How to compute $\lim_{n\rightarrow\infty}e^{-n}\left(1+n+\frac{n^2}{2!}\cdots+\frac{n^n}{n!}\right)$ [duplicate]

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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.

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

• 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. – Ian May 25 '15 at 3:35
• The expression after $\lim\limits_{n\to\infty}$ is the probability that a Poisson-distributed random variable with expected value $n$ is $\le n$. ${}\qquad{}$ – Michael Hardy May 25 '15 at 3:55
• 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 $1$. – Ian May 25 '15 at 4:43