# Homework Help: Radius of convergence

1. Jan 30, 2012

### fauboca

$$\sum_{n=2}^{\infty}z^n\log^2(n), \ \text{where} \ z\in\mathbb{C}$$

$$\sum_{n=2}^{\infty}z^n\log^2(n) = \sum_{n=0}^{\infty}z^{n+2}\log^2(n+2)$$

By the ratio test,

$$\lim_{n\to\infty}\left|\frac{z^{n+3}\log^2(n+3)}{z^{n+2}\log^2(n+2)}\right|$$

$$\lim_{n\to\infty}\left|z\left(\frac{\log(n+3)}{ \log (n+2)}\right)^2\right| = |z|$$

if $|z|<1$, then the sum converges, and if $|z|>1$, then the sum diverges.

Does this mean that $R=1$?

Last edited: Jan 30, 2012
2. Jan 30, 2012

### LCKurtz

Yes, and there was no need to shift the indices.

3. Jan 30, 2012

Thanks.