$\sum^{\infty}_{n=1} \frac {x^n}{1+x^{2n}}$ interval of convergence I need to find interval of convergence for this series:
$$\sum^{\infty}_{n=1} \frac {x^n}{1+x^{2n}}$$
I noticed that if $x=0$ then series is convergent. Unfortunately, that’s it.
 A: If $\left| x\right|<1$ then
$$
\left| \frac{x^n}{1+x^{2n}} \right|\leq \left| x\right|^n
$$
because $1+x^{2n}\geq1$. Now $\sum_{n=1}^{\infty} \left| x\right|^n$ is a finite geometric series.
Similarly, if $\left| x\right|>1$ then
$$
\left| \frac{x^n}{1+x^{2n}} \right|\leq \frac{1}{ \left| x\right|^n}
$$
because $1+x^{2n}\geq x^{2n}$. Now $\sum_{n=1}^{\infty} \frac{1}{\left| x\right|^n}$ is a finite geometric series.
If $x=1$, or $x=-1$ the series is divergent.
A: When $|x| < 1$, we have, for $n$ sufficiently large, $\left|x^{2n}\right| < \frac 12$. By the triangle inequality,
$$1 = \left|1 + x^{2n} - x^{2n}\right| \le \left|1 + x^{2n}\right| + \left|x^{2n}\right| < \left|1 + x^{2n}\right| + \frac 12,$$
so
$$\frac{1}{\left|1 + x^{2n}\right|} < 2.$$
Therefore,
$$
\left|\frac{x^n}{1 + x^{2n}}\right| < 2|x^n|,
$$
and the series converges absolutely by comparison with the geometric series.
When $|x| > 1$, we have, for $n$ sufficiently large, $\left|x^{2n}\right| > 2$.
By the triangle inequality,
$$
\left|x^{2n}\right| = \left|1 + x^{2n} - 1\right| \le \left|1 + x^{2n}\right| + 1
< \left|1 + x^{2n}\right| + \frac 12\left|x^{2n}\right|,
$$
so
$$
\frac 1{\left|1 + x^{2n}\right|} < \frac{2}{\left|x^{2n}\right|}.
$$
Therefore,
$$
\left|\frac{x^n}{1 + x^{2n}}\right| < \frac{2}{\left|x^n\right|},
$$
and the series converges absolutely by comparison with the geometric series.
