What's the series of $\sum_{n\geqslant1} \dfrac{\zeta(2n)}{n2^{2n}}$.

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}

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.