Holomorphic function on unit disc has no continuous extension to boundary

Question

The following is a qual study question:

Let $$f(z) = \sum_{n=1}^\infty \sqrt{n}z^n$$ Having proven that the radius of convergence is 1, I'm asked to show that this function cannot be extended to a continuous complex-valued function on the closed unit disc.

I'm not sure what sort of techniques to use here. I know a statement from Stein and Shakarchi,

$f$ cannot be continued analytically past the unit circle if no point of $\partial \mathbb D$ is regular for $f$. A point $w$ is regular for $f$ if there is an open neighborhood $U$ of $w$ and an analytic function $g$ on $U$ such that $f = g$ on $\mathbb D \cap U$.

but I haven't figured out how to make use of it. Any hints or helpful tools/theorems would be appreciated!

Answer 1

If $x\in(0,1)$ and $N\in\mathbb N$,$$f(x)\geqslant\sum_{n=1}^N\sqrt nx^n$$and therefore$$\lim_{x\to1}f(x)\text{(if it exists)}\geqslant\sum_{n=1}^N\sqrt n.$$Since the series $\sum_{n=1}^\infty\sqrt n$ diverges, $\lim_{x\to1}f(x)$ cannot exist.

Answer 2

It is enough to show that $\lim_{z\to 1^-}f(z)=+\infty$ and we have an interesting technique for proving much more: squaring. By Cauchy product, for any $z\in(-1,1)$ we have

$$ f(z)^2 = \sum_{m\geq 2} z^m \sum_{n=1}^{m-1}\sqrt{n(m-n)}=\sum_{m\geq 2}z^m\underbrace{m\sum_{n=1}^{m-1}\sqrt{\frac{n}{m}\left(1-\frac{n}{m}\right)}}_{\approx m^2\int_{0}^{1}\sqrt{x(1-x)}\,dx=\frac{\pi}{8}m^2} $$ hence $f(z)$ behaves like something $\geq z\sqrt{\frac{\pi}{8(1-z)}}$ in a left neighbourhood of $z=1$.
A more accurate approximation is $f(z)\approx\frac{z\sqrt{\pi}}{2(1-z)^{3/2}}$ as $z\to 1^-$.