We need to find out the limit of,

lim$_{n \to \infty} \sum _{ k =0}^ n \frac{e^{-n}n^k}{k!}$

One can see that $\frac{e^{-n}n^k}{k!}$ is the cdf of Poisson distribution with parameter $n$.

Please give some hints on how to find out the limit.


marked as duplicate by heropup, Did, N. F. Taussig, Martin Sleziak, Lucian Nov 24 '15 at 14:17

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  • $\begingroup$ As interesting as your question is, you need to show some of your own efforts to solve the problem before I will consider furnishing any hints. $\endgroup$ – heropup Nov 24 '15 at 5:46
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    $\begingroup$ Central limit theorem (you should get $½$) $\endgroup$ – Julien Godawatta Nov 24 '15 at 5:46
  • $\begingroup$ @heropup, I am not able to understand how to proceed! $\endgroup$ – gamma Nov 24 '15 at 5:58
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    $\begingroup$ A downvote to this problem is unfair, at least the OP observed something. $\endgroup$ – Zhanxiong Nov 24 '15 at 6:00
  • $\begingroup$ This looks like a duplicate of Evaluating $\lim_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$ $\endgroup$ – robjohn Nov 24 '15 at 6:57

It's a good start to try to solve it in a probabilistic way: notice that the Poisson random variable has the reproducibility property, that is, if $X_{k} \sim \text{Poisson}(1)$, $k = 1, 2, \ldots, n$ independently, then $$S_n = \sum_{k = 1}^n X_{k} \sim \text{Poisson}(n),$$ whose distribution function $F_{S_n}$ satisfies: $$F_{S_n}(n) = P[S_n \leq n] = \sum_{k = 0}^n e^{-n} \frac{n^k}{k!},$$ which is exactly the expression of interest. Hence this suggests linking this problem to central limit theorem.

By the classic CLT, we have $$\frac{S_n - n}{\sqrt{n}} \Rightarrow \mathcal{N}(0, 1).$$ Hence $$P[S_n \leq n] = P\left[\frac{S_n - n}{\sqrt{n}} \leq 0\right] \to P[Z \leq 0] = \frac{1}{2}$$ as $n \to \infty$.


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