# An Infinite Product

1. May 28, 2006

### benorin

Prof. Putinar,

Guillera & Sondow gave

$$e^{x}=\prod_{n=1}^{\infty}\left(\prod_{k=1}^{n} (1+kx)^{(-1)^{k+1}\left(\begin{array}{c}n\\k\end{array}\right)} \right) ^{\frac{1}{n}}$$​

for $$x\geq 0$$, to which I add

$$\boxed{\frac{e^{u}\Gamma (u)}{\sqrt{2\pi e}}=\prod_{n=0}^{\infty}\left(\prod_{k=0}^{n} (k+u)^{(-1)^{k}\left(\begin{array}{c}n\\k\end{array}\right) (k+u)} \right) ^{\frac{1}{n+1}}}$$​

for $$\mbox{Re} \geq 0$$.

-Ben Orin

2. May 28, 2006

### Zurtex

Cool.

:uhh:

3. May 29, 2006

### benorin

4. Jun 27, 2011

### benorin

An Infinite Product for e^x

Edit: Fixing my post for TeX and updating link to paper.

Prof. Putinar,

Guillera & Sondow1 gave $e^{x} = \prod_{n=1}^{\infty}\left( \prod_{k=1}^{n} (kx+1) ^{(-1)^{k+1} \left( \begin{array}{c}n\\k\end{array}\right) } \right) ^{\frac{1}{n}}\mbox{ for }x\geq 0,$
to it's company I add $\frac{e^{u}\Gamma (u)}{\sqrt{2\pi e}}=\prod_{n=0}^{\infty}\left(\prod_{k=0}^{n}(k+u)^{ (-1)^{k}\left(\begin{array}{c}n\\k\end{array}\right)(k+u)}\right)^{\frac{1}{n+1}}\mbox{ for }\mbox{Re} \geq 0.$

-Ben Orin

benorin@umail.ucsb.edu

1 The infinite product for $e^{x}$