# Radius of Convergence of this Series

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?

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Seems legit. I would have done the same. –  echoone Aug 25 '12 at 14:54
Correct (provided $\lt$ is replaced by $\leqslant$, twice, in the Attempt). –  Did Aug 25 '12 at 14:58

Applying the root rest: $$\limsup_{n \to \infty} \left( \frac{n! x^{2n}}{n^n (1+x^{2n})} \right)^{1/n} = \limsup_{n \to \infty} \frac{x^2}{n \cdot \max(1,x^2)} \left( n! \right)^{1/n} = \mathrm{e}^{-1} \cdot \frac{x^2}{ \max(1,x^2)}$$ The right-hand-side is bounded from about by $\mathrm{e}^{-1} < 1$ for all $x$, that is it converges for all $x \in \mathbb{R}$.
\begin{align*} \left|\lim_{n\to\infty}\frac{\dfrac{(n+1)!x^{2n+2}}{(n+1)^{n+1}(1+x^{2n+2})}}{\dfrac{n!x^{2n}}{n^n(1+x^{2n})}}\right|&=\lim_{n\to\infty}\frac{n^n(n+1)!x^{2n+2}(1+x^{2n})}{(n+1)^{n+1}n!x^{2n}(1+x^{2n+2})}\\ &=\lim_{n\to\infty}\frac{n^n(n+1)x^2(1+x^{2n})}{(n+1)^{n+1}(1+x^{2n+2})}\\ &=\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n\frac{1+x^{2n}}{1+x^{2n+2}}\\ &=\lim_{n\to\infty}\left(1-\frac1{n+1}\right)^n\frac{1+x^{2n}}{1+x^{2n+2}}\\ &=\lim_{n\to\infty}\left(\left(1-\frac1{n+1}\right)^{n+1}\right)^{\frac{n}{n+1}}\frac{1+x^{2n}}{1+x^{2n+2}}\\ &=\frac1e\lim_{n\to\infty}\frac{1+x^{2n}}{1+x^{2n+2}}\\ &=\begin{cases} \frac1e,&\text{if }|x|\le 1\\\\ 0,&\text{if }|x|>1 \end{cases}\\ &<1\;. \end{align*}