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

Question

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

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

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

Use the property of limits. Break it into two limits, since they both converge. Your answer should be less than 1. — ReverseFlowControl, Jul 18 '14 at 04:25
@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 04:29
Some testing with WolframAlpha suggests that the limit is indeed $1$. Can find a bound $\displaystyle\sum_{k = n+1}^{\infty}\dfrac{n^k}{k!}$? — JimmyK4542, Jul 18 '14 at 04:31
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 04:49
Yeah, I messed up. The limit is infact $1/2$. See this: https://www.dropbox.com/s/s4dtr78rs9gnobk/On%20The%20Limit%20of%20A%20Sequence.pdf# — JimmyK4542, Jul 18 '14 at 04:54
@Genomeme that property only applies to when both limits exist — Kamster, Jul 18 '14 at 06:12
The both do, I just realized however, that things get a bit tricky but the idea is essentially the same. — ReverseFlowControl, Jul 18 '14 at 06:16
@Lucian: Why not cast a dupe vote if you know a question already exists? — Aryabhata, Jul 18 '14 at 06:37

marty cohen · Answer 1 · 2014-07-23T00:14:38.030

0

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:

This is called Ramanujan's Q-function.

edited Jul 23 '14 at 00:14

answered Jul 18 '14 at 04:57

marty cohen

While Ramanujan being able to sum this series is impressive, that hardly explains why it converges... – Semiclassical Jul 18 '14 at 05:47
He did not sum it. He showed that the result is $\frac12 + O(\frac1{n})$. – marty cohen Jul 22 '14 at 23:31

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