# Is the sequences$\{S_n\}$ convergent? [duplicate]

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

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

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

• Use the property of limits. Break it into two limits, since they both converge. Your answer should be less than 1. – ReverseFlow Jul 18 '14 at 4:25
• @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. – Jonas Meyer Jul 18 '14 at 4:29
• 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 $1/2$. – Silynn Jul 18 '14 at 4:49
• Yeah, I messed up. The limit is infact $1/2$. See this: dropbox.com/s/s4dtr78rs9gnobk/… – JimmyK4542 Jul 18 '14 at 4:54
• This question has been asked at least three times before. – Lucian Jul 18 '14 at 5:50

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.

• He did not sum it. He showed that the result is $\frac12 + O(\frac1{n})$. – marty cohen Jul 22 '14 at 23:31