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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.2012-07-04

2 Answers 2

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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 $x$.