numbers (see problem 5092), prove the inequality: 4nBn(2n)!<15n\frac{4^{n}B_{n}}{(2n)!} < \frac{1}{5^{n}}, showing that the series for the function xex+e−xex−e−xx\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}} converges for ∣x∣<5\left|x\right| < \sqrt{5}.

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Problem 5094

Setting

1cos⁡x=∑n=0∞En(2n)!x2n, \frac{1}{\cos x} = \sum _{n=0}^{\infty } \frac{E_{n}}{(2n)!}x^{2n},

prove the recurrence formula for the Euler numbers EnE_{n}:

En(2n)!−En−12!(2n−2)!+En−24!(2n−4)!−…+(−1)nE0(2n)!=0. \frac{E_{n}}{(2n)!} - \frac{E_{n-1}}{2!(2n-2)!} + \frac{E_{n-2}}{4!(2n-4)!} - \ldots + (-1)^{n}\frac{E_{0}}{(2n)!} = 0.
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Problem 5095

Starting from the formula for the numbers EnE_{n} (see problem 5094), prove the inequality: ∣En(2n)!∣<ρn\left|\frac{E_{n}}{(2n)!}\right| < \rho^{n}, where ρ=ch⁡1−1=0,543…\rho = \operatorname {ch}1-1 = 0{,}543\ldots, and show that the power series for 1cos⁡x\frac{1}{\cos x} converges for ∣x∣<1ρ\left|x\right| < \frac{1}{\sqrt{\rho }}.

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Problem 5096

From the result of problem 5092, obtain the expansion:

11−e−x−1x−12=∑n=1∞(−1)n−1Bnx2n−1(2n)!. \frac{1}{1-e^{-x}} - \frac{1}{x} - \frac{1}{2} = \sum _{n=1}^{\infty } \frac{(-1)^{n-1}B_{n}x^{2n-1}}{(2n)!}.
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Problem 5097

Comparing the results of problems 5080 and 5096, obtain the equality ∑ν=1∞1ν2n=Bn(2π)2n2(2n)!\sum_{\nu =1}^{\infty } \frac{1}{\nu^{2n}} = \frac{B_{n}(2\pi )^{2n}}{2(2n)!} and prove that the series of problem 5096 converges for ∣x∣<2π\left|x\right| < 2\pi and diverges for ∣x∣>2π\left|x\right| > 2\pi.

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Problem 5098

Comparing problems 5079 and 5094, obtain the equality

∑ν=0∞(−1)ν(2ν+1)2n+1=En⋅π2n+122n+2(2n)! \sum _{\nu =0}^{\infty } \frac{(-1)^{\nu }}{(2\nu +1)^{2n+1}} = \frac{E_{n}\cdot \pi ^{2n+1}}{2^{2n+2}(2n)!}

and prove that the series of problem 5094 converges for ∣x∣<π2\left|x\right| < \frac{\pi }{2} and diverges for ∣x∣>π2\left|x\right| > \frac{\pi }{2}.

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Problem 5099

By forming a differential equation for the corresponding function, prove the following series expansion (∣x∣<1\left|x\right| < 1):

12(ln⁡(1+x)x)2=12−(1+12)x3+(1+12+13)x24−… \frac{1}{2}\left(\frac{\ln (1+x)}{x}\right)^{2} = \frac{1}{2}-\left(1+\frac{1}{2}\right)\frac{x}{3}+\left(1+\frac{1}{2}+\frac{1}{3}\right)\frac{x^{2}}{4}-\ldots
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Problem 5100

By forming a differential equation for the corresponding function, prove the following series expansion (∣x∣<1\left|x\right| < 1):

(arctg⁡xx)2=1−(1+13)x22+(1+13+15)x43−… \left(\frac{\operatorname {arctg}x}{x}\right)^{2} = 1-\left(1+\frac{1}{3}\right)\frac{x^{2}}{2}+\left(1+\frac{1}{3}+\frac{1}{5}\right)\frac{x^{4}}{3}-\ldots
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Problem 5101

By forming a differential equation for the corresponding function, prove the following series expansion (∣x∣<1\left|x\right| < 1):

12arctg⁡xln⁡(1+x2)=(1+12)x33−(1+12+13+14)x55+… \frac{1}{2}\operatorname {arctg}x\ln (1+x^{2}) = \left(1+\frac{1}{2}\right)\frac{x^{3}}{3}-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\frac{x^{5}}{5}+\ldots
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Problem 5102

By forming a differential equation for the corresponding function, prove the following series expansion (∣x∣<1\left|x\right| < 1):

cos⁡(marcsin⁡x)=1−m22!x2+m2(m2−4)4!x4−m2(m2−4)(m2−16)6!x6+… \cos (m\arcsin x) = 1-\frac{m^{2}}{2!}x^{2}+\frac{m^{2}(m^{2}-4)}{4!}x^{4} - \frac{m^{2}(m^{2}-4)(m^{2}-16)}{6!}x^{6}+\ldots
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Problem 5103

By forming a differential equation for the corresponding function, prove the following series expansion (∣x∣<1\left|x\right| < 1):

sin⁡(marcsin⁡x)=mx−m(m2−1)3!x3+m(m2−1)(m2−9)5!x5−… \sin (m\arcsin x) = mx-\frac{m(m^{2}-1)}{3!}x^{3}+\frac{m(m^{2}-1)(m^{2}-9)}{5!}x^{5}-\ldots
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Problem 5104

By forming a differential equation for the corresponding function, prove the following series expansion (∣x∣<1\left|x\right| < 1):

cos⁡(marccos⁡x)=Acos⁡mπ2+Bsin⁡mπ2, \cos (m\arccos x) = A\cos \frac{m\pi }{2}+B\sin \frac{m\pi }{2},

where AA and BB are the series of the two preceding problems, namely:

A=1−m22!x2+m2(m2−4)4!x4−m2(m2−4)(m2−16)6!x6+…(the series of problem 5102), A = 1-\frac{m^{2}}{2!}x^{2}+\frac{m^{2}(m^{2}-4)}{4!}x^{4} - \frac{m^{2}(m^{2}-4)(m^{2}-16)}{6!}x^{6}+\ldots \quad (\text{the series of problem 5102}), B=mx−m(m2−1)3!x3+m(m2−1)(m2−9)5!x5−…(the series of problem 5103). B = mx-\frac{m(m^{2}-1)}{3!}x^{3}+\frac{m(m^{2}-1)(m^{2}-9)}{5!}x^{5}-\ldots \quad (\text{the series of problem 5103}).
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Problem 5105

Prove that in the equalities of problems 5102 and 5103, both sides can be expanded in powers of the variable mm for ∣x∣<1\left|x\right| < 1. Equating the coefficients of corresponding powers of mm, obtain the expansion for ∣x∣<1\left|x\right| < 1:

arcsin⁡x=2∑n=0∞(2n)!(n!)2(2n+1)⋅(x2)2n+1. \arcsin x = 2\sum _{n=0}^{\infty } \frac{(2n)!}{(n!)^{2}(2n+1)} \cdot \left(\frac{x}{2}\right)^{2n+1}.
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Problem 5106

Equating the coefficients of corresponding powers of mm in the equalities of problems 5102 and 5103 (see problem 5105), obtain the expansion for ∣x∣<1\left|x\right| < 1:

(arcsin⁡x)2=12∑n=1∞((n−1)!)2(2n)!(2x)2n. (\arcsin x)^{2} = \frac{1}{2}\sum _{n=1}^{\infty } \frac{((n-1)!)^{2}}{(2n)!}(2x)^{2n}.
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Problem 5107

Equating the coefficients of corresponding powers of mm in the equalities of problems 5102 and 5103 (see problem 5105), obtain the expansion for ∣x∣<1\left|x\right| < 1:

(arcsin⁡x)3=6∑n=1∞(2n)!(n!)2⋅22n+1(1+132+152+…+1(2n−1)2)(x2)2n+1. (\arcsin x)^{3} = 6\sum _{n=1}^{\infty } \frac{(2n)!}{(n!)^{2}} \cdot \frac{2}{2n+1}\left(1+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\ldots +\frac{1}{(2n-1)^{2}}\right)\left(\frac{x}{2}\right)^{2n+1}.
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Problem 5108

Equating the coefficients of corresponding powers of mm in the equalities of problems 5102 and 5103 (see problem 5105), obtain the expansion

(arcsin⁡x)4=6∑n=2∞4n−1((n−1)!)2(2n)!x2n(1+122+132+…+1(n−1)2). (\arcsin x)^{4} = 6\sum _{n=2}^{\infty } \frac{4^{n-1}((n-1)!)^{2}}{(2n)!}x^{2n}\left(1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots +\frac{1}{(n-1)^{2}}\right).
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Remark.
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The expansions of (arcsin⁡x)n(\arcsin x)^{n} for n>4n > 4 are of a much more complicated form.

Problem 5109

Prove the equality ∑n=0∞(n!)2(2n)!=36+2π327\sum_{n=0}^{\infty } \frac{(n!)^{2}}{(2n)!} = \frac{36+2\pi \sqrt{3}}{27}.

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Problem 5110

Prove the equality ∑n=0∞((2n)!)2(4n)!=1615+π327+255ln⁡5−12\sum_{n=0}^{\infty } \frac{((2n)!)^{2}}{(4n)!} = \frac{16}{15}+\frac{\pi \sqrt{3}}{27}+\frac{2}{5\sqrt{5}}\ln \frac{\sqrt{5}-1}{2}.

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Problem 5111

Prove the equality ln⁡2(1−1−x)x=12⋅x2+1⋅32⋅4⋅x24+1⋅3⋅52⋅4⋅6⋅x36+…\ln \frac{2(1-\sqrt{1-x})}{x} = \frac{1}{2}\cdot \frac{x}{2}+\frac{1\cdot 3}{2\cdot 4}\cdot \frac{x^{2}}{4}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\cdot \frac{x^{3}}{6}+\ldots; ∣x∣<1\left|x\right| < 1.

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Problem 5112

Prove the equality

(2(1−1−x)x)m=1+m1(x4)+m(m+3)2!(x4)2+m(m+3)(m+5)3!(x4)3+…;∣x∣<1. \left(\frac{2(1-\sqrt{1-x})}{x}\right)^{m} = 1+\frac{m}{1}\left(\frac{x}{4}\right)+\frac{m(m+3)}{2!}\left(\frac{x}{4}\right)^{2}+\frac{m(m+3)(m+5)}{3!}\left(\frac{x}{4}\right)^{3}+\ldots ; \quad \left|x\right| < 1.
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Problem 5113

Writing the integral for 0<x<π0 < x < \pi of the identity

cos⁡x+cos⁡2x+cos⁡3x+…+cos⁡nx=−12+12sin⁡(n+12)xsin⁡x2 \cos x+\cos 2x+\cos 3x+\ldots +\cos nx = -\frac{1}{2}+\frac{1}{2}\frac{\sin \left(n+\frac{1}{2}\right)x}{\sin \frac{x}{2}}

and applying integration by parts, obtain the equality:

∑ν=1nsin⁡νxν=π−x2+cos⁡(n+12)x(2n+1)sin⁡x2+12n+1∫xπcos⁡(n+12)ucos⁡u22sin⁡2u2 du. \sum _{\nu =1}^{n} \frac{\sin \nu x}{\nu } = \frac{\pi -x}{2}+\frac{\cos \left(n+\frac{1}{2}\right)x}{(2n+1)\sin \frac{x}{2}} + \frac{1}{2n+1}\int _{x}^{\pi } \frac{\cos \left(n+\frac{1}{2}\right)u\cos \frac{u}{2}}{2\sin ^{2}\frac{u}{2}}\, du.
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Problem 5114

Prove that

sin⁡x+sin⁡2x2+sin⁡3x3+…={π−x2,for 0<x<π;−π+x2,for −π<x<0. \sin x+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+\ldots = \begin{cases} \frac{\pi -x}{2}, & \text{for } 0 < x < \pi ; \\ -\frac{\pi +x}{2}, & \text{for } -\pi < x < 0. \end{cases}
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Problem 5115

Applying the identity in a similar way

cos⁡x+cos⁡3x+…+cos⁡(2n−1)x=12sin⁡2nxsin⁡x, \cos x+\cos 3x+\ldots +\cos (2n-1)x = \frac{1}{2}\frac{\sin 2nx}{\sin x},

prove, for 0<x<π0 < x < \pi, the expansion:

sin⁡x+sin⁡3x3+sin⁡5x5+…=π4. \sin x+\frac{\sin 3x}{3}+\frac{\sin 5x}{5}+\ldots = \frac{\pi }{4}.
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Problem 5116

Starting from the expansion of problem 5115 (sin⁡x+sin⁡3x3+sin⁡5x5+…=π4\sin x+\frac{\sin 3x}{3}+\frac{\sin 5x}{5}+\ldots = \frac{\pi }{4} for 0<x<π0 < x < \pi), prove Steklov's formula:

f(x+π)−f(x)=−4π∑m=0∞∫xx+πf(u)cos⁡((2m+1)(u−x)) du, f(x+\pi )-f(x) = -\frac{4}{\pi }\sum _{m=0}^{\infty } \int _{x}^{x+\pi } f(u)\cos \left((2m+1)(u-x)\right)\, du,

where f(x)f(x) is a function differentiable on the interval (x,π+x)(x, \pi +x).

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Problem 5117

Find the sum of the series ∑n=1∞sin⁡nxn\sum_{n=1}^{\infty } \frac{\sin nx}{n}, noting that

1n=∫01xn−1 dx. \frac{1}{n} = \int _{0}^{1} x^{n-1}\, dx.
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Problem 5118

Find the sum of the series ∑n=1∞cos⁡nxn\sum_{n=1}^{\infty } \frac{\cos nx}{n}; 0<x<2π0 < x < 2\pi.

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Problem 5119

Find the sum of the series ∑n=0∞sin⁡(2n+1)x2n+1\sum_{n=0}^{\infty } \frac{\sin (2n+1)x}{2n+1}; 0<x<π0 < x < \pi.

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Problem 5120

Find the sum of the series ∑n=0∞cos⁡(2n+1)x2n+1\sum_{n=0}^{\infty } \frac{\cos (2n+1)x}{2n+1}; 0<x<π0 < x < \pi.

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Problem 5121

Find the sum of the series ∑n=2∞cos⁡nxn2−1\sum_{n=2}^{\infty } \frac{\cos nx}{n^{2}-1}; 0<x<2π0 < x < 2\pi.

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Problem 5122

Find the sum of the series ∑n=2∞sin⁡nxn2−1\sum_{n=2}^{\infty } \frac{\sin nx}{n^{2}-1}; 0<x<2π0 < x < 2\pi.

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Problem 5123

Find the sum of the series ∑n=1∞sin⁡nxn(n2+a2)\sum_{n=1}^{\infty } \frac{\sin nx}{n(n^{2}+a^{2})}; 0<x<π0 < x < \pi.

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Problem 5124

Find the sum of the series ∑n=1∞(−1)nsin⁡nxn(n2+a2)\sum_{n=1}^{\infty } \frac{(-1)^{n}\sin nx}{n(n^{2}+a^{2})}; −π<x<π-\pi < x < \pi.

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Problem 5125

Find the sum of the series ∑n=1∞(−1)ncos⁡nxn4−4\sum_{n=1}^{\infty } \frac{(-1)^{n}\cos nx}{n^{4}-4}; −π<x<π-\pi < x < \pi.

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