Does there exist the following limitation? If the answer is yes, could you show me how to find that? $$\lim_{n\to\infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}$$

In the following, I'm going to write what I've done.

By Taylor expansion, $$e^x=\sum_{k=0}^n \frac{x^k}{k!}+\frac{1}{n!}\int_{0}^x (x-t)^ne^tdt.$$

Letting $x=n$,
$$e^n=\sum_{k=0}^n \frac{n^k}{k!}+\frac{1}{n!}\int_{0}^n (n-t)^ne^tdt.$$

Dividing the both sides by $e^n$, $$1=e^{-n}\sum_{k=0}^n \frac{n^k}{k!}+\frac{e^{-n}}{n!}\int_{0}^n (n-t)^ne^tdt.$$

Hence, I know that what I need to do is to find the following limitation: $$\lim_{n\to\infty}\frac{e^{-n}}{n!}\int_{0}^n (n-t)^ne^tdt$$

However, I'm facing difficulty. I need your help.


merged by user642796 Oct 1 '13 at 17:15

This question was merged with Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$ because it is an exact duplicate of that question.

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