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Let $$S_n=e^{-n}\sum_{k=0}^n\frac{n^k}{k!}$$

Is the sequences$\{S_n\}$ convergent?

The following is my answer,but this is not correct. please give some hints.

For all $x\in\mathbb{R}$, $$\lim_{n\rightarrow\infty}\sum_{k=0}^n\frac{x^k}{k!}=e^x.$$ then

$$\lim_{n\rightarrow\infty}e^{-n}\sum_{k=0}^n\frac{n^k}{k!}=1.$$

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marked as duplicate by Aryabhata, jdoicj, Gerry Myerson, Claude Leibovici, Did Jul 18 '14 at 6:53

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  • $\begingroup$ Use the property of limits. Break it into two limits, since they both converge. Your answer should be less than 1. $\endgroup$ – ReverseFlow Jul 18 '14 at 4:25
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    $\begingroup$ @Genomeme: One limit is $0$, the other is $\infty$. One cannot determine the limit of the product from the product of the limits in this case. $\endgroup$ – Jonas Meyer Jul 18 '14 at 4:29
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    $\begingroup$ Are you sure @JimmyK4542, because I wrote it in terms for the incomplete gamma function and the limit after factoring out the $e^n$ looks to be $1/2$. $\endgroup$ – Silynn Jul 18 '14 at 4:49
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    $\begingroup$ Yeah, I messed up. The limit is infact $1/2$. See this: dropbox.com/s/s4dtr78rs9gnobk/… $\endgroup$ – JimmyK4542 Jul 18 '14 at 4:54
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    $\begingroup$ This question has been asked at least three times before. $\endgroup$ – Lucian Jul 18 '14 at 5:50
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Ramanujan showed this limit to be $\frac12$.

I gave a reference to this result in a previous answer of mine, but I can't find it right now.

Added later:

Here is a reference:

http://citeseerx.ist.psu.edu/viewdoc/download;jsessionid=B04924E79B2361751E7AE86C5AF43688?doi=10.1.1.217.7589&rep=rep1&type=pdf

This is called Ramanujan's Q-function.

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  • $\begingroup$ While Ramanujan being able to sum this series is impressive, that hardly explains why it converges... $\endgroup$ – Semiclassical Jul 18 '14 at 5:47
  • $\begingroup$ He did not sum it. He showed that the result is $\frac12 + O(\frac1{n})$. $\endgroup$ – marty cohen Jul 22 '14 at 23:31

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