A closed form for $\sum _{j=0}^{\infty } -\frac{\zeta (-j)}{\Gamma (j)}$

Question

Is there a closed form for

$$\sum _{j=0}^{\infty } -\frac{\zeta (-j)}{\Gamma (j)}$$

where $\zeta (-j)$ Zeta function and $\Gamma (j)$ Gamma function.

I tried everything, but I still can not solve it. Any Ideas?

Answer 1

Reflection formula for $\zeta(s)$ transforms the sum into $\sum_{n=1}^{\infty}\left(2\pi i\right)^{-2n}\left(2-4n\right)\zeta\left(2n\right)$. The latter can be computed by differentiating the well-known generating function $\sum_{n=0}^{\infty}\zeta\left(2n\right)z^{2n}=-\frac{\pi z\cot \pi z}{2}$, with the result $$1-\frac{1}{2\cosh1-2}.$$

Answer 2

Using this identity:

$$\sum _{j=0}^{\infty } -\frac{x^j \zeta (-j)}{\Gamma (1+j-n)}=\frac{(-1)^{1+n} \Gamma (1+n)}{x}+(-1)^{1+2 n} x^n \text{Li}_{-n}\left(e^x\right)$$ $ n\geq 1$

where:$\text{Li}_n(x)$ is polylogarithm function.

for $x=1$, and $n=1$

$$\sum _{j=0}^{\infty } -\frac{\zeta (-j)}{\Gamma (j)}=1-\frac{e}{(1-e)^2}$$