find region of convergence of the series

Question

I tried to solve this question

Find domains of convergence of the series
$$\sum_{n=1}^ \infty \frac{z^n}{n (\log n)^2 }$$

How can I do this .

Answer 1

$$\sum_{n=1}^ \infty \frac{z^n}{n (\log n)^2 }\\ \lim_{n\to\infty}\left|\frac{z^{n+1}}{(n+1) (\log (n+1))^2 }\frac{n (\log n)^2 }{z^n}\right|<1\\ \implies |z|\lim_{n\to\infty}\left|\frac{n (\log n)^2}{(n+1) (\log (n+1))^2 }\right|<1$$ See that $\lim_{n\to\infty}\left|\frac{n (\log n)^2}{(n+1) (\log (n+1))^2 }\right|\to1$, and thus, the sum converges for $$|z|<1\\ \implies z\in(-1,1)$$ Note however, that if $z=1$, then $$\sum_{n=1}^ \infty \frac{1}{n (\log n)^2 }$$ Converges. Thus, the sum converges for $$\boxed{x\in[-1,1]}$$

Answer 2

For $z=1$ the Bertrand series $$\sum_{n\ge2}\frac1{n(\log n)^2}$$ is convergent hence $R\ge1$ and if $|z|>1$ then the sequence $\left(\frac{|z|^n}{n(\log n)^2}\right)_n$ doesn't converge to $0$ the the given series is divergent and then $R\le1$ hence $R=1$. Moreover for $|z|=1$ the series is convergent then the domain of convergence is the closed unit disc.