Avoiding the analytic continuation of extended binomial theorem, $$ \sum_{n=0}^{\infty}\frac{\Gamma(n+z)}{n!}\,x^n = \frac{\Gamma(z)}{(1-x)^z} \quad\colon\space |x|\lt1 $$

How to prove: $$ \sum_{n=0}^{\infty}\frac{\Gamma(n+s)}{n!} = 0 \quad\Rightarrow\, \frac{s}{1!}+\frac{s(s+1)}{2!}+\cdots = -1 \quad\colon\space Re\{s\}\lt0 $$


We find that

$$ \sum_{k=0}^{n} \frac{\Gamma(k+s)}{k!} = \frac{\Gamma(n+1+s)}{n!s} \tag{*}$$

for all $n = 0, 1, 2, \cdots$. Indeed, this is easily proved by the mathematical induction:

  1. When $n = 0$, it boils down to the equality $\Gamma(s) = \frac{\Gamma(1+s)}{s}$, which is of course true.

  2. Assuming that $\text{(*)}$ is true for $n \geq 0$, then

    \begin{align*} \sum_{k=0}^{n+1} \frac{\Gamma(k+s)}{k!} &= \frac{\Gamma(n+1+s)}{n!s} + \frac{\Gamma(n+1+s)}{(n+1)!} \\ &= \frac{(n+1+s)\Gamma(n+1+s)}{(n+1)!s} = \frac{\Gamma(n+2+s)}{(n+1)!s} \end{align*}

Therefore $\text{(*)}$ is true for all $n \geq 0$. Now the conclusion follows by taking $n\to\infty$. (Stirling's formula is enough for this purpose.)

Remark. The identity $\text{(*)}$ becomes more natural once we recognize it as a disguise of the famous formula

$$ \sum_{k=0}^{n} \binom{k+s-1}{k} = \binom{n+s}{n}. $$

When $s$ is a positive integer, this indeed follows from the hockey-stick argument.

  • $\begingroup$ Imho that is a bit too short $\endgroup$ – mick Feb 7 '17 at 1:56
  • $\begingroup$ @mick, Is there anything you want me to improve? $\endgroup$ – Sangchul Lee Feb 7 '17 at 1:59
  • $\begingroup$ More details about the induction etc $\endgroup$ – mick Feb 7 '17 at 2:04
  • $\begingroup$ Beauty is Simplicity (+1). However, for the limit, $\{\,\lim\frac{\Gamma(N+s)}{s\,\Gamma(N)}=0\,\colon{\small Re\{s\}\lt0}\,\}$ , I would use the identity $\{\,\lim\frac{\Gamma(N+s)}{N^s\,\Gamma(N)}=1\,\colon{\small s\in\mathbb{C}}\,\}$. Interestingly you mentioned other method, please expound. Thanks. $\endgroup$ – Hazem Orabi Feb 7 '17 at 20:54

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\Re\pars{s} < 0}$:

\begin{align} \left.\sum_{n = 0}^{\infty}{\Gamma\pars{n + s} \over n!} \,\right\vert_{\ \Re\pars{s}\ <\ 0} & = \pars{s - 1}!\sum_{n = 0}^{\infty}{n + s - 1 \choose n} = \pars{s - 1}!\sum_{n = 0}^{\infty}{-s \choose n}\pars{-1}^{n} \\[5mm] & = \pars{s - 1}!\,\bracks{1 + \pars{-1}}^{\,-s} = \bbx{\ds{0}} \end{align}


\begin{eqnarray*} \Gamma(z) = \int_0 ^{\infty} x^{z-1} e^{-x} dx \end{eqnarray*} So \begin{eqnarray*} \sum_{n=0}^{\infty} \frac{ \Gamma(n+s)}{n!} =\sum_{n=0}^{\infty} \int_0 ^{\infty} \frac{x^{n+s-1} e^{-x}}{n!} dx \end{eqnarray*} Now invert the sum & integral \begin{eqnarray*} \int_0 ^{\infty} x^{s-1} \sum_{n=0}^{\infty} \frac{x^{n} }{n!} e^{-x}dx =\int_0 ^{\infty} x^{s-1} dx = \left[ \frac{x^s}{s} \right]_0 ^{\infty} \end{eqnarray*} Now $x^s$ will tend to zero provided $ Re(s)< 0$.

\begin{eqnarray*} \Gamma(s+1) = s \Gamma(s) \\ \Gamma(s+2) = s(s+1) \Gamma(s) \\ \Gamma(s+n) = s(s+1) \cdots (s+n-1) \Gamma(s) \end{eqnarray*} Divide the equation by $\Gamma(s)$ and move the first term to the right hand side.

  • $\begingroup$ The integral is only valid for $Re\{z\}>0$ $\endgroup$ – Hazem Orabi Feb 7 '17 at 1:29
  • $\begingroup$ I had a feeling that might be a validity range issue. I also would have need to justify inverting the sum & integral. $\endgroup$ – Donald Splutterwit Feb 7 '17 at 1:33

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