Convergence of a Power Series

Let $a_n$ be a convergent series with limit of the terms $a_n$ equal to $0$. I have to show $\sum_n{a_n z^n}$ does not have a pole on Unit circle. Well I was thinking whether it is enough to show that the power series converges for $|z|=1,$ $z\neq 1$.

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What do you mean by "convergent series with limit 0"? That the sequence $(a_n)$ is summable, and $\sum_{n=0}^\infty a_n = 0$? –  mt_ Mar 10 '12 at 14:17
no $a_n\rightarrow 0$ –  Tojamaru Mar 10 '12 at 14:21
singularity at 1 could be essential as well... –  Tojamaru Mar 10 '12 at 14:26
As stated the theorem is false. Take for example $a_n = (-1)^n/n$, which gives $\sum a_n z^n$ a pole at $z=-1.$ –  Ragib Zaman Mar 10 '12 at 14:27
@RagibZaman it isn't a pole at z=-1, though it is a singularity. –  mt_ Mar 10 '12 at 14:33
Show that for each $\varepsilon > 0$ there exists some $M \geq 0$ (depending on $\varepsilon$) such that for all $z$ with $|z| < 1$
$$|f(z)| \leq M + \frac{\varepsilon}{1-|z|}.$$ Therefore $\lim_{|z| \rightarrow 1} (1-|z|) \cdot |f(z)| = 0$. This cannot hold if $f$ has a pole on the unit circle and $z$ moves towards that pole.