Evaluate $\lim_{n\to\infty} \exp(-n^2) \sum_{j=n}^{4n} \frac{n^j}{j! }$.

Question

Question : Evaluate $\lim_{n\to\infty} \exp(-n^2) \sum_{j=n}^{4n} \frac{n^j}{j! }$.

We can not separate the limits here as there are chances of the resultant to be some indeterminate. Let $a_n=\sum_{j=n}^{4n}\frac{n^j}{j!}$ (a) , on expanding we have $a_n=\frac{n^n}{n!}+\frac{n^{n+1}}{(n+1)!}+\frac{n^{n+2}}{(n+2)!}+\dots+\frac{n^{4n}}{(4n)!}$ (this is not a gp). I know that $\sum_{n=0}^{\infty}\frac{x^n}{n!}=\exp(x)$. Using it here didn't made any sense to me. How about using https://en.wikipedia.org/wiki/Stirling%27s_approximation, but didn't simplified. Taking $\log$ on both sides of (a) we have $\log a_n=\sum_{j=n}^{4n}\log \frac{n^j}{j!}$,this will include another sum (not easy for "ME" to handle).

Thanks.

Answer 1

$\sum_{j=n}^{4n} \frac{n^j}{j! }\leq \sum_{j=1}^{\infty} \frac{n^j}{j! }=e^{n}$ and $e^{-n^{2}+n} \to 0$