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! 
 A: 
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.
