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For which values of $x$ does the following series $$\sum \frac{x^n}{x^{2n}-1}$$ converge?

The author in the solution says that for $-1<x<1$ the series converges because $\sum x^n$ is convergent and $x^{2n}\to 0$ as $n\to \infty.$ I am not sure that I understand this reasoning. Is he using the fact that if $\sum a_n$ is a convergent series and the $b_n$ is a bounded sequence then $\sum a_n b_n$ is convergent provided $a_n\geq0.$ This may work for $0\leq x< 1,$ but I how can he use this fact when the terms of $x^n$ may be negative?

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    $\begingroup$ He uses the fact that if $\sum a_n$ is an absolutely convergent series and $b_n$ is bounded then $\sum a_n b_n$ is also an absolutely convergent series $\endgroup$ – charmd Jan 20 '18 at 12:23
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$\sum{x^n}$ is absolutely convergent. This should resolve your difficulties, I think.

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In fact the series converges for $x\ne1,-1$.

If $-1<x<1$ we've seen that the series converges.

Now suppose $|x|>1$. There exists $N$ so that $$x^{2n}-1\ge\frac12 x^{2n},\quad(n>N).$$Hence for $n>N$ we have$$\left|\frac{x^n}{x^{2n}-1}\right|\le 2|x|^{-n}.$$

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    $\begingroup$ One may also exploit symmetry: as soon as $f(x)=\sum_{n\geq 1}\frac{x^n}{x^{2n}-1}$ is well defined, we have $f\left(\frac{1}{x}\right)=-f(x)$. $\endgroup$ – Jack D'Aurizio Jan 20 '18 at 17:13

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