I am currently reading Sarason's complex analysis text. I am reading the proof that every Cauchy integral can be represented as power series.
The part I don't understand is why if $\phi$ is bounded and continuous, then $$ \sum_{n = 0}^\infty \frac{\phi(\zeta)(z - z_0)^n}{(\zeta - z_0)^{n + 1}} d\zeta$$ will uniformly converge on $\gamma$.
The idea can be seen as following from the Weierstrass $M$-test. Given the series $\sum_{n=1}^\infty f_n(x)$, provided that $|f_n(x)| \le M_n$ for some sequence of real $M_n$ and $\sum_{n=1}^\infty M_n$ converges, then $\sum_{n=1}^\infty f_n(x)$ converges uniformly on the domain of the $f_n$.
We note that, since $\phi$ is assumed bounded and continuous, then there is a constant $M$ such that $|\phi(\zeta)| \le M$. We also note that $$ \left| \frac{(z-z_0)^n}{(\zeta-z_0)^{n+1}} \right| = \frac{1}{|\zeta-z_0|} \cdot \left| \frac{ z-z_0 }{ \zeta-z_0 } \right|^n $$ We will let $\rho$ denote the ratio, $$ \rho := \left| \frac{ z-z_0 }{ \zeta-z_0 } \right| $$ We have seen that $0 \le \rho<1$. Note that the remaining factor, $|\zeta-z_0|^{-1}$, is independent of the index of summation.
We thus see that $$ \left| \sum_{n = 0}^\infty \frac{\phi(\zeta)(z - z_0)^n}{(\zeta - z_0)^{n + 1}} \right| \le \sum_{n=0}^\infty \left| \frac{\phi(\zeta)(z - z_0)^n}{(\zeta - z_0)^{n + 1}} \right| \le \frac{M}{|\zeta-z_0|} \sum_{n=0}^\infty \rho^n $$ Since $\rho \in [0,1)$, the summation converges, and so the original series converges uniformly by the $M$ test.