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.

  • $\begingroup$ 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. $\endgroup$
    – Ian
    Commented May 25, 2015 at 3:35
  • 3
    $\begingroup$ This question is a duplicate, and has been asked many times before on this very site. $\endgroup$
    – Lucian
    Commented May 25, 2015 at 3:48
  • $\begingroup$ The expression after $\lim\limits_{n\to\infty}$ is the probability that a Poisson-distributed random variable with expected value $n$ is $\le n$. ${}\qquad{}$ $\endgroup$ Commented May 25, 2015 at 3:55
  • $\begingroup$ 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 $1$. $\endgroup$
    – Ian
    Commented May 25, 2015 at 4:43
  • 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. $\endgroup$ Commented May 25, 2015 at 12:12

1 Answer 1


Also related:

  • 8
    $\begingroup$ Just for you, I will ask the same question tomorrow :P $\endgroup$
    – ParaH2
    Commented May 25, 2015 at 4:25
  • 5
    $\begingroup$ Oh wow, this is awesome. $\endgroup$
    – Asinomás
    Commented Jul 7, 2015 at 6:05
  • $\begingroup$ hahaha, damn... $\endgroup$ Commented Sep 23, 2016 at 15:49

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