# Math Help - z

1. ## z

prove that the series summation from n=1 to infinity from (n+1) /(n+2) z^n converges if lzl<1 and diverges if lzl>1

thanks

2. You can either use one of the many ways to find radius of convergence (e.g., root or ratio tests), or you might notice similarity to the geometric series.

For example, $\sum_{n=0}^\infty z^n$ is known to converge for $\lvert z\rvert <1$ and to diverge for $\lvert z\rvert >1$. You could say that since $\frac{1}{2}\leq\frac{n+1}{n+2}\leq1$, then $\frac{1}{2}\sum_{n=0}^\infty z^n\leq\sum_{n=0}^\infty \frac{n+1}{n+2}z^n\leq\sum_{n=0}^\infty z^n$. One of these inequalities shows convergence at $\lvert z\rvert <1$, and the other shows divergence at $\lvert z\rvert >1$ (it’s not hard to figure out which is which).