Find the range of convergence of the series$\,\,\sum_{n=0}^\infty {\frac{z^n}{1+z^{2n}}}$

The series I have is $$\displaystyle\sum_{n=0}^\infty {\dfrac{z^n}{1+z^{2n}}}$$

The same series with absolute values is: $$\displaystyle\sum_{n=0}^\infty {\dfrac{|z|^n}{1+|z|^{2n}}}$$

Using D'Alembert's principle, $$\displaystyle\lim {\dfrac{a_{n+1}}{a_n}} = {\dfrac{|z|^n \cdot |z|}{1+|z|^{2n} \cdot |z|}} \cdot {\dfrac{1+|z|^{2n}}{|z|^n}} = |z|$$

The convergence range is when $$|z| < 1$$. But the book answer is $$|z| \ne 1$$.

• What's $Z_n$? Is it actually $Z^n$? – xbh Nov 18 '18 at 9:33
• Yes, sorry. Updated – user3132457 Nov 18 '18 at 9:34
• I don't think the calculation of $\lim a_{n+1}/a_n$ is correct. – xbh Nov 18 '18 at 9:36
• So what is wrong? – user3132457 Nov 18 '18 at 9:38
• Wouldn't $|z|>1$ and $|z|<1$ lead to different results? – xbh Nov 18 '18 at 9:41

If $$|z|=r<1$$, and $$n\ge 1$$, then $$\left|\frac{z^n}{1+z^{2n}}\right|\le \frac{|z|^n}{1-|z|^{2n}}=\frac{r^n}{1-r^{2n}} <\frac{r^n}{1-r}$$ and hence the series $$\sum_{n=0}^\infty\frac{z^n}{1+z^{2n}}$$ converges, due to Comparison Test.
If $$|z|=1$$, and in particular $$z=i$$, then the series is not even definable.
Note. This is not a power series, and hence finding the radius of convergence is out of question. Clearly, there exist values of $$z$$, with $$|z|>1$$, for which the series converges absolutely, i.e., all $$z\in\mathbb R$$, with $$|z|>1$$. Meanwhile, the unit circle is a natural boundary of the series, since, for the points $$z=\exp(ik/2^\ell)$$ are singularities (not isolated) of the series, for all $$k,\ell\in\mathbb N$$.
• I don't get how $$\left|\frac{z^n}{1+z^{2n}}\right|\le \frac{|z|^n}{1-|z|^{2n}}$$? If $z^{2n}$ is greater than 1, then the inequality doesn't hold. – user3132457 Nov 18 '18 at 9:56
• This is true when $|z|=r<1$, as mentioned in the beginning of the answer. – Yiorgos S. Smyrlis Nov 18 '18 at 10:00
• I know, but the answer is $|z| \ne 1$ which means $|z| > 1$ is possible to have. – user3132457 Nov 18 '18 at 10:02
$$a_n\left(\frac1z\right)=\frac{1/z^n}{1+1/z^{2n}}\\ =\frac{z^n}{z^{2n}+1}\\ =a_n(z)$$ So it converges for $$z$$ whenever it converges for $$1/z$$.