A closed form for $\sum _{j=0}^{\infty } -\frac{\zeta (-j)}{\Gamma (j)}$ 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?
 A: 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
$$1-\frac{1}{2\cosh1-2}.$$
A: 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 $n=1$

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

A: The reflection formula for the Riemann zeta function gives
$$
\begin{align}
\zeta(-k)
%&=\zeta(k+1)\frac{\Gamma\!\left(\frac{k+1}2\right)}{\pi^{\frac{k+1}2}}\frac{\pi^{-\frac{k}2}}{\Gamma\!\left(-\frac{k}2\right)}\frac{\Gamma\!\left(1+\frac{k}2\right)}{\Gamma\!\left(1+\frac{k}2\right)}\\
%&=\zeta(k+1)\frac{\Gamma\!\left(\frac{k+1}2\right)\Gamma\!\left(\frac{k+2}2\right)}{\pi^{\frac{2k+3}2}}\sin(-k\pi/2)\\
&=\zeta(k+1)\frac{\Gamma(k+1)}{2^k\pi^{k+1}}\sin(-k\pi/2)\tag1
\end{align}
$$
If $k$ is an even integer, $\sin(-k\pi/2)=0$. This matches the zeroes of the zeta function at the negative even integers. Let $k=2j-1$.
$$
\zeta(1-2j)
=(-1)^j\frac{\zeta(2j)}{\pi^{2j}}\frac{(2j-1)!}{2^{2j-1}}\tag2
$$
Therefore,
$$
\begin{align}
-\sum_{j=1}^\infty\frac{\zeta(1-2j)}{\Gamma(2j-1)}
&=\sum_{j=1}^\infty(-1)^{j-1}\frac{\zeta(2j)}{\pi^{2j}}\frac{2j-1}{2^{2j-1}}\\
&=\sum_{k=1}^\infty\sum_{j=1}^\infty(-1)^{j-1}\frac{\color{#C00}{4j}-\color{#090}{2}}{(2k\pi)^{2j}}\\
&=\sum_{k=1}^\infty\left(\color{#C00}{4\frac{4k^2\pi^2}{\left(4k^2\pi^2+1\right)^2}}-\color{#090}{2\frac1{4k^2\pi^2+1}}\right)\\
&=\sum_{k=1}^\infty\left(2\frac1{4k^2\pi^2+1}-4\frac1{\left(4k^2\pi^2+1\right)^2}\right)\tag3
\end{align}
$$
Using the formula from this answer
$$
\sum_{k=1}^\infty\frac1{k^2+x^2}
=\frac{\pi x\coth(\pi x)-1}{2x^2}\tag4
$$
and its derivative times $-\frac1{2x}$
$$
\sum_{k=1}^\infty\frac1{\left(k^2+x^2\right)^2}
=\frac{\pi^2x^2\operatorname{csch}^2(\pi x)+\pi x\coth(\pi x)-2}{4x^4}\tag5
$$
we get
$$
\sum_{k=1}^\infty\frac1{4k^2\pi^2+1}
=\frac{\frac12\coth\left(\frac12\right)-1}{2}\tag6
$$
and
$$
\sum_{k=1}^\infty\frac1{\left(4k^2\pi^2+1\right)^2}
=\frac{\frac14\operatorname{csch}^2\left(\frac12\right)+\frac12\coth\left(\frac12\right)-2}{4}\tag7
$$
Applying $(6)$ and $(7)$ to $(3)$ yields
$$
\bbox[5px,border:2px solid #C0A000]{-\sum_{k=1}^\infty\frac{\zeta(-k)}{\Gamma(k)}=1-\frac14\operatorname{csch}^2\left(\frac12\right)}\tag8
$$
