if $f_n(z)\rightarrow f(z)$ uniformly, then $ \frac{f'_n(z)}{f_n(z)}\rightarrow\frac{f'(z)}{f(z)}$ uniformly?

Question

I am reading "The Taylor Series: An Introduction to the Theory of Functions of a Complex Variable", Dienes P.N., Dover (1957).

In the proof of a theorem (Hurwitz) on page 351 it says that, if $f_n(z)\rightarrow f(z)$ uniformly, and given a circle C in wich $f(z)\neq 0$ then $$ \frac{f'_n(z)}{f_n(z)}\rightarrow\frac{f'(z)}{f(z)}$$ uniformly on C. And moreover, $$\frac{1}{2\pi i}\oint_C \frac{f'_n(z)}{f_n(z)}dz\rightarrow\frac{1}{2\pi i}\oint_C \frac{f'(z)}{f(z)}dz$$

This statements are not exactly clear to me, is there any small proof to convince me?

Answer 1

This uses the fact that when a sequence $(f_n)_{n\in\mathbb N}$ of analytic functions converges uniformly to $f$ on each compact set, then $(f_n{\,'})_{n\in\mathbb N}$ also converges uniformly to $f'$ on each compact. This, together with the fact that $f$ has no zeros on $C$ (which implies that $f_n$ has no zeros there if $n\gg1$) is enough to prove that statement.