Let $f:\Bbb{D} \rightarrow G\subset\Bbb{C}$ where $f$ is analytic and also $f$ is an isomorphism between $\Bbb{D}$ and $G$. $f$ has the series expansion $f(z)=\sum_{n=0}^{\infty}c_n z^n$.
I must show that the area of $G=:S$ is
$$S = \pi \sum_{n=1}^{\infty} n|c_n|^2$$
My try
I've already proved that $S = \iint_{\Bbb{D}} |f'(x+iy)|^2dxdy$. Now making the change $z=\rho e^{i\theta}$ the integral becomes
$$S=\int_0^1 \rho d\rho \int_{-\pi}^{\pi} |f'(\rho e^{i\theta})|^2 d\theta $$
Using the Parseval's Identity and with some little manipulations I get that the integral becomes
$$S =2\pi \int_0^1 \sum_{n=1}^{\infty} n^2 |c_n|^2 \rho^{2n-1} d\rho$$
Now if could just change the sum and integrate sign the result will follow inmediatly but I don't know right now how can I justify that $S$ can be written as
$$S = 2\pi \sum_{n=1}^{\infty} n^2 |c_n|^2 \int_0^1 \rho^{2n-1} d\rho$$
Any ideas??