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This question already has an answer here:

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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marked as duplicate by Winther, JimmyK4542, Lucian, Joel Reyes Noche, user99914 May 25 '15 at 4:46

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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. $\endgroup$ – Ian May 25 '15 at 3:35
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    $\begingroup$ This question is a duplicate, and has been asked many times before on this very site. $\endgroup$ – Lucian May 25 '15 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 $\le n$. ${}\qquad{}$ $\endgroup$ – Michael Hardy May 25 '15 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 May 25 '15 at 4:43
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    $\begingroup$ 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 Fischer May 25 '15 at 12:12
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Also related:

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    $\begingroup$ Just for you, I will ask the same question tomorrow :P $\endgroup$ – ParaH2 May 25 '15 at 4:25
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    $\begingroup$ Oh wow, this is awesome. $\endgroup$ – Jorge Fernández Hidalgo Jul 7 '15 at 6:05
  • $\begingroup$ hahaha, damn... $\endgroup$ – jeremy radcliff Sep 23 '16 at 15:49

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