Proving infinite series involving the zeta-function How do I prove that
$$\sum_{n=1}^{\infty} \frac{\zeta(2n)}{n(2n+1)} = \ln\frac{2\pi}{e}$$
and $$\sum_{n=1}^∞ \frac{\zeta(2n)}{n(2n+1)4^n} = \ln\frac{\pi}{e}$$
I tried to use bernoulli numbers but failed. Could someone please help? Thanks a lot
 A: Let, for $\displaystyle x\in [0,1), \varphi(x)=\sum_{n=1}^{+\infty}\frac{\zeta(2n)}{n(2n+1)}x^{2n+1}$, then
$$ \begin{aligned} \varphi''(x)&=2\sum_{n=1}^{+\infty}\zeta(2n)x^{2n-1} \\
&=\frac{2}{x}\sum_{n=1}^{+\infty}\sum_{k=1}^{+\infty}\frac{x^{2n}}{k^{2n}} \\
&=\frac{2}{x}\sum_{k=1}^{+\infty}\sum_{n=1}^{+\infty}\frac{x^{2n}}{k^{2n}} \\
&=\frac{2}{x}\sum_{k=1}^{+\infty}\frac{\frac{x^2}{k^2}}{1-\frac{x^2}{k^2}} \\
&=2x\sum_{k=1}^{+\infty}\frac{1}{k^2-x^2} \\
&=\frac{1}{x}-\pi\cot(\pi x)
\end{aligned} $$
But $\displaystyle \int \cot(\pi x)dx=\frac{\log\sin(\pi x)}{\pi}+c$ thus
$$ \varphi'(x)=\log(x)-\log(\sin(\pi x))+c $$
Since $\varphi'(0)=0$ and $\log(\sin(\pi x))=\log(\pi x+o(x))=\log(\pi x)+\log(1+o(1))=\log(\pi x)+o(1)$, we have $c=\log\pi$ and thus, using $\varphi(0)=0$, we get
$$ \varphi(x)=x\log(x)+\log\left(\frac{\pi}{e}\right)x-\int_0^x\log(\sin(\pi x))dx $$
It only remains to show that
$$ \int_0^1\log(\sin(\pi x))dx=-\log 2 \text{ and } \int_0^{\frac{1}{2}}\log(\sin(\pi x))dx=-\frac{\log 2}{2} $$
However using the identity $\sin(\pi-x)=\sin(x)$, the substitution $x\mapsto 1-x$ gives that
$$ \int_{\frac{1}{2}}^1\log(\sin(\pi x))dx=\int_0^{\frac{1}{2}}\log(\sin(\pi x))dx $$
Let $\displaystyle I=\int_0^{\frac{1}{2}}\log(\sin(\pi x))dx$, then because of what said above, $\displaystyle \int_0^1\log(\sin(\pi x))dx=2I$, thus it only remains to show that $I=-\frac{\log 2}{2}$. Let $\displaystyle J=\int_0^{\frac{1}{2}}\log(\cos(\pi x))dx$, the substitution $x\mapsto\frac{1}{2}-x$ leads to $I=J$, thus
$$ 2I=I+J=\int_0^{\frac{1}{2}}\log(\sin(\pi x)\cos(\pi x))dx=\int_0^{\frac{1}{2}}\log(\sin(2\pi x))dx-\frac{\log 2}{2} $$
Substituting $x\mapsto 2x$ gives
$$ \int_0^{\frac{1}{2}}\log(\sin(2\pi x))dx=\frac{1}{2}\int_0^1 \log(\sin(\pi x))dx=I $$
because of what said above. Thus $2I=I-\frac{\log 2}{2}$ and $I=-\frac{\log 2}{2}$. This ends the proof.
