What is the interval of convergence of : $\sum_{n=1}^\infty\frac{n^n}{n!}x^n$?

Question

$x+ \frac{2^2x^2}{2!}+ \frac{3^3x^3}{3!}+ \frac{4^4x^4}{4!}+...$

Possible answers-

1.($0,1/e$)

2.(1/e, $\infty$)

3.(2/e, 3/e)

4.(3/e, 4/e)

My answer is not matching

$u_n= \frac{n^nx^n}{n!}$,$u_{n+1}= \frac{{(n+1)}^{n+1}x^{n+1}}{(n+1)!}$

Applying ratio test,we get

$\vert \frac{u_n}{u_{n+1}} \vert$=$\vert $ $(1-\frac{1}{n+1})^n $($\frac{1}{x} $)$ \vert$,

from this we get $\vert ex \vert<1$ i.e.,$\vert x \vert<1/ e$ as $n $ tends to infinity

need help!!

Answer 1

As suggested by @lab bhattacharjee, one may use the ratio test, as $n \to \infty$, giving $$ \frac{u_{n+1}}{u_n}=\frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n}=\left(1+\frac1n\right)^n \to e. $$

Can you take it from here?