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
2026-04-06 00:12:53.1775434373
Power series difficulty
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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.