Does there exist a pole on the unit circle?

Question

Show that a power series $\sum _{n\ge 0} a_n z^n$ where $a_n \to 0 $ as $n \to \infty $ cannot have a pole on the unit circle.

Is the statement true with the hypothesis that $(a_n )$ is a bounded sequence?

As $a_n\to 0$ so the radius of convergence of the power series is greater than or equal to $1$. So the power series converges on the unit disc and hence defines an analytic function say $f$ therein.

Hence the power series can't have any isolated singularity on the unit disc and hence neither can have a pole.

Though I feel the second answer is NO ,I am unable to find a counter-example

I would have been happy to find some help from someone.

Answer 1

Consider the constant sequence $a_n=1$.

(I assume the question means assume $a_n$ bounded instead of the assumption $a_n\to0$.)

Answer 2

What you need to show is this: Let $D$ be the open unit disc. Suppose $z_0\in \partial D, r > 0,$ and $f$ is analytic on $\{z|<1\} \cup D(z_0,r).$ Suppose further that in $D,$ $f(z) = \sum_{n=0}^{\infty}a_nz^n$, with $a_n \to 0.$ Then $f$ does not have a pole at $z_0.$

Here's a hint: WLOG, $z_0=1.$ Show $\lim_{x\to 1^-}(1-x)f(x)=0.$ Could any function with a pole satisfy this?