Consider the power series $\sum a_n z^n$.Given that $a_n$ converges to $0$, prove that
$f(z)$ cannot have pole on the unit circle, where $f(z)$ is the function represented by the power series in the question.
EDIT
I have thought an answer for it. Since $a_n$ converges to $0$, we can write $\lvert a_n \rvert <1$ for all $n >N_0$. From here, we can say radius of convergence of the power series is bigger than or equal to $1$. If the radius of convergence is bigger than $1$, the series converges on the unit circle. If it is equal to $1$, then points on the unit circle cannot be an isolated singularity. But I am not sure of my answer.
Best Answer
A power series whose coefficients tend to $0$, and whose radius of convergence is therefore at least $1$, can define a function that has isolated sigularities on the unit circle; only those isolated singularities cannot be poles. So your reasoning is not correct, and this claim needs to be changed.
Take for instance the power series with $a_0=0$ and $a_n=\frac1n$ for $n>0$. This defines the function $f: z\mapsto\ln(\frac1{1-z})$ which has an isolated singularity at $1$.
So instead of singularities, you should be thinking about what having a pole means.