What's the series of $\sum_{n\geqslant1} \dfrac{\zeta(2n)}{n2^{2n}}$. 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!
 A: 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}
A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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$\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 $\ds{\Gamma\pars{1/2} = \root{\pi}}$.

