Computing $\sum_{n=1}^{\infty}\frac{n \zeta(n+1)}{2}=\frac{\pi^2}{4}$ I saw this result and tried to proof, but I probably made a mistake throughout the computation and I cannot spot where it is. Can someone please help me?
$$S=\sum_{n=1}^{\infty}\frac{n \zeta(n+1)}{2}=\frac{\pi^2}{4}$$
I started by the integral representation
$$\frac{1}{k^{n+1}}=\frac{1}{\Gamma(n+1)}\int_{0}^{\infty}e^{-ku}u^n du $$
$$S=\frac{1}{2}\sum_{n=1}^{\infty}n\sum_{k=1}^{\infty}\frac{1}{\Gamma(n+1)}\int_{0}^{\infty}e^{-ku}u^n du $$
Supposing I can swap integration and summation
$$S=\frac{1}{2}\sum_{k=1}^{\infty}\int_{0}^{\infty}e^{-ku}\sum_{n=1}^{\infty}\frac{n}{n!}u^n du\\
=\frac{1}{2}\sum_{k=1}^{\infty}\int_{0}^{\infty}e^{-ku}u\sum_{n=1}^{\infty}\frac{u^{n-1}}{(n-1)!} du\\
=\frac{1}{2}\sum_{k=1}^{\infty}\int_{0}^{\infty}ue^{-ku}e^udu\\
=\frac{1}{2}\sum_{k=1}^{\infty}\int_{0}^{\infty}ue^{-(k-1)u}du\\
=\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{(k-1)^2}\\$$
which clearly is not the right answer!
 A: As it stands, the series does not converge (note K. defaoite's observation) as $\zeta(n)\to1$ for $n\to\infty$ and consequently $n\zeta(n+1)\to\infty$. If, however, we use $\zeta(n+1)-1$ instead we obtain a converging series (found by Mason). Then there also is an easier way than using integrals though:
$$
\sum_{n\ge1} n(\zeta(n+1)-1)=\sum_{n\ge1}\sum_{k\ge2}\frac n{k^{n+1}}=\sum_{k\ge2}\frac1{(k-1)^2}=\frac{\pi^2}6
$$
Your computations are fine. Assuming not given convergence the same argument as above applies and gives you the same final series as you have obtained via integration.
A: The sum does not converge: For $n>0$ one has $\zeta(n+1)>1$, therefore $\sum_{n=1}^{\infty}\frac{n\zeta(n+1)}{2}>\sum_{n=1}^{\infty}\frac{n}{2}=\infty\neq\frac{\pi^{2}}{4}$.
Your calculations however are perfectly fine.
There are several formulae involving fractional multiples of $\pi^{2}$ such as
\begin{equation*}
    \begin{matrix}\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}\text{,} &  
    \sum_{n=1}^{\infty}\frac{1}{(2n+1)^{2}}=\frac{\pi^{2}}{8}\text{,} & \text{...}\end{matrix}
\end{equation*}
but yours is wrong, unless $\zeta$ is supposed to mean something different than usual. Do not believe something must be true just because someone said it on the internet!
