This is a problem from a previous complex analysis qualifying exam that I'm working through to study for my own upcoming exam. I'm feeling pretty clueless on this one.
Problem:
Suppose $f(z)$ is an analytic function on the unit disk $\mathbb{D}$ and that there exists a sequence of points $z_n \in \mathbb{D}$ such that $|f(z_n)| \to \infty$ as $n \to \infty$. Prove that the radius of convergence of the power series for $f$ about the origin is equal to 1.
My thoughts so far:
If $f(z)$ is analytic on the open unit disk, then there exists a power series centered at the origin. If it's analytic, this power series has only exponents $\geq 0$ and should be convergent everywhere it's analytic. With no singularities, I don't understand how we'd have $|f(z_n)| \to \infty$ as $n \to \infty$. I'm guessing this is a failure to understand how sequences work in the complex plane.
I could use a clear and thorough explanation of what's going on here and direct references to a theorem or topic I can look up. Thanks!
Let $\rho$ be radius of convergence. We have to have $\rho\geqslant1$, since the radius of convergence is always at least as large as the radius of the largest open disk centered at the origin and contained in $\Bbb D$.
Now, suppose that $\rho>1$ and, for each $z\in D_\rho(0)$, let$$F(z)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}z^n.$$Then $|z|<1\implies F(z)=f(z)$. On the other hand, the sequence $(z_n)_{n\in\Bbb Z_+}$ is bounded, and therefore it has a convergent subsequence $\left(z_{n_k}\right)_{k\in\Bbb Z_+}$; let $\omega$ be its limit. Then$$\lim_{k\to\infty}\left|f\left(z_{n_k}\right)\right|=|F(\omega)|,$$which is impossible, since $\lim_{n\to\infty}|f(z_n)|=\infty$.