Value of $z$ so that the series converges

Question

$$ \begin{align} \sum_{n=0}^\infty \frac{1}{n^2}\left(z^n+\frac{1}{z^n}\right) \end{align} $$

Detrmine the value of $z$ so that the series converges

I believe that the series converges when $|z|<1$, but taking the $\lim_{n\to \infty}|\left(\frac{1}{n^2}(z^n+\frac{1}{z^n})\right)^{\frac{1}{n}}|<1$ I couldn't come to a conclusion.

Answer 1

I believe the series only converges when $|z|=1$, since we have $$\sum_{n=0}^\infty\frac{2}{n^2}$$ which converges, but for $|z|\ne 1$, consider $|z|\lt 1$, then we can see that $$\sum_{n=0}^\infty\frac{1}{n^2}(\frac{1}{z^n})=\infty$$ also by analysing the limit: $\lim_{n\to\infty}|(\frac{1}{n^2}\frac{1}{z^n})^\frac{1}{n}|=\lim_{n\to\infty}\frac{1}{n^\frac{2}{n}}|\frac{1}{z}|=|\frac{1}{z}|\gt 1$

Answer 2

If $|z|\neq 1$, the absolute value of $\frac{1}{n^2}\left(z^n+\frac{1}{z^n}\right)$ tends to $\infty$, hence the series cannot converge. In the other case, $z=e^{i\theta}$, the series is convergent since the absolute value of $z^n+\frac{1}{z^n}$ is bounded by $2$, and we have: $$\sum_{n=1}^{+\infty}\frac{z^n+z^{-n}}{n^2}=\operatorname{Li}_2(z)+\operatorname{Li}_2(1/z).$$