Region of convergence of the complex series $\sum \frac{z^n}{1+z^{2n}}?$ How to find region of convergence of the complex series $\sum_1^\infty\frac{z^n}{1+z^{2n}}$? I only know that for $|z|\geq \sqrt{2}$ series is convergent by comparison test and also for $|z|=1$ it’s not convergent as nth term doesn’t tend to zero . Please suggest me for others portion . Thanks .   
 A: Using the ratio test,
$$\left|\frac{z^{n+1}}{1 + z^{2n+2}} \div \frac{z^n}{1 + z^{2n}}\right| = |z|\frac{|1 + z^{2n}|}{|1 + z^{2n+2}|}$$
When $|z| < 1$, then this limits to $|z| < 1$, hence the series converges. For $|z| = 1$, as you mention, the series diverges, as
$$\left|\frac{z^n}{1 + z^{2n}}\right| = \frac{1}{|1 + z^{2n}|} \ge \frac{1}{|1| + |z^{2n}|} = \frac{1}{2}.$$
If $|z| > 1$, then let $w = \frac{1}{z}$. Then we get
$$\frac{z^n}{1 + z^{2n}} = \frac{w^{-n}}{1 + w^{-2n}} = \frac{w^n}{w^{2n} + 1},$$
which is convergent by the above arguments, as $|w| < 1$.
A: We can use ratio test where $a_n=z^n/(1+z^{2n})$
$$\begin{align}
\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| &=
\lim_{n\to\infty} \frac{z^{n+1}}{1+z^{2(n+1)}} \frac{1+z^{2n}}{z^n}\\
&= \lim_{n\to\infty} z\frac{1+z^{2n}}{1+z^{2(n+1)}}\\
&= \lim_{n\to\infty} \frac{z+z^{2n+1}}{1+z^{2(n+1)}}\\
&= \lim_{n\to\infty} \frac{z^{-2n-1}+z^{-1}}{z^{-2n-2}+1}\quad\text{(Dividing numerator and denominator by $z^{2n+2}$)}
\end{align}$$
So series converges for all $|z|\neq 1$
