2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

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}

$\endgroup$
0
$\begingroup$

\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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.