I have trouble with the following excercise:
Show that the series converges absolutely for $ -e < x < e $ $$ \sum_{n=1}^{\infty}\frac{n! x^n}{n^n} $$
I tried to find the interval of convergence with the ratio-test, but it didn't work, and it is probably not the correct method.
Help would be appreciated!
Hint:
$$\dfrac{\dfrac{(n+1)!x^{n+1}}{(n+1)^{n+1}}}{\dfrac{n! x^n}{n^n}}=\dfrac{x(n+1)n^n}{(n+1)^{n+1}}=\dfrac x{\left(1+\dfrac1n\right)^n}$$
$\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=?$