Im trying to find the radius of convergence for $\sum_{n=1}^\infty \frac{\log n}{n^2}z^n$. Applying the ratio test $\frac{C_{n+1}}{C_n}$, I simplified $\frac{\log n}{n^2}$ to $\frac{\log(n+1)}{\log n}\cdot\frac{n^2}{(n+1)^2}$.
Would I be correct in saying the radius of convergence is 1, by applying the squeeze theorem: $\frac{n^2}{(n+1)^2} < \frac{\log(n+1)}{\log n}\cdot\frac{n^2}{(n+1)^2} < \frac{\log(n+1)}{\log n}$ because the limit as $n$ tends to infinity of both $\frac{\log(n+1)}{\log n}$ and $\frac{n^2}{(n+1)^2}$ is 1?
Thanks.