sum of an infinite series $\sum_{k=1}^\infty \left( \prod_{m=1}^k\frac{1}{1+m\gamma}\right) $ I am trying to find a closed form expression of
$$
\sum_{k=1}^\infty \left( \prod_{m=1}^k\frac{1}{1+m\gamma}\right)
$$
where $\gamma>1$.
I've been trying this for a long time. Is there an easy way to tackle this problem? Any hint will be extremely helpful. Thanks!
Edit: changing variable name because $i$ stands form the imaginary constant.
 A: $$
\begin{align}
\sum_{k=1}^\infty\left(\prod_{m=1}^k\frac1{1+m\gamma}\right)
&=\sum_{k=1}^\infty\gamma^{-k}\frac{\Gamma\!\left(\frac1\gamma+1\right)}{\Gamma\!\left(k+\frac1\gamma+1\right)}\\
&=\frac1\gamma\sum_{k=1}^\infty\gamma^{1-k}\frac{\Gamma\!\left(\frac1\gamma+1\right)}{\Gamma\!\left(k+\frac1\gamma+1\right)}\frac{\Gamma(k)}{(k-1)!}\\
&=\frac1\gamma\sum_{k=1}^\infty\frac{\gamma^{1-k}}{(k-1)!}\int_0^1t^{k-1}(1-t)^{1/\gamma}\,\mathrm{d}t\\[3pt]
&=\frac1\gamma\int_0^1e^{t/\gamma}(1-t)^{1/\gamma}\,\mathrm{d}t\\[6pt]
&=\frac{e^{1/\gamma}}\gamma\int_0^1e^{-t/\gamma}t^{1/\gamma}\,\mathrm{d}t\\
&=(e\gamma)^{1/\gamma}\int_0^{1/\gamma}e^{-t}t^{1/\gamma}\,\mathrm{d}t\\[6pt]
&=(e\gamma)^{1/\gamma}\Gamma\!\left(\tfrac1\gamma+1,\tfrac1\gamma\right)
\end{align}
$$
where the two variable $\Gamma(a,b)=\int_0^be^{-t}t^{a-1}\,\mathrm{d}t$ is the Lower Incomplete Gamma Function.
A: First let's see the product: 
$P=\prod\limits_{m=1}^{k}\big({1+m\gamma}\big)^{-1}=\gamma^{-k}\prod\limits_{m=1}^k\big(\frac{1}{\gamma}+m\big)^{-1}=\gamma^{-(k+1)}\prod\limits_{m=0}^k\big(\frac{1}{\gamma}+m\big)^{-1}$
Using the fact that $\Gamma(z)= \frac{\Gamma(z+n+1)}{z(z+1)....(z+n)}$ we have:
$P=\gamma^{-(k+1)}\frac{\Gamma(\frac{1}{\gamma})}{\Gamma(\frac{1}{\gamma}+k+1)}$
Return to the original expression and use the factorial form inside the sum: 
$S=\sum\limits_{k=1}^\infty \gamma^{-(k+1)}\frac{\Gamma(\frac{1}{\gamma})}{\Gamma(\frac{1}{\gamma}+k+1)}=\Gamma(1+\frac{1}{\gamma})\sum\limits_{k=1}^\infty \frac{(\frac{1}{\gamma})^k}{\Gamma({1+\frac{1}{\gamma}+k})}$
$S=-1+\Gamma(1+\frac{1}{\gamma})\sum\limits_{k=0}^\infty \frac{(\frac{1}{\gamma})^k}{\Gamma({1+\frac{1}{\gamma}+k})}$
On the other hand we have to know that: 
$\Gamma_L(s,x)=x^s \Gamma(s) e^{-x}\sum\limits_{k=0}^\infty \frac{x^k}{\Gamma({1+s+k})}$, where $\Gamma_L(s,x)$ is the lower incomplete gamma function.
Let $s=\frac{1}{\gamma}$ and $x=\frac{1}{\gamma}$ we can express the: 
$\sum\limits_{k=0}^\infty \frac{(\frac{1}{\gamma})^k}{\Gamma({1+\frac{1}{\gamma}+k})}=\gamma^{\frac{1}{\gamma}}e^{\frac{1}{\gamma}} \frac{\Gamma_L(\frac{1}{\gamma},\frac{1}{\gamma})}{\Gamma(\frac{1}{\gamma})}$
Substiture back into the last expression of S we get:
$S=(\gamma e)^{\frac{1}{\gamma}}\frac{\Gamma_L(\frac{1}{\gamma},\frac{1}{\gamma})}{\Gamma(\frac{1}{\gamma})}\Gamma(1+\frac{1}{\gamma})-1$
We could reach further simplifications if we use the recurrance relation of lower incomplete gamma function: 
$\Gamma_L(s+1,x)=s\Gamma_L(s,x)-x^s e^{-x}$
Finally we have that: 
$S=(\gamma e)^{\frac{1}{\gamma}}\Gamma_L(1+\frac{1}{\gamma},\frac{1}{\gamma})$
We can check it in $\gamma=1$ and $\infty$ places:


*

*If $\gamma \rightarrow \infty$ then easy to see $S \rightarrow 0$ $\big(\Gamma_L(1,0)=0\big)$

*If $\gamma \rightarrow 1$ then $S\rightarrow e\Gamma_L(2,1)$
$\Gamma_L(2,1)$ can be calculated by the definition of lower incomplete gamma function: 
$\Gamma_L(2,1)=\int \limits_0^1 te^{-t}dt=-2e^{-1}+1$
So $S\rightarrow e-2$
