Determining: $\lim_{n\rightarrow \infty}e^{-n}\sum_{k=n}^\infty n^k/k!$ What is the value of $\lim_{n\rightarrow \infty}e^{-n}\sum_{k=n}^\infty n^k/k!$ ?
I have tried initially but could not proceed any further. What I have tried is:
$$\lim_{n\rightarrow \infty}e^{-n}\sum_{k=n}^\infty{n^k \over k!}\\
=\lim_{n\rightarrow \infty}e^{-n}\left[e^n-\sum_{k=0}^{n-1}{n^k \over k!}\right]$$
I got no clue after this. I am not sure if how that limit can be be determined.
Any kind of help will be welcome.


we have  $$e^{-n}\sum_{k=n}^\infty \frac{n^k}{k!}
=e^{-n}\left[e^n-\sum_{k=0}^{n-1} \frac{n^k}{k!}\right]= \left[1+ \frac{e^{-n}n^n}{n!}-e^{-n}\sum_{k=0}^{n} \frac{n^k}{k!}\right] $$
By Stirling formula $$ \frac{e^{-n}n^n}{n!}\sim \frac{1}{\sqrt{2n\pi}}\to0$$
from and from here:Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$ we have 
$$\lim_{n\rightarrow \infty}\left[e^{-n}\sum_{k=0}^{n} \frac{n^k}{k!}\right] = \frac12$$
thus $$\lim_{n\rightarrow \infty}e^{-n}\sum_{k=n}^\infty \frac{n^k}{k!}=1-\frac12$$
