There is a probabilistic method to solve it. But I am not familiar with probability. I am trying to compute it by analytic method, such as using L Hospital's rule or Stolz formula, but they are not working.
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$\begingroup$ Well, $\sum_{k=0}^n \frac{n^k}{k!}=e^n-\sum_{k=n+1}^\infty \frac{n^k}{k!}$. That seems like a place to start. $\endgroup$– IanMay 25, 2015 at 3:35
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3$\begingroup$ This question is a duplicate, and has been asked many times before on this very site. $\endgroup$– LucianMay 25, 2015 at 3:48
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$\begingroup$ The expression after $\lim\limits_{n\to\infty}$ is the probability that a Poisson-distributed random variable with expected value $n$ is $\le n$. ${}\qquad{}$ $\endgroup$– Michael HardyMay 25, 2015 at 3:55
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$\begingroup$ An interesting related fact: $\lim_{n \to \infty} e^{-n} \left ( 1+1/n \right )^{n^2} = e^{-1/2}$. This is probably easier to prove (you can just use L'Hospital's rule), and hints that this problem is not that simple (i.e. it will need something relatively tight), since the answer is neither $0$ nor $1$. $\endgroup$– IanMay 25, 2015 at 4:43
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2$\begingroup$ There's no need to delete this question. Duplicates happen, this question as such isn't terrible, and it has a very useful list of duplicates now. $\endgroup$– Daniel FischerMay 25, 2015 at 12:12
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1 Answer
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Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$
Finding $\lim_{n\to\infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}$ if it exists
Show that $\lim\limits_{n\rightarrow\infty} e^{-n}\sum\limits_{k=0}^n \frac{n^k}{k!}=\frac{1}{2}$
Does n power of e grow much more faster than its Maclaurin polynomial?
Limits of sequences connected with real and complex exponential
The limit $\lim_{n\to \infty}\frac{T_n(n)}{e^n}$ where $T_n(x)$ is the Taylor polynomial of $e^x$
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8$\begingroup$ Just for you, I will ask the same question tomorrow :P $\endgroup$– ParaH2May 25, 2015 at 4:25
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