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I encountered the following problems when I was learning about the central limit theorem. Prove $$ e^{-n}\sum_{k=0}^{n}\frac{n^k}{k!}\rightarrow\frac{1}{2} \tag1\\ n\rightarrow\infty $$ This result can be easily obtained by using the central limit theorem of the Poisson distribution. But I don’t know how to use the following results to prove (1). $$ \sum_{\frac{n}{2}-x\sqrt{n}\leq k\leq \frac{n}{2}+x\sqrt{n}} \left( \begin{array}{c} n\\ k\\ \end{array} \right) \sim 2^n\frac{1}{\sqrt{2\pi}}\int_{-2x}^{2x}e^{-\frac{y^2}{2}}dy\tag2\\ n\rightarrow\infty\\ \forall x>0 $$ I use Stirling's formula to prove (2), but I still will not use (2) to prove (1). I am here to ask everyone for help.

mathhahaha
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    What is “this conclusion” meant to refer to? You use “this conclusion” twice, but I’m not sure you are referring to the same thing each time. – Thomas Andrews Nov 11 '21 at 16:12
  • I’ve added tags $(1)$ and $(2)$ so you can be more specific about what “this conclusion” means. – Thomas Andrews Nov 11 '21 at 16:14
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    As the series is a truncated version of $e^n$, one might expect the limit to be $1$. But as $\frac{n^n}{n!}$ is a large number, the tail of the infinite series is not neglectable. –  Nov 11 '21 at 16:27
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    There are many questions (and answers) about this limit already; see this one and the ones linked to it. Maybe you can find what you're looking for there. – Hans Lundmark Nov 11 '21 at 18:18

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