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I want to compute the following limit $$ \lim_{n\rightarrow\infty} \sum_{k = 0}^{n} \cfrac{n^{k}}{k!}e^{-n} $$ It is clear that $\sum_{k = 0}^{n} \cfrac{n^{k}}{k!}e^{-n} = e^{-n}\sum_{k = 0}^{n} \cfrac{n^{k}}{k!}$ for every $n\in\mathbb{N}$. Maybe the sum $\sum_{k = 0}^{n} \cfrac{n^{k}}{k!}$ could be computed with the Stirling formula, but I don't know.

I want to related this with Limit Theorems in Probability, but I'm totally stuck.

Maybe it is related with the use of Gamma distribution and its characteristic function.

CantorEuler1996
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  • One definition is $\exp(x)= \sum \limits_{k=0}^\infty \frac{x^k}{k!}$ which makes this easy if you let $x=n$ and use $\exp(x)=e^x$ – Henry Dec 01 '22 at 10:45
  • This leads us to compute $\sum_{j=0}^{\infty}\sum_{k=0}^{n}\cfrac{(-1)^{j}n^{j+k}}{j! k!}$. I don't know how to manage this series. – CantorEuler1996 Dec 01 '22 at 10:50
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    I had misread the limits, so my previous comment may not help. Instead you can see (though it is not a proof) this is essentially the left hand side of a Poisson distribution and so you might expect it to have a limit of $\frac12$. With suitable scale/location adjustment, you can use a normal distribution argument. Empirically, the sum seems close to something like $\frac12 +\frac{0.26596}{\sqrt{n}}$ for large $n$ – Henry Dec 01 '22 at 11:28
  • Absolutely. Thank you very much for your comments. I didn't remember the Poison distribution hehe. – CantorEuler1996 Dec 01 '22 at 12:11
  • From one of the answers in the linked question, $0.26596 \approx \sqrt{\frac{2}{9\pi}}$ – Henry Dec 01 '22 at 13:48

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