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.
HINT (if you're familiar with the gamma function):
$$x=\frac{1}{\sum_{n=1}^{m}\frac{e^{-m}m^n}{n!}}=\frac{1}{\frac{\Gamma(m+1,m)}{m\Gamma(m)}-e^{-m}}=\frac{e^mm\Gamma(m)}{-m\Gamma(m)+e^m\Gamma(1+m,m)}=$$ $$-\frac{e^mm!}{m!-e^m\Gamma(m+1,m)}$$
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}) $.
The problem here is that I don't know how to proceed. I am not looking for an answer to this specific problem, I am looking more for any tips or help as to how we proceed with such questions.
– Nikos Koukos Nov 29 '15 at 21:45