Uniform convergence of series of holomorphic functions implies uniform convergence of derivative series on compact subsets. Suppose $f_{n}$
  is a sequence of holomorphic functions in the ball $B\left(0,r\right)$
  such that $\sum_{n=1}^{\infty}\left|f_{n}\left(z\right)\right|$
  converges uniformly on $B\left(0,r\right)$.
  Show that $\sum_{n=1}^{\infty}\left|f_{n}^{'}\left(z\right)\right|$
  converges uniformly on compact subsets of of $B\left(0,r\right)$.
My line of thought was to try using Cauchy's formula to represent the values of the derivatives through the values of the functions themselves and maybe use Weirstrass M-Test but I didn't quite manage to make it work. I'm also a bit confused why the requirement is convergence only on compact subsets and not on the entire ball.
Help would be appreciated! 


My line of thought was to try using Cauchy's formula to represent the values of the derivatives through the values of the functions themselves

That's the right idea.

I'm also a bit confused why the requirement is convergence only on compact subsets and not on the entire ball.

In general, the series/sequence of the derivatives will not converge uniformly on the entire disk, even when the series/sequence of functions does. Consider for a trivial example
$$\sum_{n=1}^\infty \frac{z^n}{n^2}.$$
The series converges uniformly on the closed unit disk, but the series of derivatives,
$$\sum_{n=1}^\infty \frac{z^{n-1}}{n}$$
does not, as the limit function is unbounded for $z\to 1$.
Cauchy's integral formula for the derivatives gives you an estimate
$$\begin{align}
\lvert f'(z) \rvert &= \left\lvert \frac{1}{2\pi i}\int_{\lvert z\rvert = \rho} \frac{f(\zeta)}{(\zeta-z)^2}\,d\zeta\right\rvert\\
&= \frac{1}{2\pi} \left\lvert \int_0^{2\pi} \frac{f(\rho e^{i\varphi})}{(\rho e^{i\varphi}-z)^2} \rho e^{i\varphi}\,d\varphi\right\rvert\\
&\leqslant \frac{\rho}{2\pi} \int_0^{2\pi} \frac{\lvert f(\rho e^{i\varphi})\rvert}{\lvert \rho e^{i\varphi}-z\rvert^2}\,d\varphi\\
&\leqslant \frac{\rho}{(\rho - \lvert z\rvert)^2}\cdot \max_{\lvert \zeta\rvert = \rho} \lvert f(\zeta)\rvert.
\end{align}$$
That estimate grows to infinity as $\lvert z\rvert \to \rho$, but it gives a finite uniform bound on every disk of smaller radius than $\rho$. Thus on every disk $D_{r_1}(0)$ with $r_1 < r$, for every $r_1 < \rho < r$, we have a uniform bound
$$\lvert f'(z)\rvert \leqslant \frac{\rho}{(\rho - r_1)^2} \cdot \sup_{\lvert \zeta\rvert \leqslant \rho} \lvert f(\zeta)\rvert$$
valid for all holomorphic functions on $D_r(0)$. For fixed $r_1$ and $\rho$, the factor with which the supremum of the moduli of $f$ is multiplied is constant, and thus uniform convergence of a series/sequence $f_n$ of functions holomorphic on $D_r(0)$ implies the uniform convergence of the series/sequence of the $f_n'$ on the smaller disk $D_{r_1}(0)$. Since every compact subset of $D_r(0)$ is contained in such a smaller disk, that is the locally uniform or compact convergence of the series/sequence of the derivatives.
