Question

$$\frac {\zeta(2)}{e^2} + \frac {\zeta(3)}{e^3} + \frac {\zeta(4)}{e^4} + \frac {\zeta(5)}{e^5}.....?$$ 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 $\frac {\zeta(2)}{e^2} + \frac {\zeta(3)}{e^3} + \frac {\zeta(4)}{e^4} + \frac {\zeta(5)}{e^5}.....$

Answer

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 $$ S = \sum_{n = 2}^{\infty} \frac{\zeta(n)}{e^n} $$

Let us consider the Taylor Series form of Digamma function:

$$ \begin{aligned} \psi_0(z) &= -\gamma + \sum_{n = 1}^{\infty} (-1)^{n+1} \zeta(n+1)(z-1)^{n} \\ &= -\gamma + \sum_{n = 2}^{\infty} (-1)^{n} \zeta(n)(z-1)^{n-1} \\ &\text{Let us multiply by } (z-1) \text{ where } (z-1) = -e^{-1}, \\ &\implies -(z-1)\gamma + \sum_{n=2}^{\infty} (-1)^{n} \zeta(n)(z-1)^{n} = (z-1)\times \psi_0(z), \\ &\implies \sum_{n=2}^{\infty} (-1)^{n} \zeta(n)(-e^{-n}) = \color{blue}{\sum_{n = 2}^{\infty} \frac{\zeta(n)}{e^n}} \\ &= \color{blue}{-e^{-1}(\psi_0(1-e^{-1}) + \gamma)}. \end{aligned} $$