Krdytkyu

I am trying to find a closed form expression of
$$
sum_{k=1}^infty left( prod_{m=1}^kfrac{1}{1+mgamma}right)
$$
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.

First let's see the product:

$P=prodlimits_{m=1}^{k}big({1+mgamma}big)^{-1}=gamma^{-k}prodlimits_{m=1}^kbig(frac{1}{gamma}+mbig)^{-1}=gamma^{-(k+1)}prodlimits_{m=0}^kbig(frac{1}{gamma}+mbig)^{-1}$

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:

$S=sumlimits_{k=1}^infty gamma^{-(k+1)}frac{Gamma(frac{1}{gamma})}{Gamma(frac{1}{gamma}+k+1)}=Gamma(1+frac{1}{gamma})sumlimits_{k=1}^infty frac{(frac{1}{gamma})^k}{Gamma({1+frac{1}{gamma}+k})}$

$S=-1+Gamma(1+frac{1}{gamma})sumlimits_{k=0}^infty frac{(frac{1}{gamma})^k}{Gamma({1+frac{1}{gamma}+k})}$

On the other hand we have to know that:

$Gamma_L(s,x)=x^s Gamma(s) e^{-x}sumlimits_{k=0}^infty frac{x^k}{Gamma({1+s+k})}$, 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:

$sumlimits_{k=0}^infty frac{(frac{1}{gamma})^k}{Gamma({1+frac{1}{gamma}+k})}=gamma^{frac{1}{gamma}}e^{frac{1}{gamma}} frac{Gamma_L(frac{1}{gamma},frac{1}{gamma})}{Gamma(frac{1}{gamma})}$

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:

$Gamma_L(s+1,x)=sGamma_L(s,x)-x^s e^{-x}$

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 $S rightarrow 0$ $big(Gamma_L(1,0)=0big)$




• If $gamma rightarrow 1$ then $Srightarrow eGamma_L(2,1)$

$Gamma_L(2,1)$ 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 $Srightarrow e-2$

$$
begin{align}
sum_{k=1}^inftyleft(prod_{m=1}^kfrac1{1+mgamma}right)
&=sum_{k=1}^inftygamma^{-k}frac{Gamma!left(frac1gamma+1right)}{Gamma!left(k+frac1gamma+1right)}\
&=frac1gammasum_{k=1}^inftygamma^{1-k}frac{Gamma!left(frac1gamma+1right)}{Gamma!left(k+frac1gamma+1right)}frac{Gamma(k)}{(k-1)!}\
&=frac1gammasum_{k=1}^inftyfrac{gamma^{1-k}}{(k-1)!}int_0^1t^{k-1}(1-t)^{1/gamma},mathrm{d}t\[3pt]
&=frac1gammaint_0^1e^{t/gamma}(1-t)^{1/gamma},mathrm{d}t\[6pt]
&=frac{e^{1/gamma}}gammaint_0^1e^{-t/gamma}t^{1/gamma},mathrm{d}t\
&=(egamma)^{1/gamma}int_0^{1/gamma}e^{-t}t^{1/gamma},mathrm{d}t\[6pt]
&=(egamma)^{1/gamma}Gamma!left(tfrac1gamma+1,gammaright)
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.