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During revision, I came across this problem:

The set of real numbers $x$ for which the series $$\sum_{n=1}^{\infty}{\frac{n!x^{2n}}{n^n(1+x^{2n})}}$$ converges is __.

I tried using the ratio test, but got stuck in the process of simplification.

(The answer is the series converges for $\mathbb{R}$.)

Sincere thanks for any help.

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Note the $x$ in the denominator. That isn't a power series. –  Cameron Buie Jul 4 '12 at 15:45
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2 Answers

up vote 7 down vote accepted

For every $x \in \mathbb{R}$ and $n \in \mathbb{N}$ we have $$ \frac{n!x^{2n}}{n^n(1+x^{2n})} \le a_n:=\frac{n!}{n^n}, $$ and $$ \frac{a_{n+1}}{a_n}=\left(\frac{n}{n+1}\right)^n=\left(1+\frac{1}{n}\right)^{-n} \to 1/e. $$ Therefore the series converges for every $x \in \mathbb{R}$.

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Hint: The terms in $x$ are not a problem. For note that $\frac{x^{2n}}{1+x^{2n}} \lt 1$. So compare with $\sum \frac{n!}{n^n}$. If you can show this converges you will be able to conclude that your original series converges for all $x$.

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