How can be evaluated this limit: $$ \lim_{n\to\infty} e^{-n}\sum_{k=1}^n \frac{n^k}{k!} .$$ Thank you.
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1the result is, somewhat amazingly, $1/2$. you can find a discussion at http://math.stackexchange.com/questions/160248/evaluating-lim-n-to-infty-e-n-sum-limits-k-0n-fracnkk – Jonathan Dec 24 '13 at 18:11
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I would guess 1? – Ali Caglayan Dec 24 '13 at 18:12
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The reason it's not trivial is that $ \sum_{k=1}^{m}\frac{n^k}{k!} $ approaches $e^n-1$ as $m\rightarrow\infty$ for fixed $n$, but diverges as $n\rightarrow\infty$ for fixed $m$; so the behavior as both $m$ and $n$ go to infinity is tricky. – mjqxxxx Dec 24 '13 at 18:18