Show that if a power-series converges for any value of $z_{0}$ of $z$, it will be absolutely convergent for all values of $z$ whose representation points are within a circle which passes through $z_{0}$ and has its center at the origin.
Proof Attempt Let z be such a point, so we have $\left| z\right| < \left| z_{0}\right| $. Now, since $\sum _{n=0}^{\infty }a_{n}z_{0}^{n}$ converges, $a_{n}z_{0}^{n}\rightarrow 0$ as $n\rightarrow \infty $, so we can find M(independent of n) such that $\left| a_{n}z_{0}^{n}\right| < M$ and we observe that $\left| a_{n}z^{n}\right| < M\left| \dfrac {z} {z_{0}}\right| ^{n}$. So every term in the series $\sum _{n=0}^{\infty }\left| a_{n}z^{n}\right| $ is less than the corresponding term in the convergent geometric series $\sum _{n=0}^{\infty }M\left| \dfrac {Z} {Z_{0}}\right| ^{n}$ the series is therefore convergent and so the power-series is absolutely convergent , as the series of moduli of its terms is a convergent series. I am unsure how to tackle the second part(converse) of the problem. Any help would be much appreciated.