The series $\sum_{k=1}^\infty\sum_{n=1}^k k^{-3}/(2n-1)$

Question

$$\sum_{k=1}^\infty\sum_{n=1}^k\frac{1}{(2n-1)k^3}$$

Can anyone help me find this series? I tried to use Cauchy product but I don't know how I can complete it.

Answer 1

We have: $$ \sum_{k\geq 1}\frac{x^{2k-1}}{2k-1}=\text{arctanh}(x),\qquad \sum_{k\geq 1}\left(\sum_{n=1}^{k}\frac{1}{2n-1}\right)x^{2k-1}=\frac{\text{arctanh}(x)}{1-x^2} $$ hence $$ \sum_{k\geq 1}\left(\sum_{n=1}^{k}\frac{1}{2n-1}\right)\frac{1}{k^3}=4\int_{0}^{1}\frac{\text{arctanh}(x)\log^2(x)}{1-x^2}\,dx.$$ Can you compute the last integral? Hint: if you apply integration by parts, you are left with an integral related with $\int_{0}^{1}\text{arctanh}^2(x)\frac{\log x}{x}\,dx$. You may consider that $$ \zeta(s) = \sum_{n\geq 1}\frac{1}{n^s} $$ through Laplace inverse transform takes the following integral representation: $$ \zeta(s) = \frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x-1}\,dx. $$ In the region $\text{Re}(s)>1$ we also have: $$ \eta(s)\stackrel{\text{def}}{=}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s} = \left(1-\frac{2}{2^s}\right)\zeta(s), $$ but the abscissa of convergence of the series defining $\eta(s)$ is $\text{Re}(s)>0$. Given that, the analytic continuation of the $\zeta$ function to the half-plane $\text{Re}(s)>0$ is given by $$\begin{eqnarray*} \zeta(s) &=& \left(1-\frac{2}{2^s}\right)^{-1}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s} \\ (\text{through }\mathcal{L}^{-1})\quad &=& \frac{1}{\Gamma(s)}\left(1-\frac{2}{2^s}\right)^{-1}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x+1}\,dx \\ (\text{by parts})\quad &=& \frac{4^s}{2(2^s-2)\Gamma(s+1)}\int_{0}^{+\infty}\frac{x^s\,dx}{\cosh^2(x)}\\ (\text{by substitution})\quad &=&\color{red}{\frac{4^s}{2(2^s-2)\Gamma(s+1)}\int_{0}^{1}\operatorname{arctanh}(x)^s\,dx}\\ &=& \frac{2^{s-1}}{(2^s-2)\Gamma(s+1)}\int_{0}^{1}\log^s\left(\frac{1+x}{1-x}\right)\,dx. \end{eqnarray*}$$ The original integral depends on the red integral at $s=2$. It follows (by Feynman's trick) that its closed form has the term $\psi\left(\frac{1}{2}\right)\zeta(3)$ and a term depending on $\zeta'(3)$.

---

Alternative approach: we want to compute $\sum_{n\geq 1}\frac{H_{2n}-H_n/2}{n^3}$.
Since $\sum_{n\geq 1}H_n x^n = -\frac{\log(1-x)}{1-x}$, we have: $$\frac{\pi^4}{180}=\int_{0}^{1}\frac{-\log(1-x)\log^2(x)}{1-x}\,dx = \sum_{n\geq 1}\frac{2H_n}{(n+1)^3}\tag{A}$$ $$ \sum_{n\geq 1}\frac{H_n}{2n^3}=\frac{\pi^4}{144}\tag{B}$$ $$ \int_{0}^{1}\left(\frac{-\log(1-x)}{1-x}+\frac{-\log(1+x)}{1+x}\right)\frac{\log^2(x)}{x}\,dx = \sum_{n\geq 1}\frac{H_{2n}}{2n^3}\\ = -\frac{\pi ^4}{30}-\frac{\pi^2\log^2(2)}{6}+\frac{\log^4(2)}{6}+4 \text{Li}_4\left(\frac{1}{2}\right)+\frac{7\log(2)\zeta(3)}{2}\tag{C}$$

and:

$$\boxed{ \sum_{n\geq 1}\frac{1}{n^3}\sum_{m=1}^{n}\frac{1}{2m-1} = \color{red}{-\frac{53 \pi^4}{720}-\frac{\pi^2\log^2(2)}{3}+\frac{\log^4(2)}{3}+8\text{Li}_4\left(\frac{1}{2}\right)+7\log(2)\zeta(3)}}$$

Answer 2

Just to give another approach, you may consider that $$ \eqalign{ & S = \sum\limits_{1\, \le \,k\,} {\sum\limits_{1\, \le \,n\, \le \,k} {{1 \over {\left( {2n - 1} \right)}}} {1 \over {k^{\,3} }}} = \sum\limits_{1\, \le \,k\,} {\left( {{1 \over 2}\sum\limits_{1\, \le \,n\, \le \,k} {{1 \over {\left( {n - 1/2} \right)}}} } \right){1 \over {k^{\,3} }}} = \cr & = \sum\limits_{1\, \le \,k\,} {\left( {{{\psi \left( {k + 1/2} \right) - \psi \left( {1/2} \right)} \over 2}} \right){1 \over {k^{\,3} }}} = \cr & = \left( {{\gamma \over 2} + \ln 2} \right)\sum\limits_{1\, \le \,k\,} {{1 \over {k^{\,3} }}} + {1 \over 2}\sum\limits_{1\, \le \,k\,} {\psi \left( {k + 1/2} \right){1 \over {k^{\,3} }}} = \cr & = \left( {{\gamma \over 2} + \ln 2} \right)\zeta (3) + {1 \over 2}\sum\limits_{1\, \le \,k\,} {\psi \left( {k + 1/2} \right){1 \over {k^{\,3} }}} \cr} $$

which computed numerically gives: $$ S \approx 1.298176 $$

Also, we have $$ \eqalign{ & S = \sum\limits_{1\, \le \,k\,} {\sum\limits_{1\, \le \,n\, \le \,k} {{1 \over {\left( {2n - 1} \right)}}} {1 \over {k^{\,3} }}} = \sum\limits_{1\, \le \,n\,} {{1 \over {\left( {2n - 1} \right)}}\sum\limits_{n\, \le \,k} {{1 \over {k^{\,3} }}} } = \cr & = - {1 \over 2}\sum\limits_{1\, \le \,n\,} {{{\psi ^{\,\left( 2 \right)} \left( n \right)} \over {\left( {2n - 1} \right)}}} = - {1 \over 4}\sum\limits_{1\, \le \,n\,} {{{\psi ^{\,\left( 2 \right)} \left( n \right)} \over {\left( {n - 1/2} \right)}}} \cr} $$ which when computed numerically provides the same result.

However at present I did not succeed and profitably introduce in the sum above the integral representation for the polygamma functions.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\sum_{k = 1}^{\infty} \sum_{n = 1}^{k}{1 \over \pars{2n - 1}k^{3}}} = \sum_{k = 1}^{\infty}{1 \over k^{3}} \pars{\sum_{n = 1}^{2k}{1 \over n} - \sum_{n = 1}^{k}{1 \over 2n}} \\[5mm] = &\ \sum_{k = 1}^{\infty}{1 \over k^{3}}\pars{H_{2k} - {1 \over 2}H_{k}} = 8\sum_{k = 1}^{\infty} {H_{2k} \over \pars{2k}^{3}} - {1 \over 2}\sum_{k = 1}^{\infty}{H_{k} \over k^{3}} \\[5mm] = &\ 8\sum_{k = 1}^{\infty}{H_{k} \over k^{3}}\,{1 + \pars{-1}^{k} \over 2} - {1 \over 2}\sum_{k = 1}^{\infty}{H_{k} \over k^{3}} \\[5mm] = &\ {7 \over 2}\ \underbrace{\sum_{k = 1}^{\infty}{H_{k} \over k^{3}}} _{\ds{\color{#f00}{\S}:\ =\ {\pi^{4} \over 72}}} + 4\sum_{k = 1}^{\infty}\pars{-1}^{k}\,{H_{k} \over k^{3}} \\[5mm] = &\ {7 \over 144}\,\pi^{4} + 4\sum_{k = 1}^{\infty}\pars{-1}^{k}\,{H_{k} \over k^{3}}\label{1}\tag{1} \end{align}

$\ds{\color{#f00}{\S}}$: The sum was evaluated by Borwein & Borwein-$\ds{1995}$. See $\ds{\pars{19}}$ in this link.

I already evaluated the 'remaining sum' $\ds{\sum_{k = 1}^{\infty}\pars{-1}^{k}\,{H_{k} \over k^{3}}}$ in a previous answer. Namely, $$ \sum_{k = 1}^{\infty}\pars{-1}^{k}\,{H_{k} \over k^{3}} = -\,{11 \over 360}\,\pi^{4} - {1 \over 12}\ln^{2}\pars{2}\pi^{2} + {1 \over 12}\,\ln^{4}\pars{2} + 2\,\mrm{Li}_{4}\pars{1 \over 2} + {7 \over 4}\,\ln\pars{2}\zeta\pars{3} $$ such that \eqref{1} becomes: \begin{align} &\sum_{k = 1}^{\infty}\sum_{n = 1}^{k}{1 \over \pars{2n - 1}k^{3}} \\[5mm] = &\ \bbx{-\,{53 \over 720}\,\pi^{4} - {1 \over 3}\ln^{2}\pars{2}\pi^{2} + {1 \over 3}\,\ln^{4}\pars{2} + 8\,\mrm{Li}_{4}\pars{1 \over 2} + 7\,\ln\pars{2}\zeta\pars{3}} \end{align}