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

\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 $\pars{1}$.
A: Do not worry about the downvote. It is just an attack on my answers.
The series can have the following integral representation

$$ \sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k-1}} = \int_{0}^{1}\frac{(t^2+1)\ln t}{t^2-1}  dt = \frac{\pi^2}{4}-1.$$

