Convergence of $\sum_{n=1}^\infty \frac{z^{n!}}{n^2}$ where $\vert z\vert = 1$?

Question

We were asked to study the convergence of this series $$\sum_{n=1}^\infty \frac{z^{n!}}{n^2}$$ on the boundary of its convergence disc i.e.: $\vert z\vert = 1$

And a friend argumented something like this: As $\sum_{n=1}^\infty \frac{1}{n^2}$ converges, by Abel's theorem(click here for the link of its wikipedia page), we can argue: $$ \lim_{n\to\infty} \biggl(\sum_{n=1}^\infty \frac{z^{n!}}{n^2}\biggr) = \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$

But I don't really know how that answers the question of how the series behaves on its convergence disc boundary anyways? Or have I understood something wrong about Abel's theorem?

Thank you!

Answer 1

If $\lvert z\rvert=1$, then $\left\lvert\dfrac{z^{n!}}{n^2}\right\rvert=\dfrac1{n^2}$. Since the series $\displaystyle\sum_{n=1}^\infty\frac1{n^2}$ converges, your series converges absolutely. In particular, it converges. You don't need Abel's theorem at all.

Answer 2

Your friend's argument is nonsense. First, the term

$$ \lim_{n\to\infty} \biggl(\sum_{n=1}^\infty \frac{z^{n!}}{n^2}\biggr)$$

does not make sense, as $n$ is no longer a free variable outside the sum, so there is no $n$ to take the limit of. But this may just be a typo by you or them.

More importantly, the Abel theorem says that your sum has some continuity properties at $z=1$, it doesn't say anything about any other place on the boundary of the unit disc.