# Power series difficulty

How would I find the region of convergence of the series of $\frac{1}{n^3}(\frac{z+1}{z-1})^n$. I thought about rewriting $\frac{z+1}{z-1}$ as $\frac{2}{z-1}+1$ but I don't think that helps.

Thanks

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Consider the linear fractional transformation $\frac{z+1}{z-1}$ as mapping some regions onto the open disks centered at the origin, and work backwards from $\sum \frac{x^n}{n^3}$ to find the region of convergence for variable $z$. – hardmath Jul 21 '14 at 11:26
That's not a power series, so it doesn't have a radius of convergence. – Thomas Andrews Jul 21 '14 at 11:31
@Thomas: I read "region" of convergence in the problem statement. – Yves Daoust Jul 21 '14 at 11:40
Yeah, I either misread it or it was quickly edited. Still, it is not a power series :) @YvesDaoust – Thomas Andrews Jul 21 '14 at 11:40

Let $w = \frac{z+1}{z-1}$. Then you have a power series in $w$, centered at $0$. Find its radius of convergence, call that $R$. Then find which $z$ correspond to $\lvert w\rvert < R$. The map $z \mapsto \frac{z+1}{z-1}$ can be explicitly inverted.
There's a pretty obvious way to express $|z-(-1)|<|z-1|$, without inverting that transformation... – Thomas Andrews Jul 21 '14 at 11:37
@Daniel Fischer I get that the series converges when $|\frac{z+1}{z-1}|<1$. How do I now check on the boundary of that ? Thanks – user137090 Jul 21 '14 at 11:38
@user137090 $\sum \frac{1}{n^3}$ converges, so it converges everywhere on the boundary. – Thomas Andrews Jul 21 '14 at 11:39
@ThomasAndrews Yes, but what if $R\neq 1$? (whistles) – Daniel Fischer Jul 21 '14 at 11:50