No analytic continuations of $\sum_{n = 1}^\infty \frac{a_n}{z - e^{i \pi \lambda n}}$ beyond unit disk. I am working on the following complex analysis problem:

Let $\lambda$ be irrational and let $\left\{a_n\right\} \subset \mathbf{C}$, $\sum |a_n| < \infty$ and $a_n \neq 0$.  Letting $D = \left\{ z \in \mathbf{C} \colon |z| < 1 \right\}$, define $$ f(z) = \sum_{n = 1}^\infty \frac{a_n}{z - e^{i \pi \lambda n}} \;\;\;\;\; \left( z \in D \right). $$  Show that $f(z)$ is analytic on $D$ and cannot be analytically continued to any larger domain containing $D$.  

It is clear that the series absolutely converges uniformly on compact sets, hence $f(z)$ is analytic.  It is also clear that the series diverges for all $z$ on the unit circle $S^1$.   
I have been having trouble determining whether $f(z_n)$ diverges if $z_n \to z_0 \in S^1$ (because the convergence of the series is not necessarily uniform on $D$).  If this were generally true, then $f(z)$ couldn't be extended to a continuous function on any larger (connected) domain.  But I am concerned, since the question specifically addresses analytic continuations, that this may not be the right way to go about it.  Perhaps I need to use some complex-analytic-specific theorem (it is kind of an "anti-Abel's theorem" statement).  But I don't know which one.  I am not that comfortable with the theory of analytic continuations.
Thanks for your help.  

EDIT:  Now that I think about it, why is it so clear that the series diverges on $S^1$?  Since the $a_n$ go to $0$, what if, say, $|a_n| \leq 2^{-n} |z - e^{i \lambda \pi n}|$?  
 A: We expect the unit circle to be the natural boundary of $f$ because the set $\{ e^{i\pi \lambda n} : n \in \mathbb{N}\}$ of points where one of the terms of the series has a pole is dense in the unit circle, and "there should not be enough cancellation possible to make such a point regular". The poles of the terms should remain singularities - not poles, of course, since they are not isolated - of $f$, and we expect "pole-like" behaviour of $f$ at $w_k = e^{i\pi \lambda k}$.
It is more or less evident that only the terms with $w_n = e^{i\pi \lambda n}$ close to $w_k$ can have a major influence on the behaviour. More precisely, given $\varepsilon > 0$, the function
$$f_{k,\varepsilon}(z) = \sum_{\substack{n = 0\\\lvert w_n - w_k\rvert \geqslant \varepsilon}}^\infty \frac{a_n}{z - w_n}$$
is holomorphic on $\mathbb{C}\setminus \{z : \lvert z\rvert = 1, \lvert z - w_k\rvert \geqslant \varepsilon\}$. The locally uniform convergence of that series on the indicated domain is not difficult to show.
So to prove the singular behaviour of $f$ at $w_k$, it suffices to show that we can choose $\varepsilon > 0$ so that $\lim\limits_{r\to 1} \lvert s_{k,\varepsilon}(r\cdot w_k)\rvert = \infty$, where
$$s_{k,\varepsilon}(z) = \sum_{\substack{n = 0\\ \lvert w_n - w_k \rvert < \varepsilon}}^\infty \frac{a_n}{z-w_n}$$
and $r\in [0,1)$.
By the convergence of $\sum \lvert a_n\rvert$, for every $k\in \mathbb{N}$ there is an $n_k\in \mathbb{N}$ (necessarily $n_k > k$) such that
$$\sum_{n = n_k}^\infty \lvert a_n\rvert < \frac{\lvert a_k\rvert}{2}.\tag{$\ast$}$$
Set $\varepsilon = \min \{ \lvert w_n - w_k\rvert : n < n_k, n \neq k\}$. Since all $w_n$ are distinct, $\varepsilon > 0$, and we have
\begin{align}
\lvert s_{k,\varepsilon}(z)\rvert &\geqslant \frac{\lvert a_k\rvert}{\lvert z - w_k\rvert} - \sum_{\substack{n = n_k\\ \lvert w_n - w_k\rvert < \varepsilon}}^\infty \frac{\lvert a_n\rvert}{\lvert z - w_n\rvert}\\
&\geqslant \frac{\lvert a_k\rvert}{\lvert z - w_k\rvert} - \sum_{n = n_k}^\infty \frac{\lvert a_n\rvert}{\lvert z - w_n\rvert}
\end{align}
for $z\in D$. Now if $z = r\cdot w_k$ for $r\in [0,1)$, then $\lvert z - w_n\rvert \geqslant \lvert z - w_k\rvert$ for all $n\in \mathbb{N}$, and therefore
\begin{align}
\lvert s_{k,\varepsilon}(r\cdot w_k)\rvert &\geqslant \frac{\lvert a_k\rvert}{\lvert r\cdot w_k - w_k\rvert} - \sum_{n = n_k}^\infty \frac{\lvert a_n\rvert}{\lvert r\cdot w_k - w_n\rvert}\\
&\geqslant \frac{\lvert a_k\rvert}{\lvert r\cdot w_k - w_k\rvert} - \sum_{n = n_k}^\infty \frac{\lvert a_n\rvert}{\lvert r\cdot w_k - w_k\rvert}\\
&= \frac{1}{1-r}\biggl(\lvert a_k\rvert - \sum_{n = n_k}^\infty \lvert a_n\rvert\biggr)\\
&> \frac{\lvert a_k\rvert}{2(1-r)}.
\end{align}
This shows that $f$ has radial limit $\infty$ at every $w_n$, and therefore there is no open set $V$ with $V\cap \partial D \neq \varnothing$ and $g$ meromorphic on $V$ with $g \lvert_{V\cap D} \equiv f\lvert_{V\cap D}$. For if such a $V$ and $g$ existed and $\zeta \in V\cap \partial D$, then for all small enough $\delta > 0$ we'd have $A(\zeta; 0, \delta) := \{ z : 0 < \lvert z - \zeta\rvert < \delta\} \subset V$ and $g$ would be holomorphic on $A(\zeta; 0, \delta)$. But then the radial limit of $g$ at $w$ would be finite at every $w \in A(\zeta; 0, \delta) \cap \partial D$, so $\{ n : w_n \in A(\zeta; 0, \delta)\} = \varnothing$, but that contradicts the denseness of $\{ w_n : n\in \mathbb{N}\}$ in $\partial D$.
