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.

  • $\begingroup$ Where did this come up? Do you have reason to believe it has a closed form? Also, what techniques have you tried to use to solve this problem? $\endgroup$ – Carl Schildkraut Nov 14 '18 at 18:04
  • $\begingroup$ Thanks! This came up when I was trying to calculate the steady state probability of a Markov Chain. I am not sure whether the closed form exists, and I am pretty new to combinatorics/series. To be honest I tried to break down the terms or find some links to the binomial coefficients, but all efforts failed. Any help will be extremely appreciated. @CarlSchildkraut $\endgroup$ – huighlh Nov 14 '18 at 18:19
  • 1
    $\begingroup$ A continued fraction representation can be obtained from Euler's continued fraction formula $\endgroup$ – Paul Enta Nov 14 '18 at 18:25
  • 3
    $\begingroup$ The sum equals to ${}_1F_1\left(1;\frac{1}{\gamma}+1;\frac{1}{\gamma}\right) - 1$. where ${}_1F_1(a;b;z)$ is a generialized hypergeometric function (also known as confluent hypergeometric function of the first kind $M(a;b;z)$ ). For the few cases I tested, WA suggest this can be expressed in terms of gamma and incomplete gamma function. $\endgroup$ – achille hui Nov 14 '18 at 18:30
  • $\begingroup$ Thanks so much! @achillehui $\endgroup$ – huighlh Nov 14 '18 at 18:41

First let's see the product:


Using the fact that $\Gamma(z)= \frac{\Gamma(z+n+1)}{z(z+1)....(z+n)}$ we have:


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 $S\rightarrow e-2$

  • $\begingroup$ (+1) I didn't notice $S=(\gamma e)^{\frac{1}{\gamma}}\Gamma_L(1+\frac{1}{\gamma},\frac{1}{\gamma})$ in your answer until I searched more carefully. $\endgroup$ – robjohn Nov 22 '18 at 19:57
  • $\begingroup$ @robjohn Thank you, $\endgroup$ – JV.Stalker Nov 22 '18 at 21:07

$$ \begin{align} \sum_{k=1}^\infty\left(\prod_{m=1}^k\frac1{1+m\gamma}\right) &=\sum_{k=1}^\infty\gamma^{-k}\frac{\Gamma\!\left(\frac1\gamma+1\right)}{\Gamma\!\left(k+\frac1\gamma+1\right)}\\ &=\frac1\gamma\sum_{k=1}^\infty\gamma^{1-k}\frac{\Gamma\!\left(\frac1\gamma+1\right)}{\Gamma\!\left(k+\frac1\gamma+1\right)}\frac{\Gamma(k)}{(k-1)!}\\ &=\frac1\gamma\sum_{k=1}^\infty\frac{\gamma^{1-k}}{(k-1)!}\int_0^1t^{k-1}(1-t)^{1/\gamma}\,\mathrm{d}t\\[3pt] &=\frac1\gamma\int_0^1e^{t/\gamma}(1-t)^{1/\gamma}\,\mathrm{d}t\\[6pt] &=\frac{e^{1/\gamma}}\gamma\int_0^1e^{-t/\gamma}t^{1/\gamma}\,\mathrm{d}t\\ &=(e\gamma)^{1/\gamma}\int_0^{1/\gamma}e^{-t}t^{1/\gamma}\,\mathrm{d}t\\[6pt] &=(e\gamma)^{1/\gamma}\Gamma\!\left(\tfrac1\gamma+1,\tfrac1\gamma\right) \end{align} $$ where the two variable $\Gamma(a,b)=\int_0^be^{-t}t^{a-1}\,\mathrm{d}t$ is the Lower Incomplete Gamma Function.


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