If ${a_k}$ is bounded, then $\sum_{k=0}^{\infty}{a_k}z^{k}$ defines an analytic function on the open unit disk.

Question

Is this conclusion true, If ${a_k}$ is bounded, then $\sum_{k=0}^{\infty}{a_k}z^{k}$ defines an analytic function on the open unit disk.

I tried to construct counterexamples but looks like the statement is true.

How can I prove it?

Answer 1

It converges for $|z|<1$ by comparison with the geometric series $\sum_k|z|^k$. So its radius of convergence is at least one. Now use the theorem that a power series is analytic in the interior of its disc of convergence.

Answer 2

Use M-test. $|z| <1$ implies $\sum |a_k||z^{k}|\leq M\sum|z|^{k} =M\frac 1 {1-|z|}<\infty$ where $M=\sup \{|a_k|:k\geq 1\}$. This implies that the sum is analytic in the open unit disk.

Answer 3

If $M>0$ is such that $(\forall n\in\mathbb Z^+):\lvert a_n\rvert\leqslant M$ and if $\lvert z\rvert<1$, then $(\forall n\in\mathbb Z^+):\lvert a_nz^n\rvert\leqslant M\lvert z\rvert^n$ and therefore the series $\sum_{n=0}^\infty a_nz^n$ converges. And the sum of a convergent power series on an open set is always an analytic function.