Radius of convergence for this series

Question

Im trying to find the radius of convergence of

$\sum_{n=0}^{\infty} \frac{n}{z^n}$

since this is a Laurent series and not a power series, we couldn't directly use the ratio test, so I modified it a little and got the radius of convergence as |z| > 1, is that correct?

Answer 1

The concept of radius of convergence applies only to power series, but, yes,$$\left\{z\in\Bbb C\,\middle|\,\text{the series }\sum_{n=0}^\infty\frac n{z^n}\text{ converges}\right\}=\{z\in\Bbb C\mid|z|>1\}.$$In fact:

• if $|z|>1$, then the series converges absolutely by the root test, since$$\lim_{n\to\infty}\left|\frac{\frac{n+1}{z^{n+1}}}{\frac n{z^n}}\right|=\lim_{n\to\infty}\frac{n+1}n\cdot\frac1{|z|}=\frac1{|z|}<1.$$
• otherwise, $(\forall n\in\Bbb Z_+):\left|\frac n{z^n}\right|\geqslant n$, and therefore we don't have $\lim_{n\to\infty}\frac n{z^n}=0$; so, the series $\sum_{n=0}^\infty\frac n{z^n}$ diverges.