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

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. 
