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 $|z| \ne 1$.
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 $k,\ell\in\mathbb N$.