This is a question from a GRE math subject test practice material.
$ \sum^{\infty}_{n=1} \frac{n!x^{2n}}{n^n(1+x^{2n})} $
The set of real numbers $x$ for which the series converges is: $\{0\}$, $\{-1 \leq x \leq 1\}$, $\{-1 < x < 1\}$, $\{-\sqrt{e} \leq x \leq \sqrt{e}\}$ or $\mathbb{R}$?
Attempt:
$\frac{n!}{n^n} = \frac{1}{n} \frac{2}{n} \frac{3(4)(5)...(n-1)(n)}{n^{n-2}} < (2/n^2)$
So by comparison test to $(2/n^2)$, $\frac{n!}{n^n}$ converges. Furthermore, $0 < \frac{x^{2n}}{(1+x^{2n})} < 1$ for all $x \in \mathbb{R}$. So this series converges for all real number $x$? The answer is $\mathbb{R}$?
Questions:
Is my logic and answer correct? Is there an easier way?