Find x so that summation equal to 1 I am trying to find the value of x so that this equation is true:
$$x = \frac{1}{\sum_{i=1}^{p} \dfrac{e^{-p} p^{i}}{i!}}$$
Another condition is that $$\frac{x e^{-p}p^{i}}{i!}$$
Should be between 0 and 1 (inclusive).
I have tried some things but I really don't get how to proceed.
Any tips are welcome. Thanks.
 A: HINT (if you're familiar with the gamma function):
$$x=\frac{1}{\sum_{n=1}^{m}\frac{e^{-m}m^n}{n!}}=\frac{1}{\frac{\Gamma(m+1,m)}{m\Gamma(m)}-e^{-m}}=\frac{e^mm\Gamma(m)}{-m\Gamma(m)+e^m\Gamma(1+m,m)}=$$
$$-\frac{e^mm!}{m!-e^m\Gamma(m+1,m)}$$
A: As been mentioned here
a number of times,
a result of Ramanujan
(isn't everything?)
states that
$e^{-p}\sum_{i=0}^{p} \dfrac{ p^{i}}{i!}
\approx \frac12
$
for large $p$.
More precisely,
$\frac12 e^n
=\sum_{k=0}^{n-1} \frac{n^k}{k!}+t(n)\frac{n^n}{n!}
$
where
$t(n)
\approx \frac13 +O(1/n)
$.
Since
$\frac{n^n}{n!}
\approx \frac{n^n}{\sqrt{2\pi n}n^n/e^n}
\approx \frac{e^n}{\sqrt{2\pi n}}
$,
$\begin{array}\\
\sum_{k=1}^{n} \frac{n^k}{k!}
&=-1+\sum_{k=0}^{n-1} \frac{n^k}{k!}+\frac{n^n}{n!}\\
&\approx -1+\frac12 e^n-t(n)\frac{n^n}{n!}+\frac{n^n}{n!}\\
&\approx -1+\frac12 e^n+(1-t(n))\frac{n^n}{n!}\\
&\approx -1+\frac12 e^n+\frac23\frac{e^n}{\sqrt{2\pi n}}\\
\text{so that}\\
e^{-n}\sum_{k=1}^{n} \frac{n^k}{k!}
&\approx -e^{-n}+\frac12+\frac23\frac{1}{\sqrt{2\pi n}}\\
&\approx \frac12+\frac23\frac{1}{\sqrt{2\pi n}}\\
\end{array}
$
So you want
$x 
\approx \frac1{\frac12+O(1/\sqrt{p})}
= \frac{2}{1+O(1/\sqrt{p})}
\approx 2+O(1/\sqrt{p})
$. 
