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I would like to show

$$\sum_{k = 0}^{\infty}\frac{1}{1 + |x|^{k}} $$

converges if and only if $|x| > 1$. I think that the best way to show the backwards direction is to assume we havve $|x| \leq 1$ then maybe doing the integral test? But I didn't get anywhere with this. I don't relaly have an idea of how to do it with the other direction either.

Any help is appreciated.

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Note that for $|x|>1$ $$ \lim_{k\to\infty}\frac{1+|x|^{k}}{1+|x|^{k+1}} =\lim_{k\to\infty}\frac{|x|^{-k}+1}{|x|^{-k}+|x|}=\frac{1}{|x|}<1. $$ Now apply the ratio test.

For $|x|\leqslant 1$, $$ \lim_{k\to\infty}\frac{1}{1+|x|^{k}} \neq0. $$

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  • $\begingroup$ don't we need to show two directions? like, first assume it converges, then show it requires $|x| > 1$. then assume $|x| > 1$ and show it converges ? $\endgroup$ – joseph Dec 10 '18 at 6:42
  • $\begingroup$ @joseph: "first assume it converges, then show it requires $|x|>1$" is equivalent to "assume that $|x|\leqslant1$, and show it diverges". $\endgroup$ – user587192 Dec 10 '18 at 13:23

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