- #1
Euge
Gold Member
MHB
POTW Director
- 2,057
- 215
Find, with proof, the sum of the alternating series $$\sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}$$
That's quite a lengthy solution! It's also the correct solution according to Wolfram :)julian said:First write:
\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^3} = - \sum_{n=1}^\infty \dfrac{(-1)^n \sin \dfrac{\pi n}{2}}{n^3}
\end{align*}
The method of evaluation employs the fact that the function
\begin{align*}
\frac{\sin \dfrac{\pi z}{2}}{\sin \pi z}
\end{align*}
has simple poles at all integer values except when ##\sin \dfrac{\pi z}{2} = 0##. This allows us to write
\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^3} & = - \sum_{n=1}^\infty \dfrac{(-1)^n \sin \dfrac{\pi n}{2}}{n^3}
\nonumber \\
& = - \frac{1}{2i} \oint_C \dfrac{\sin \dfrac{\pi z}{2}}{z^3 \sin \pi z} dz
\end{align*}
where the contour ##C## is defined in Fig 1. Note
\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^3} & = - \sum_{n=1}^\infty \dfrac{(-1)^n \sin \dfrac{\pi n}{2}}{n^3}
\nonumber \\
& = - \frac{1}{2} \sum_{n=1}^\infty \dfrac{(-1)^n \sin \dfrac{\pi n}{2}}{n^3} - \frac{1}{2} \sum_{n=-1}^{-\infty} \dfrac{(-1)^n \sin \dfrac{\pi n}{2}}{n^3}
\nonumber \\
& = - \frac{1}{4i} \oint_{C+C'} \dfrac{\sin \dfrac{\pi z}{2}}{z^3 \sin \pi z} dz
\end{align*}
where the contour ##C'## is defined in Fig 1.
View attachment 344267
Fig 1.
Consider the square contour shown in Fig 2 with corners at ##(N+\frac{1}{2}) (1+i)##, ##(N+\frac{1}{2}) (-1+i)##, ##(N+\frac{1}{2}) (-1-i)## and ##(N+\frac{1}{2}) (1-i)##. This contour encloses all the poles at ##-N, - (N-1) , \dots, -1,0,1, \dots , N-1,N##. We will prove that the integral
\begin{align*}
\oint_{C_N} \dfrac{\sin \dfrac{\pi z}{2}}{z^3 \sin \pi z} dz \qquad (*)
\end{align*}
vanishes ##N \rightarrow \infty##. This will imply
\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^3} & = - \sum_{n=1}^\infty \dfrac{(-1)^n \sin \dfrac{\pi n}{2}}{n^3}
\nonumber \\
& = - \frac{1}{4i} \oint_{C_0} \dfrac{\sin \dfrac{\pi z}{2}}{z^3 \sin \pi z} dz
\nonumber \\
& = - \frac{1}{8i} \oint_{C_0} \dfrac{1}{z^3 \cos \dfrac{\pi z}{2}} dz \qquad (**)
\end{align*}
where ##C_0## is an infinitesimal clockwise contour around the origin.
View attachment 344279
Fig 2.
We start by showing that the value of ##\left| \sin \dfrac{\pi z}{2} \csc (\pi z) \right|## around the square ##C_N## is bounded by a constant that is independent of ##N##.
We write ##z=x+iy##
Case 1: ##y > \frac{1}{2}##, we have
\begin{align*}
\left| \dfrac{\sin \dfrac{\pi z}{2}}{\sin \pi z} \right| & = \left| \dfrac{e^{i \pi z/2} - e^{-i \pi z/2}}{e^{i \pi z} - e^{-i \pi z}} \right|
\nonumber \\
& \leq \dfrac{|e^{i \pi z/2}| + |e^{-i \pi z/2}|}{|e^{-i \pi z}| - |e^{i \pi z}|}
\nonumber \\
& = \dfrac{|e^{i \pi x/2 - \pi y/2}| + |e^{-i \pi x + \pi y}|}{|e^{-i \pi x + \pi y}| - |e^{i \pi x - \pi y}|}
\nonumber \\
& = \dfrac{e^{\pi y/2} + e^{-\pi y/2}}{e^{\pi y} - e^{-\pi y}}
\nonumber \\
& = \dfrac{e^{-\pi y/2} + e^{-3\pi y/2}}{1 - e^{- 2 \pi y}}
\nonumber \\
& \leq \dfrac{e^{-\pi y/4} + e^{-3\pi y/4}}{1 - e^{- \pi}} \qquad \text{ as we are taking } y > \frac{1}{2}
\nonumber \\
& =: A_1
\end{align*}
Case 2: ##y < - \frac{1}{2}##, we have
\begin{align*}
\left| \dfrac{\sin \dfrac{\pi z}{2}}{\sin \pi z} \right| & \leq \dfrac{|e^{i \pi x/2 - \pi y/2}| + |e^{-i \pi x + \pi y}|}{|e^{i \pi x - \pi y}| - |e^{-i \pi x + \pi y}|}
\nonumber \\
& = \dfrac{e^{\pi y/2} + e^{-\pi y/2}}{e^{-\pi y} - e^{\pi y}}
\nonumber \\
& = \dfrac{e^{\pi y/2} + e^{3\pi y/2}}{1 - e^{2 \pi y}}
\nonumber \\
& \leq \dfrac{e^{-\pi y/4} + e^{-3\pi y/4}}{1 - e^{- \pi}} \qquad \text{ as we are taking } y < - \frac{1}{2}
\nonumber \\
& =: A_1
\end{align*}
Case 3: ##-\frac{1}{2} \leq y \leq \frac{1}{2}##. We consider ##z = N +\frac{1}{2} + iy##. We make repeated use of ##\sin (\alpha + \beta) = \cos \alpha \sin \beta + \sin \alpha \cos \beta## and ##\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta##. In particular we will use:
\begin{align*}
\sin (\pi (N + \frac{1}{2} + i y)) = \cos (\pi N) \sin (\pi/2 + i \pi y) = (-1)^N \cosh (\pi y)
\end{align*}
We have:
\begin{align*}
& \; \left| \dfrac{\sin \dfrac{\pi z}{2}}{\sin \pi z} \right|
\nonumber \\
& = \left| \dfrac{\sin (\pi (N + \frac{1}{2} + i y)/2)}{\sin (\pi (N + \frac{1}{2} + i y))} \right|
\nonumber \\
& = \dfrac{| \cos (\pi N /2) \sin (\pi/4+iy/2) + \sin (\pi N /2) \cos (\pi/4+iy/2) |}{\cosh (\pi y)}
\nonumber \\
& \leq \dfrac{| \cos (\pi N/2) \sin (\pi/4+iy/2)}{\cosh (\pi y)} + \dfrac{| \sin (\pi N/2) \cos (\pi /4+iy/2) |}{\cosh (\pi y)}
\nonumber \\
& \leq \dfrac{| \sin (\pi/4+iy/2) |}{\cosh (\pi y) |} + \dfrac{| \cos (\pi/4+iy/2) |}{\cosh (\pi y) |}
\nonumber \\
& \leq \dfrac{| \cos (\pi/4) \sin (iy/2) + \sin (\pi/4) \cos (iy/2) |}{\cosh (\pi y)} + \dfrac{| \cos (\pi/4) \cos (iy/2) - \sin (\pi/4) \sin (iy/2) |}{\cosh (\pi y)}
\nonumber \\
& \leq 2 \dfrac{| \sinh (y/2) |}{\cosh (\pi y)} + 2 \dfrac{\cosh (y/2)}{\cosh (\pi y)}
\nonumber \\
& = 2 \dfrac{| e^{y/2} - e^{-y/2} |}{e^{\pi y} + e^{-\pi y}} + 2 \dfrac{e^{y/2} + e^{-y/2}}{e^{\pi y} + e^{-\pi y}}
\nonumber \\
& \leq 4 \dfrac{e^{y/2} + e^{-y/2}}{e^{\pi y} + e^{-\pi y}}
\nonumber \\
& = 4 \dfrac{e^{y (1/2 - \pi)} + e^{-y (1/2 + \pi)}}{1 + e^{-\pi y}}
\nonumber \\
& \leq \dfrac{8 (e^{1/4} + e^{-1/4}) e^{\pi/2}}{1 + e^{-\pi / 2}} \qquad \text{ as we are taking } -\frac{1}{2} \leq y \leq \frac{1}{2}
\nonumber \\
& =: A_2
\end{align*}
When ##z = -N - \frac{1}{2} + iy##, we have similarly:
\begin{align*}
\left| \dfrac{\sin \dfrac{\pi z}{2}}{\sin \pi z} \right| & = \left| \dfrac{\sin ((-N - \frac{1}{2} + i y)/2)}{\sin (\pi (-N - \frac{1}{2} + i y))} \right|
\nonumber \\
& \leq A_2
\end{align*}
So choose ##A## such that ##A > \max \{ A_1 , A_2 \}##. Then we have ##\left| \sin \dfrac{\pi z}{2} \csc (\pi z) \right| < A## on ##C_N## with an ##A## independent of ##N##. Then
\begin{align*}
\left| \oint_{C_N} \csc (\pi z) \sin \dfrac{\pi z}{2} \frac{1}{z^3} dz \right| \leq \frac{\pi A}{N^3} (8N+4)
\end{align*}
as ##(8N+4)## is the length of the curve ##C_N##. Letting ##N \rightarrow \infty## we get that the integral, ##(*)##, vanishes. This establishes ##(**)##. We now use ##(**)## to evaluate the sum. The residue is obtained from
\begin{align*}
\dfrac{1}{z^3 \cos \dfrac{\pi z}{2}} & = \dfrac{1}{z^3 \left( 1 - \dfrac{1}{2!} \dfrac{\pi^2 z^2}{4} \right) + \cdots}
\nonumber \\
& = \frac{1}{z^3} \left( 1 + \dfrac{\pi^2 z^2}{8} \right) + \cdots
\nonumber \\
& = \frac{1}{z^3} + \dfrac{\pi^2}{8 z} + \cdots
\end{align*}
So that
\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^3} & = (-2 \pi i) \times - \frac{1}{8i} \times \dfrac{\pi^2}{8}
\nonumber \\
& = \frac{\pi^3}{32}
\end{align*}
Thanks for pointing out the difficulty. From my last linejulian said:@anuttarasammyak. Is the sum Re∑n=1∞e−i(p−1)nπ2 related to the periodic delta function though? With period 4? And so the integral ∫−∞∞dp⋯ wouldn't pick out a single value of p?
You mentioned your Insight here. Reading your Insight is where I got the idea to rewrite an alternating sum over odd integers using ##\sin \dfrac{n \pi}{2}##, an idea which has opened up fruitful avenues.Svein said:As I have noted in an Insight, just create the Fourier serien from the function [itex]\pi ^{2}\cdot x - x^{3} [/itex]. This ends up in [itex]12\cdot \sum_{n=1}^{\infty}\frac{\sin(n(\pi - x))}{n^{3}} [/itex]. Evaluate both expressions at π/2. and you have the desired proof.
Excellent. Let me follow you.Svein said:As I have noted in an Insight, just create the Fourier serien from the function [itex]\pi ^{2}\cdot x - x^{3} [/itex]. This ends up in [itex]12\cdot \sum_{n=1}^{\infty}\frac{\sin(n(\pi - x))}{n^{3}} [/itex]. Evaluate both expressions at π/2. and you have the desired proof.
The correct formula isanuttarasammyak said:Excellent. Let me follow you.
[tex]T(x):=\sum_{n=1}^{\infty}\frac{\sin(n(\pi-x))}{n^3}[/tex] for ##-\pi < x < \pi##. Regarding this as Fourier series, ##n^{3}## in denominator shows that T(x) is a tertial function of x thinking of integration by parts for the calculation of Fourier components. We can write T(x) with constant a as
[tex]T(x)=ax(x-\pi)(x+\pi)[/tex] because [tex]T(0)=T(\pi)=T(-\pi)=0[/tex] To determine a, let us compare T'(0) of the both forms
[tex]1-2^{-2}+3^{-2}-4^{-2}+...=\frac{1}{2}\zeta(2)=-\pi^2 a[/tex]
[tex]a=-\frac{1}{12}[/tex]
Thus the value we want is
[tex]T(\frac{\pi}{2})=-\frac{1}{12}(\frac{\pi}{2})(-\frac{\pi}{2})\frac{3\pi}{2}=\frac{\pi^3}{32}[/tex]
With wolfram.
View attachment 344990
It shows [tex]\frac{\pi^3}{ 32}=0.9674...[/tex]View attachment 344995
Similary say
[tex]T_+ (x):=\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^3}[/tex] for ##-\pi < x < \pi##.
[tex]T_+ (x)=-\frac{1}{6}x(x-\pi)(x+\pi)=2T(x)[/tex] But from the summation
[tex]T_+ (\frac{\pi}{2})=T (\frac{\pi}{2})=1-3^{-3}+5^{-3}-...[/tex] which is confirmed as below shown.
View attachment 345002
View attachment 345031
Where I went wrong ? We may have to choose the basic area for repetition where Y of sin Y for n=1 does not change sign, e.g. ##0 < x < 2\pi## for
[tex]T_+ (x)=\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^3}[/tex]