Proving that $\left(\frac{\pi}{2}\right)^{2}=1+\sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k-1}}$.

Wolfram$\alpha$ says that we have the following identity $$ \left(\frac{\pi}{2}\right)^{2}=1+2\sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k}} $$

but, how does one prove such identity?

Solution 1:

It's well known that $$ \pi\cot(\pi x)=\frac 1x-2\sum_{n=1}^\infty\zeta(2n)x^{2n-1}; $$ taking derivatives we get $$ \frac{\pi^2}{\sin^2\pi x}=x^{-2}+2\sum_{n=1}^\infty(2n-1)\zeta(2n)x^{2n-2}; $$ in particular for $x=\frac12$ $$ \pi^2=4+2\sum_{n=1}^\infty(2n-1)\zeta(2n)\frac1{2^{2n-2}}; $$ this is your formula (multiplied by 4).

Solution 2:

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$\pars{\pi \over 2}^{2} = 1 + 2\sum_{k=1}^{\infty}{\pars{2k - 1}\zeta\pars{2k} \over 2^{2k}}:\ {\Large ?} \tag{1} $$

\begin{align} &\color{#00f}{\large\sum_{k=1}^{\infty}{\pars{2k - 1}\zeta\pars{2k} \over 2^{2k}}}= \sum_{k=1}^{\infty}{2k - 1 \over 2^{2k}}\sum_{\ell = 1}^{\infty}{1 \over \ell^{2k}} =\sum_{\ell = 1}^{\infty}\sum_{k=1}^{\infty}\pars{2k - 1}\pars{1 \over 2\ell}^{2k} \\[3mm]&=\sum_{\ell = 1}^{\infty}\bracks{x^{2} \totald{}{x}\sum_{k=1}^{\infty}x^{2k - 1}}_{x = 1/\pars{2\ell}} =\sum_{\ell = 1}^{\infty}\braces{x^{2}\, \totald{}{x}\bracks{x \over 1 - x^{2}}}_{x = 1/\pars{2\ell}} \\[3mm]&=\sum_{\ell = 1}^{\infty}\bracks{x^{2}\, {1 + x^{2} \over \pars{1 - x^{2}}^{2}}}_{x = 1/\pars{2\ell}} =\sum_{\ell = 1}^{\infty}\bracks{ {1/x^{2} + 1 \over \pars{1/x^{2} - 1}^{2}}}_{x = 1/\pars{2\ell}} \\[3mm]&=\sum_{\ell = 1}^{\infty}\bracks{ {1 \over 1/x^{2} - 1} + {2 \over \pars{1/x^{2} - 1}^{2}}}_{x = 1/\pars{2\ell}} =\left.\pars{2\,\totald{}{\mu} + 1}\sum_{\ell = 1}^{\infty}{1 \over 4\ell^{2} - \mu} \right\vert_{\mu = 1} \\[3mm]&=\pars{2\,\totald{}{\mu} + 1}\bracks{% {2 - \root{\mu}\pi\cot\pars{\root{\mu}\pi/2}} \over 4\mu}_{\mu = 1} =\color{#00f}{\large\half\bracks{\pars{\pi \over 2}^{2} - 1}}\tag{2} \end{align}

Replace $\pars{2}$ in $\pars{1}$.