Proving $\sum_{k = 0}^{\infty} \frac{1}{1 + |x|^{k}}$ converges if and only if $|x| > 1$

Question

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.

Answer 1

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