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?

  • $\begingroup$ I have tried with math software and it can't either. What makes you think that it has a closed form? First digits: $0.07932640579$ $\endgroup$ – ajotatxe Feb 7 '17 at 15:07
  • $\begingroup$ Just a point of interest, but you can replace the Gamma function and zeta function with the factorial and Bernoulli numbers respectively. $\endgroup$ – Simply Beautiful Art Feb 7 '17 at 15:07
  • $\begingroup$ It's interesting that the three answers so far, which give the same value, have such different appearances. $$1-\frac{1}{2\cosh1-2}$$ $$1-\frac{e}{(1-e)^2}$$ $$1-\frac14\operatorname{csch}^2\left(\frac12\right)$$ $\endgroup$ – robjohn Oct 29 '18 at 5:59

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 $$ \bbox[5px,border:2px solid #C0A000]{-\sum_{k=1}^\infty\frac{\zeta(-k)}{\Gamma(k)}=1-\frac14\operatorname{csch}^2\left(\frac12\right)}\tag8 $$


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 $$1-\frac{1}{2\cosh1-2}.$$

  • 1
    $\begingroup$ ...you skipped a few steps there sir... $\endgroup$ – Simply Beautiful Art Feb 7 '17 at 15:17
  • $\begingroup$ (It's an awful lot of steps to not include in the answer) $\endgroup$ – Simply Beautiful Art Feb 7 '17 at 15:24

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 $n=1$

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


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