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

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.

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 $$|x|\leqslant 1$$, $$\lim_{k\to\infty}\frac{1}{1+|x|^{k}} \neq0.$$
• 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 ? – joseph Dec 10 '18 at 6:42
• @joseph: "first assume it converges, then show it requires $|x|>1$" is equivalent to "assume that $|x|\leqslant1$, and show it diverges". – user587192 Dec 10 '18 at 13:23