Limit of $n \left(\frac{1}{2}-e^{-n}\left(1+n+\frac{n^2}{2!}+...+\frac{n^n}{n!}\right)\right)$

Question

It is known around here that $$\lim_{n\to\infty }e^{-n}\left(1+n+\frac{n^2}{2!}+...+\frac{n^n}{n!}\right)=\frac12$$ So what about the following limit: $$L=\lim_{n \to \infty}n \left(\frac{1}{2}-\frac{1}{e^n}\left(1+n+\frac{n^2}{2!}+...+\frac{n^n}{n!}\right)\right)$$ My thought was to apply Stolz-Cesaro theorem, after rewritting as $$L=\lim_{n \to \infty}\frac{ \left(\frac{1}{2}-\frac{1}{e^n}\left(1+n+\frac{n^2}{2!}+...+\frac{n^n}{n!}\right)\right)}{\frac1n}$$ would produce: $$L=\lim_{n \to \infty}\frac{e^{-(n+1)}\left(1+(n+1)+\frac{(n+1)^2}{2!}+...+\frac{(n+1)^{n+1}}{(n+1)!}\right)-e^{-n}\left(1+n+\frac{n^2}{2!}+...+\frac{n^n}{n!}\right)}{\frac1{n(n+1)}}$$ $$=\lim_{n \to \infty}\frac{e^{-n}\left(\frac{1-e}{e} +\frac1e\frac{(n+1)^{n+1}}{(n+1)!} +n\left(\frac{\left(1+\frac1n\right)}{e}-1\right) +\frac{n^2}{2!}\left(\frac{\left(1+\frac1n\right)^2}{e}-1\right)+\cdots +\frac{n^n}{n!}\left(\frac{\left(1+\frac1n\right)^n}{e}-1\right)\right)}{\frac1{n(n+1)}}$$ But I dont see how can I go further, can you help me evaluate it? Also, are limits of this form already known?

Answer 1

The follow equation is from Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$.

\begin{align} e^{-n}\sum_{k=0}^n\frac{n^k}{k!} &=\frac{1}{n!}\int_n^\infty e^{-t}\,t^n\,\mathrm{d}t\\ &=\frac12+\frac{2/3}{\sqrt{2\pi n}}+O(n^{-1})\tag{11} \end{align} Use this equation we know: $$\lim_{n\to\infty}\sqrt{n} \left(e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}-\frac{1}{2}\right)=\frac{2}{3\sqrt{2\pi }}.$$ So your limit is $\infty$.