I know with the formula $$1-\sum_{n\geq 1}2\zeta(2n)\,x^{2n}=\pi x\cot(\pi x)$$ may I find the following relation used here
$$ \sum_{n\geqslant1} \dfrac{\zeta(2n)}{n2^{2n}}=\color{blue}{\ln\dfrac{\pi}{2}} $$
hardly, since I have $$\int\sum_{n\geq 1}\zeta(2n)\,x^{n-1}dx=\int\left(\dfrac{1}{2x^{n+1}}-\dfrac{\pi}{2} \dfrac{\cot(\pi x)}{x^n}\right)dx$$ and after integration set $x=\dfrac14$, but it seems so hard.
Any suggestion, thanks in advanced!
On
Another (similar) approach, just for fun. From the integral representation of the Riemann Zeta function $$\zeta\left(s\right)=\frac{1}{\Gamma\left(s\right)}\int_{0}^{\infty}\frac{u^{s-1}}{e^{u}-1}du,\,\mathrm{Re}\left(s\right)>1$$ we have $$S=\sum_{n\geq1}\frac{\zeta\left(2n\right)}{n4^{n}}=\sum_{n\geq1}\frac{1}{n4^{n}\left(2n-1\right)!}\int_{0}^{\infty}\frac{u^{2n-1}}{e^{u}-1}du=\int_{0}^{\infty}\frac{e^{u/2}+e^{-u/2}-2}{u\left(e^{u}-1\right)}du$$ where the exchange is justified by the dominated convergence theorem. Then, by the Frullani's theorem, we get $$S=\sum_{m\geq1}\left(\int_{0}^{\infty}\frac{e^{-u\left(m-1/2\right)}-e^{-mu}}{u}dx+\int_{0}^{\infty}\frac{e^{-u\left(1/2+m\right)}-e^{-mu}}{u}dx\right)$$ $$=-\sum_{m\geq1}\log\left(1-\frac{1}{4m^{2}}\right)$$ and so the claim by the Wallis product.
On
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\sum_{n \geqslant 1}{\zeta\pars{2n} \over n\, 2^{2n}} = \color{blue}{\ln\pars{\pi \over 2}}:\ {\LARGE ?}}$.
Lets start with the identities: $$ \left\{\begin{array}{rcl} \ds{\Psi\pars{1 + z}} & \ds{=} & \ds{-\gamma + \sum_{n = 2}^{\infty}\pars{-1}^{n}\,\zeta\pars{n}z^{n - 1}} \\ \ds{\Psi\pars{1 - z}} & \ds{=} & \ds{-\gamma - \sum_{n = 2}^{\infty}\zeta\pars{n}z^{n - 1}} \end{array}\right.\,,\qquad\qquad \verts{z} < 1 $$ \begin{align} &\mbox{Then,}\quad\Psi\pars{1 + z} - \Psi\pars{1 - z} = 2\sum_{n = 1}^{\infty}\zeta\pars{2n}z^{2n - 1} \end{align} Integrate the above expression over $\ds{\pars{0,1/2}}$: $$ \ln\pars{\Gamma\pars{3 \over 2}\Gamma\pars{1 \over 2}} = 2\sum_{n = 1}^{\infty}\zeta\pars{2n}{\pars{1/2}^{2n} \over 2n} $$ $$ \sum_{n \geqslant 1}{\zeta\pars{2n} \over n\, 2^{2n}} = \ln\pars{\bracks{{1 \over 2}\,\Gamma\pars{1 \over 2}}\Gamma\pars{1 \over 2}} = \bbx{\ln\pars{\pi \over 2}} $$
because $\ds{\Gamma\pars{1/2} = \root{\pi}}$.
Using your formula we have $$ \sum \limits_{n=1}^\infty \frac{\zeta(2n)}{n 2^{2n}} = \int \limits_0^{1/2} \sum \limits_{n=1}^\infty 2 \zeta(2n) x^{2n-1} \, \mathrm{d} x = \int \limits_0^{1/2} \frac{1-\pi x \cot(\pi x)}{x} \, \mathrm{d} x \, .$$ Now let $\pi x = t$ and integrate: $$ \sum \limits_{n=1}^\infty \frac{\zeta(2n)}{n 2^{2n}} = \lim_{\varepsilon \searrow 0} \int \limits_\varepsilon^{\pi/2} \left[\frac{1}{t} - \cot(t)\right] \, \mathrm{d} t = \lim_{\varepsilon \searrow 0} \left[\ln\left(\frac{t}{\sin(t)}\right)\right]_\varepsilon^{\pi /2} = \ln \left(\frac{\pi}{2}\right) \, .$$
Alternatively you can of course compute the series directly using Wallis' product: \begin{align} \sum \limits_{n=1}^\infty \frac{\zeta(2n)}{n 2^{2n}} &= \sum \limits_{n=1}^\infty \frac{1}{n 2^{2n}} \sum \limits_{k=1}^\infty \frac{1}{k^{2n}} = \sum \limits_{k=1}^\infty \sum \limits_{n=1}^\infty \frac{1}{n (4k^2)^n} = \sum \limits_{k=1}^\infty - \ln\left(1-\frac{1}{4k^2}\right) \\ &= \sum \limits_{k=1}^\infty \ln \left(\frac{4k^2}{4k^2 -1}\right) = \ln \left(\prod \limits_{k=1}^\infty \frac{4k^2}{4k^2 -1} \right) = \ln \left(\frac{\pi}{2}\right) \, . \end{align}