The radius of convergence of $\sum \left(\frac{z^2-1}{z^2+1}\right)^n$

Question

I'm trying to find the radius of convergence of complex series $$S=\sum_0^\infty\left(\frac{z^2-1}{z^2+1}\right)^n$$ with the help of ratio test.

With simple observation that the term $a_n$ become singular when $$z^2+1=0\rightarrow z=\pm i$$ The radius of convergence should be $1$. But I want to find this with ratio test:

First, I assume $$\zeta=\frac{z^2-1}{z^2+1}\rightarrow S=\sum_0^\infty \zeta^n$$ This series have radius of convergence $$R=\lim_{n\rightarrow \infty}\left|\frac{a_n}{a_{n+1}}\right|=1$$ $$\left|\frac{z^2-1}{z^2+1}\right|<1$$ I don't how this implies that $R<1$. Also, I'm sure if this method is correct. Please help me with this.

Answer 1

The "radius of convergence" is a notion applying to power series. I don't think your series apply to this category.

$$\left|\frac{z^2-1}{z^2+1}\right|<1 \Longleftrightarrow \left|z^2-1\right| < \left|z^2+1\right|$$ which means $z^2$ is closer to $-1$ than to $1$ in the complex plane. This is equivalent to $\Re(z^2)<0$, or $|x|<|y|$ if you write $z=x+iy$.

The locus of points $z$ such that your series converges is the union of the open quarter-planes corresponding to those inequations. The shape of this locus proves that your series have nothing in common with power series.

Hope this helps.