<s>[INST] $\frac {\zeta(2)}{e^2} + \frac {\zeta(3)}{e^3} + \frac {\zeta(4)}{e^4} + \frac {\zeta(5)}{e^5}.....?$ $$\frac {\zeta(2)}{e^2} + \frac {\zeta(3)}{e^3} + \frac {\zeta(4)}{e^4} + \frac {\zeta(5)}{e^5}.....?$$
I tried to solve it by using this product formula,
$$\frac 1{\Gamma (x)}=xe^{\gamma x} \prod_{n=1}^{\infty} \left(1+\frac x{n}\right)e^{-\frac x{n}}$$
I tried by differentiating after taking  $\log$
I'm interested in finding the answer for the above summation $\frac {\zeta(2)}{e^2} + \frac {\zeta(3)}{e^3} + \frac {\zeta(4)}{e^4} + \frac {\zeta(5)}{e^5}.....$
 [/INST] Moreover,
\begin{align*}
 \sum_{n=2}^{\infty} \frac{\zeta(n)}{k^{n}}&= \frac{1}{k} \sum_{n=1}^{\infty} \frac{1}{n!k^n} \Gamma(n+1)\zeta(n+1) \\&= \frac{1}{k} \sum_{n=1}^{\infty} \frac{1}{n!k^n} \int_{0}^{\infty} \frac{x^n}{e^{x}-1} \, dx \\&= \frac{1}{k}\int_{0}^{\infty} \frac{e^{x/k}-1}{e^{x}-1} \, dx \\&= - \frac{1}{k}\left(\gamma + \psi\left(1 - \frac{1}{k}\right) \right)
\end{align*}
where $\psi$ is the digamma function.
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