Vrftjkry

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

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

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

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

If $|z|=r<1$, and $nge 1$, 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
$$
sum_{n=0}^inftyfrac{z^n}{1+z^{2n}}
$$
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 $zinmathbb R$, 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 $k,ellinmathbb N$.

$$a_nleft(frac1zright)=frac{1/z^n}{1+1/z^{2n}}\
=frac{z^n}{z^{2n}+1}\
=a_n(z)$$
So it converges for $z$ whenever it converges for $1/z$.