meromorphic function with a pole on the unit circle diverges Let $f$ be a meromorphic function in a neighborhood of the closed unit disk $\bar{\mathbb{D}}$. Suppose that $f$ is holomorphic in $\mathbb{D}$ and 
$$ f(z) = \sum_{n=0}^\infty a_n z^n $$
for $z \in \mathbb{D}$. Prove that if $f$ has a pole on the unit circle $\mathbb{T}$ then the above power series diverges at any $z \in \mathbb{T}$.
Attempt at solution :
My first observation was that if I can prove this claim when the pole is at a specific point like $z = 1$ on $\mathbb{T}$, then using rotation the claim will be valid for any arbitrary point on $\mathbb{T}$. So I assumed 
$$ \sum_{n=0}^\infty a_n z^n = \frac{g(z)}{(z-1)^k}$$ 
with $g(z)$ holomorphic in a neighborhood of $\bar{\mathbb{D}}$. Even with the further simplification of assuming k = 1, I could only prove that $\lim_{n-> \infty} a_n = g(1)$ which is finite. Can someone please help with the solution.
 A: It suffices to show that the coefficients $a_n$ cannot converge to $0$.  Now suppose that $|a_k| \leq \varepsilon$ for $k \geq m$.  Then for $z\in\mathbb{D}$ $$|f(z)|\leq \frac{\varepsilon}{1-|z|}+\sum_{k=0}^{m-1}|a_k|.$$  This implies that if $a_k$ converges to $0$ then $\lim_{|z|\to 1}(1-|z|)|f(z)|=0$. In particular $f$ cannot have a pole on $\mathbb{T}$ in this case.
A: If what you say you've proven is correct, then you are done. If the series converged at some $z$ on the boundary, then $\lim_{n\rightarrow \infty} a_n=0=g(1)$, contradicting that the order of the pole is $k$.
A: I suppose the the pole is at $z_0=1$. 
Put $\displaystyle f(z)=g(z)+\sum_{k=1}^s \frac{u_k}{(z-1)^k}$ where $\displaystyle h(z)=\sum_{k=1}^s \frac{u_k}{(z-1)^k}$ is the singular part of $f$ at $1$ ($u_s\not =0)$. Hence $\displaystyle g(z)=\sum b_n z^n$ is analytic in a neighborhood of $\overline{\mathbb{D}}$, thus its radius of convergence is $>1$. Thus $b_n\to 0$ if $n\to \infty$. Note also that $\displaystyle h(z)=\sum P(n)z^n$, where $\displaystyle P(x)=\alpha_{s-1}x^{s-1}+...$ is a polynomial of degree $s-1$ $(\alpha_{s-1}\not =0)$. If $\displaystyle f(z)=\sum_{n\geq 0} a_n z^n$, we have hence $a_n=b_n+P(n)$. We immediately get that $|a_n|\sim |\alpha_{s-1}|n^{s-1}$, and for $z\in \mathbb{T}$ that $|a_n z^n|\sim |\alpha_{s-1}|n^{s-1}$ if $n\to +\infty$. Hence $a_n z^n$ does not go to $0$ as $n\to \infty$, and the series is divergent. 
