Is there any known way to compute this sum $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{(2n-1)^3}$

Question

In a video that talks about Riemann's problem someone said that we don't know yet how to compute sums like this one $$\sum_{n=1}^{\infty}\frac{1}{n^{2k+1}}.$$

I was wandering if that concerns alternate series too like the one in my question !

Thanks in advance !

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over \pars{2n - 1}^{3}} & = \sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\ \overbrace{{1 \over 2}\int_{0}^{1}\ln^{2}\pars{x}x^{2n - 2}\,\dd x} ^{\ds{1 \over \pars{2n - 1}^{3}}} = {1 \over 2}\int_{0}^{1}\ln^{2}\pars{x}\sum_{n = 1}^{\infty} \pars{-x^{2}}^{n - 1}\,\dd x \\[5mm] & = {1 \over 2}\int_{0}^{1}{\ln^{2}\pars{x} \over 1 + x^{2}}\,\dd x = -{1 \over 2}\,\Im\int_{0}^{1}{\ln^{2}\pars{x} \over \ic - x}\,\dd x = -{1 \over 2}\,\Im\int_{0}^{-\ic}{\ln^{2}\pars{\ic x} \over 1 - x}\,\dd x \\[5mm] & = -\,\Im\int_{0}^{-\ic}{\ln\pars{1 - x} \over x}\,\ln\pars{\ic x}\,\dd x \\[5mm] & = \Im\int_{0}^{-\ic}\mrm{Li}_{2}'\pars{x}\ln\pars{\ic x}\,\dd x\qquad \pars{~\mrm{Li}_{s}:\ PolyLogarithm\ Function~} \\[5mm] & = -\,\Im\int_{0}^{-\ic}\mrm{Li}_{3}'\pars{x}\,\dd x = -\,\Im\mrm{Li}_{3}\pars{-\ic} = \Im\mrm{Li}_{3}\pars{\ic} \\[5mm] & = {\,\mrm{Li}_{\color{#f00}{3}}\expo{2\pi\ic\pars{\color{#f00}{1/4}}} + \pars{-1}^{\color{#f00}{3}}\,\mrm{Li}_{\color{#f00}{3}}\expo{-2\pi\ic\pars{\color{#f00}{1/4}}} \over 2\ic} = -{1 \over 2}\,\ic\bracks{-\,{\pars{2\pi\ic}^{\color{#f00}{3}} \over \color{#f00}{3}!}\,\mrm{B}_{\color{#f00}{3}}\pars{\color{#f00}{1 \over 4}}} \end{align} The last expression is Jonqui$\mrm{\grave{e}}$re's Inversion Formula and $\ds{\mrm{B}_{n}}$ is a Bernoulli Polynomial. Note that $\ds{\mrm{B}_{3}\pars{x} = x^{3} - {3 \over 2}\,x^{2} + {1 \over 2}\,x}$ such that $\ds{\,\mrm{B}_{\color{#f00}{3}}\pars{\color{#f00}{1 \over 4}} = {3 \over 64}}$. $$ \bbx{\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over \pars{2n - 1}^{3}} = {\pi^{3} \over 32}} \approx 0.9689 $$

Answer 2

$\beta(3)=\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^3}$ can be computed in many ways, for instance as explained here.
Here I propose a different approach through Fourier series. $$ W(x)=\sum_{n\geq 0}\frac{(-1)^n \cos((2n+1)x)}{2n+1} \tag{1} $$ is the Fourier cosine series of a $2\pi$-periodic rectangle wave, that equals $\frac{\pi}{4}$ on the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ and $-\frac{\pi}{4}$ on the interval $\left(\frac{\pi}{2},\frac{3\pi}{2}\right)$. By integrating twice $(1)$, we get that $$ P(x) = \sum_{n\geq 0}\frac{(-1)^n\cos((2n+1)x)}{(2n+1)^3}\tag{2} $$ is a uniformly convergent Fourier series for a piecewise-polynomial function, that equals $\frac{\pi^3}{32}\left(1-\frac{4x^2}{\pi^2}\right)$ on the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. By evaluating at $x=0$, $$ \beta(3)=\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^3}=\color{red}{\frac{\pi^3}{32}}\tag{3} $$ immediately follows.