For which value of $x \in R $ the following series converges $ \sum_{n=1}^\infty\frac{x^{n}}{1+x^{2n}}$

Question

For which value of $x \in R $ the following series converges

$ \sum_{n=1}^\infty\frac{x^{n}}{1+x^{2n}}$

The series of the absolute values is $ \sum_{n=1}^\infty |\frac{x^{n}}{1+x^{2n}}|=\sum_{n=1}^\infty \frac{|x|^{n}}{1+x^{2n}}$

and applying the root test: $ lim_{n\rightarrow \infty} \sqrt[n]{ \frac{|x|^{n}}{1+x^{2n}}}= lim_{n\rightarrow \infty} \frac{|x|}{\sqrt[n] {1+x^{2n}}} = |x| lim_{n\rightarrow \infty} \frac{1}{\sqrt[n] {1+x^{2n}}} = |x|* \frac{ 1}{x^2}= \frac{ 1}{|x|}$

$\frac{ 1}{|x|}<1$ $ \forall x \in R $ with $|x|>1$

while the solution should be $|x| \ne 1$ and I'm not sure about the limit.

Can someone help me to understant where is the mistake?

Answer 1

Your computation in the root test is wrong when $|x|\lt1$. It should be $$ \lim_{n\to\infty}\left|\frac{x^n}{1+x^{2n}}\right|^{1/n}=|x| $$ For $|x|\gt1$, your computation is correct: $$ \lim_{n\to\infty}\left|\frac{x^n}{1+x^{2n}}\right|^{1/n}=\frac1{|x|} $$ Now you just need to consider the case $|x|=1$.