Math GRE - #13

Determine the set of real numbers x for which the series below is convergent: \[\sum_{n=1}^{\infty}\frac{n!x^{2n}}{n^{n}\left(1+x^{2n}\right)}\]

• $\{0\}$
• $\{x: -1 \leq x \leq 1\}$
• $\{x: -1 < x < 1\}$
• $\{x: -\sqrt{e} \leq x \leq \sqrt{e}\}$
• $\mathbb{R}$

Solution :

The answer is $\mathbb{R}$.

Note that for all numbers $x\in\mathbb{R}$, \[x^{2n}\geq 0\implies \frac{x^{2n}}{1+x^{2n}}\leq \frac{x^{2n}}{x^{2n}}=1.\]
This means that \[\sum_{n = 1}^\infty \frac{n! x^{2n}}{n^n(1 + x^{2n})} < \sum_{n = 1}^\infty \frac{n!}{n^n}.\] Thus, if we can prove that \[\sum_{n = 1}^\infty \frac{n!}{n^n}\] is convergent, then by the comparison test our original series is convergent for all $x$.

By the ratio test,  $\sum_{n = 1}^\infty \frac{n!}{n^n}$ converges as:
\[\begin{eqnarray*}
\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_{n}} & = & \lim_{n\rightarrow\infty}\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^{n}}{n!} \\
& = & \lim_{n\rightarrow\infty}\frac{n+1}{n}\cdot\left(\frac{n}{n+1}\right)^{n+1} \\
& = & \lim_{n\rightarrow\infty}\left(\frac{n}{n+1}\right)^{n} \\
& = & \lim_{n\rightarrow\infty}\left(1-\frac{1}{n+1}\right)^{n} \\
& = & \frac{1}{e}
\end{eqnarray*}\]
Where the last line is essentially a result of the limit definition of e.

Note that if we hadn't guessed that the answer was $\mathbb{R}$, it would have been safer to do the ratio test on the original sequence directly and obtain the same answer (the method above just saves a bit of work).