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$$\lim \limits_{n\mathop\to\infty}\frac{1}{e^n}\sum \limits_{k\mathop=0}^n\frac{n^k}{k!} $$

I thought this limit was obviously $1$ at first but approximations on Mathematica tells me it's $1/2$. Why is this?

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    $\begingroup$ Very heuristically speaking, stopping the summation at $\frac{n^n}{n!}$ is stopping at the maximal summand, so it is somewhat reasonable to expect that the contribution up to that summand might be about the same as the contribution of the remaining summands. $\endgroup$ – Hagen von Eitzen Mar 31 '14 at 9:43
  • $\begingroup$ @HagenvonEitzen. Good point, indeed ! Cheers. $\endgroup$ – Claude Leibovici Mar 31 '14 at 9:46
  • $\begingroup$ This exact same question has been asked before here on this site. $\endgroup$ – Lucian Mar 31 '14 at 9:53
  • $\begingroup$ Well if they made it easy to search for questions then that wouldn't be a problem. $\endgroup$ – user85798 Mar 31 '14 at 9:55
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    $\begingroup$ This looks like a duplicate of this question. $\endgroup$ – robjohn Mar 31 '14 at 10:06
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In this answer, it is shown that $$ \begin{align} e^{-n}\sum_{k=0}^n\frac{n^k}{k!} &=\frac{1}{n!}\int_n^\infty e^{-t}\,t^n\,\mathrm{d}t\\ &=\frac12+\frac{2/3}{\sqrt{2\pi n}}+O(n^{-1}) \end{align} $$

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You made a mistake since the summation goes to $n$ and not to infinity; then $$\sum \limits_{k\mathop=0}^n\frac{n^k}{k!} \neq e^n$$ In fact $$\sum \limits_{k\mathop=0}^n\frac{n^k}{k!}=\frac{e^n \Gamma (n+1,n)}{\Gamma (n+1)}$$ and so the limit you look for is then the limit of $\frac{\Gamma (n+1,n)}{ \Gamma (n+1)}$ when $n$ goes to infinity and this limit is $\frac{1}{2}$ as Mathematica told you.

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  • $\begingroup$ Thanks. I wonder if there is an elementary way, without the gamma function. $\endgroup$ – user85798 Mar 31 '14 at 9:40

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