How to prove $(\frac{1}{5^3}-\frac{1}{7^3})+(\frac{1}{11^3}-\frac{1}{13^3})+(\frac{1}{17^3}-\frac{1}{19^3})+...=(1-\frac{\pi ^3}{18\sqrt{3}})$

Question

How to prove $$ \sum_{k=1}^\infty \left[\frac{1}{(6k-1)^3} - \frac{1}{(6k+1)^3}\right] = 1 - \frac{\pi^3}{18\sqrt{3}}$$

I think this equality likes the Dirichlet Beta function. The numerical value is checked but I don't have the proving. Any help

Answer 1

Notice$\color{blue}{^{[1]}}$ $$\sum_{k=1}^\infty \left( \frac{1}{(6k-1)^3} - \frac{1}{(6k+1)^3}\right) = \sum_{\substack{k=-\infty\\ k\ne 0}}^\infty \frac{1}{(6k-1)^3} = 1 - \frac{1}{6^3}\sum_{k=-\infty}^\infty \frac{1}{(\frac16-k)^3}$$

Recall the infinite product expansion of $\sin x$ $$\sin x = x \prod_{k=1}^\infty \left( 1 - \frac{x^2}{k^2\pi^2}\right)$$

If one take logarithm and differentiate, one obtain an expansion of $\cot x$

$$\cot x = \sum_{k=-\infty}^\infty \frac{1}{x - k\pi} \quad\iff\quad \sum_{k=-\infty}^\infty \frac{1}{x-k} = \pi\cot(\pi x)\tag{*1} $$

Differentiate the expansion on the right two more times, we get$\color{blue}{^{[2]}}$

$$\sum_{k=-\infty}^\infty \frac{1}{(x-k)^3} = \frac12 \frac{d^2}{dx^2} \left[ \sum_{k=-\infty}^\infty \frac{1}{x-k} \right] = \frac{\pi}{2} \left[ \frac{d^2}{dx^2}\cot(\pi x) \right] = \frac{\pi^3 \cos(\pi x)}{\sin(\pi x)^3}$$

As a result, the sum we want is

$$\sum_{k=1}^\infty \left( \frac{1}{(6k-1)^3} - \frac{1}{(6k+1)^3}\right) = 1 - \frac{\pi^3}{6^3}\frac{\cos\frac{\pi}{6}}{\sin(\frac{\pi}{6})^3} \ = 1 - \frac{\pi^3}{18\sqrt{3}} $$

Notes

• $\color{blue}{[1]}$ - Infinite sum of the form $\sum\limits_{k=-\infty}^\infty (\cdots)$ should be interpreted as $\lim\limits_{N\to\infty}\sum\limits_{k=-N}^N (\cdots)$.
• $\color{blue}{[2]}$ - To those who are not comfortable with the use of differentiation of an expansion. An alternate approach is start from the more well known expansion $(*1)$, compute a contour integral of the form: $$\frac{1}{2\pi i}\int_{|z|=R} \frac{\pi\cot(\pi z)}{(\frac16 - z)^3} dz$$ and show it vanishes as $R \to \infty$. The sum of the residues from $z \in \mathbb{Z}$ will be equal to the sum $\sum\limits_{k=-\infty}^\infty \frac{1}{\left(\frac16 - k\right)^3}$. It will be compensate by the residue of the pole at $z = \frac16$. Since the pole at $z = \frac16$ is a triple one, its contribution will be proportional to $\frac{d^2}{dx^2}\pi\cot(\pi x)$.

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\sum_{k\ =\ 1}^{\infty}\bracks{ {1 \over \pars{6k - 1}^{3}} - {1 \over \pars{6k + 1}^{3}}} =1 - {\pi^{3} \over 18\root{3}}:\ {\large ?}}$.

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\begin{align}&\color{#66f}{\large \sum_{k\ =\ 1}^{\infty}\bracks{ {1 \over \pars{6k - 1}^{3}} - {1 \over \pars{6k + 1}^{3}}}} =\left.\half\,\partiald[2]{}{\mu} \sum_{k\ =\ 1}^{\infty}\pars{ {1 \over 6k - \mu} - {1 \over 6k + \mu}}\right\vert_{\,\mu\ =\ 1} \\[5mm]&=\left. {1 \over 12}\,\partiald[2]{}{\mu} \sum_{k\ =\ 0}^{\infty}\pars{ {1 \over k + 1 - \mu/6} - {1 \over k + 1 + \mu/6}}\right\vert_{\,\mu\ =\ 1} \\[5mm]&=\left. {1 \over 12}\,\partiald[2]{}{\mu} \bracks{\Psi\pars{1 + {\mu \over 6}} - \Psi\pars{1 - {\mu \over 6}}} \right\vert_{\,\mu\ =\ 1} ={1 \over 432}\bracks{\Psi''\pars{7 \over 6} - \Psi''\pars{5 \over 6}} \\[5mm]&={1 \over 432}\bracks{ \Psi''\pars{1 \over 6} + {2 \over \pars{1/6}^{3}}- \Psi''\pars{5 \over 6}} ={1 \over 432}\bracks{432 + \pi\,\totald[2]{\cot\pars{\pi\mu}}{\mu}}_{\mu\ =\ 5/6} \\[5mm]&=1 + {1 \over 432}\,\bracks{2\pi^{3}\cot\pars{5\pi \over 6} \csc^{2}\pars{5\pi \over 6}} =\color{#66f}{\large 1 - {\pi^{3} \over 18\root{3}}} \end{align}

$\ds{\Psi}$ is the Digamma Function and we used the identities:

\begin{align} \sum_{k\ =\ 0}{1 \over \pars{k + x}\pars{k + y}} &={\Psi\pars{x} - \Psi\pars{y} \over x - y} \\[5mm]\Psi''\pars{1 + z}&=\Psi''\pars{z} + {2 \over z^{3}} \\[5mm] \Psi''\pars{1 - z}&=\Psi''\pars{z} + \pi\,\totald[2]{\cot\pars{\pi z}}{z} =\Psi''\pars{z} + 2\pi^{3}\cot\pars{\pi z}\csc^{2}\pars{\pi z} \end{align}

Answer 3

Approach, which uses Fourier series.

Denote $$ S = \sum_{k=1}^\infty \left[\frac{1}{(6k-1)^3} - \frac{1}{(6k+1)^3}\right]. $$

Consider function $$ f(x) = \dfrac{\pi x(\pi-x)}{8}, \qquad x\in[0,\pi];\tag{1} $$ construct the odd extension of $f(x)$ to the interval $[−\pi, \pi]$: $f(-x)=-f(x), x\in [0,\pi]$;

and make it $2\pi$-periodic: copy to each segment $[\pi (2k-1), \pi(2k+1)]$, $k\in\mathbb{Z}$.

Then, function $f(x)$ is odd, $2\pi$-periodic, and it belongs to class $C^1$.

Fourier series of this function:

$$ f(x) = \sum_{k=1}^{\infty} \dfrac{\sin(2k-1)x}{(2k-1)^3}= \dfrac{\sin x}{1^3}+\dfrac{\sin 3x}{3^3}+\dfrac{\sin{5x}}{5^3}+\dfrac{\sin{7x}}{7^3}+ \ldots .\tag{2} $$

Consider $f(2\pi/3)$:

$(1)\Rightarrow$ $$ f(2\pi/3) = \dfrac{\pi^3}{36}.\tag{3} $$

$(2)\Rightarrow$ $$ f(2\pi/3) = \dfrac{\sin{2\pi/3}}{1^3}+\dfrac{\sin{6\pi/3}}{3^3}+\dfrac{\sin{10\pi/3}}{5^3}+\dfrac{\sin{14\pi/3}}{7^3}+\ldots \\ =\dfrac{\sqrt{3}}{2}\left( \dfrac{1}{1^3}+\dfrac{0}{3^3}+\dfrac{-1}{5^3}+\dfrac{1}{7^3}+\dfrac{0}{9^3}+\dfrac{-1}{11^3}+\dfrac{1}{13^3}+\ldots \right) = \dfrac{\sqrt{3}}{2}(1-S).\tag{4} $$

$(3),(4) \Rightarrow $

$$ \dfrac{\pi^3}{36} = \dfrac{\sqrt{3}}{2}(1-S), $$ $$ 1-S = \dfrac{\pi^3}{18\sqrt{3}}, $$ $$ S=1- \dfrac{\pi^3}{18\sqrt{3}}. $$