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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. $\endgroup$ Mar 31, 2014 at 9:43
  • $\begingroup$ @HagenvonEitzen. Good point, indeed ! Cheers. $\endgroup$ Mar 31, 2014 at 9:46
  • $\begingroup$ This exact same question has been asked before here on this site. $\endgroup$
    – Lucian
    Mar 31, 2014 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, 2014 at 9:55
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    $\begingroup$ This looks like a duplicate of this question. $\endgroup$
    – robjohn
    Mar 31, 2014 at 10:06

2 Answers 2

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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, 2014 at 9:40
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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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