In the chapter concerning modes of convergence, I've seen this theorem in my book :
For $(f_n)_{n\in \mathbb{N}}$ a sequence of power series defined in $\mathbb{C}$: $$f_n(x)=\sum_{m=0}^\infty a_{n,m}x^m$$ If $\forall n\in\mathbb{N}$, the radius of convergence $R_n>R>0$, and $\forall z : |z|<R$, $|f_n(z)|\le M$. Then if $f_n \to f$ pointwise in the open circle of radius $R$, $f_n$ converges to $f$, and uniformly in any closed circle in the open one.
To prove it, I first proved that for all $m$, $(a_{n,m})$ converges in $\mathbb{C}$, say to $a_m$. Then I could show that the new power series $$g(x)=\sum_{m=0}^\infty a_{m}x^m$$ also has a radius $\ge R$. I wanna show that $f_n$ converges to $g$ uniformly, and by the uniqueness of limit, I can make the conclusion that $g=f.$ But I failed to show it. I think it should involve using the condition that $|f_n(z)|\le M$, but I haven't figured out how. Would anyone give me a hint?