Prove the limit: \[ \lim_{n\to\infty}e^{-n}\Big({1+\frac n{1!}+\dots+\frac {n^n}{n!}}\Big)=\frac12. \]
There is a solution based on some knowledge of probability:
Let $X_1,\dots,X_n\stackrel{\mathrm{i.i.d}}\sim\mathrm{Poisson}(1)$ and $Y_n=X_1+\dots+X_n$. Then $Y_n\sim\mathrm{Poisson}(n)$, and by central limit theorem we have \[ \mathbb P(Y_n\leqslant n)=e^{-n}\sum_{k=0}^n \frac{n^k}{k!}\to\frac{1}{\sqrt{2\pi}}\int_{-\infty}^0 e^{-\frac12x^2}\, dx=\frac12. \]
I am looking for a more direct solution, but failed to got it yet. After using Taylor expansion and Stirling's formula, I tried to prove \[ \lim_{n\to\infty}\frac1{\sqrt{2\pi n}}\Big(\frac n{n+1}+\frac{n^2}{(n+1)(n+2)}+\dots\Big)=\frac12. \] This one has not been proved successfully. It looks quite different from the original one but I am also intersted in its solution staring from it independently.
