$$\frac {\zeta(2)}{e^2} + \frac {\zeta(3)}{e^3} + \frac {\zeta(4)}{e^4} + \frac {\zeta(5)}{e^5}.....?$$
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 $\frac {\zeta(2)}{e^2} + \frac {\zeta(3)}{e^3} + \frac {\zeta(4)}{e^4} + \frac {\zeta(5)}{e^5}.....$
You can follow the same method and use Maclaurin's Series for $\log$. eventually, you'll end up to Taylor's Series form of Digamma function
$$S = \sum_\color{red}{n = 2}^{\infty} \frac {\zeta(n)}{e^n}$$
Let's consider the Taylor Series form of Digamma function:
$$\begin{align*} \psi_0(z) &= -\gamma + \sum_\color{red}{n = 1}^{\infty} (-1)^{n+1} \zeta(n+1)(z-1)^{n}\\ & = -\gamma + \sum_\color{red}{n = 2}^{\infty} (-1)^{n} \zeta(n)(z-1)^{n-1}\\ & \text{Let multiply by } {(z-1)} \text{ & } {(z-1) = -e^{-1}}\\ & \implies -{(z-1)}\gamma + \sum_{n=2}^{\infty} (-1)^{n} \zeta(n)(z-1)^{{n}}= {(z-1)}\times \psi_0(z)\\ & \implies \sum_{n=2}^{\infty}{(-1)^{n}} \zeta(n){(-e^{-n})} = \color{blue}{\sum_{n = 2}^{\infty} \frac {\zeta(n)}{e^n}}=\color{blue}{-e^{-1}(\psi_0(1-e^{-1}) + \gamma)} \\ \end{align*}$$
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.
Too long for a comment.
Let us start from the Maclaurin series in the form of $\;\psi(1+z)=-\gamma+\sum\limits_{n=2}^\infty (-1)^n \zeta(n) z^{n-1}:$ $$\psi(1-e^{-1})=-\gamma - \sum\limits_{n=2}^\infty \dfrac{\zeta(n)}{e^{n-1}},$$ $$\sum\limits_{n=2}^\infty \dfrac{\zeta(n)}{e^n} = -e^{-1}\left(\psi(1-e^{-1})+\gamma\right).$$
