We are given $$f = \sum_{n= - \infty} ^{\infty} a_n (z-z_0)^n \in \mathcal{O} ( \text{ann} (z_0, r, R)), \ \ 0<r<R< \infty. $$
Prove that $$\frac{1}{\pi} \int _{ann (z_0, r, R)} |f(z)|^2 d \lambda(z) = \sum _{n \neq -1} \frac{R^{2n+2} - r^{2n+2}}{n+1}|a_n|^2 + 2 \log \frac{R}{r}|a_{-1}|^2.$$
I know that $$a_n = \frac{1}{2 \pi i} \int_{\partial B(z_0, \rho)} \frac{f(s)}{(s-z_0)^{n+1}}ds, \ \ \rho \in (r,R), \ n \in \mathbb{Z}$$
We know that the series above is convergent, so $R = \frac{1}{\limsup_ {n \rightarrow \infty} \sqrt[n]{|a_n|}}$ and $r = \limsup_ {n \rightarrow \infty} \sqrt[n]{|a_{-n}|}$.
In the series $$\sum _{n \neq -1} \frac{R^{2n+2}}{n+1}|a_n|^2, \ \ \sum _{n \neq -1} \frac{r^{2n+2}}{n+1}|a_n|^2$$ we have $b_n = \frac{|a_n|^2}{n+1}$ and radii of convergence are $$R'=\frac{1}{\limsup_ {n \rightarrow \infty} \sqrt[n]{\frac{|a_n|^2}{n+1}}} \ge \frac{1}{\limsup_ {n \rightarrow \infty} \sqrt[n]{|a_n|^2}} \ge \frac{1}{\limsup_ {n \rightarrow \infty} \sqrt[n]{|a_n|}}^2 = R^2,$$ so $R' \ge R^2$,
and similarly, $$r' = \limsup_ {n \rightarrow \infty} \sqrt[n]{\frac{|a_{-n}|^2}{n+1}} \le \limsup_ {n \rightarrow \infty} \sqrt[n]{|a_{-n}|^2} \le r^2.$$
So the two series have radii of convergentce $R', r'$, assuming they have the form $\sum b_n (z-z_0)^n$ with $z=R^2$ or $-r^2$.
Does that make sense? Could you tell me how to prove the equality of the integral and the series?
Assume $z_0 = 0$. I write $A_{r,R}(0)$ for your annulus.
To integrate over the annulus, we first start by parametrizing it. Let $u: [0,1]^2 \to A_{r,R}(0)$ by $ u(s,t) := (r + s(R-r))e^{2 \pi i t}$. It is easy to check that Jacobian of this parametrization is $2 \pi (R-r)( r + s(R-r))$.
Now $$\displaystyle{\int_{A_{r,R} (0)} g(z) := \int_{[0,1]^2} g(u(s,t)) \text{Jac(u(s,t))} \text{ds dt} }$$.
Which is,
$ \displaystyle{2 \pi (R-r) \int_{[0,1]} (r + s(R-r))\int _{[0,1]}\left(\sum_{n \in \mathbb{Z}} a_n \left(r + s(R-r) \right)^n e^{2 \pi i t n} \right)\cdot \overline{\left(\sum_{n \in \mathbb{Z}} a_n \left(r + s(R-r) \right)^n e^{2 \pi i t n} \right)} \text {dt ds}} $
(*)
Note that, for $k \in \mathbb{Z}$,
$\displaystyle{\int_{[0,1]} e^{2 \pi i t k} dt = 0} $ if $k \ne 0$ and equals to $1$, when $k=0$.
Which means,
$\displaystyle{\int _{[0,1]}\left(\sum_{n \in\mathbb{Z} } a_n \left(r + s(R-r) \right)^n e^{2 \pi i t n} \right)\cdot \overline{\left(\sum_{n \in\mathbb{Z}} a_n \left(r + s(R-r) \right)^n e^{2 \pi i t n} \right)} \text {dt}}$
$\displaystyle{= \int_{[0,1]}\left ( \sum _{n \in \mathbb{Z}} a_n (r + s(R-r))^n e^{2 \pi i t n} \right )\cdot \left( \sum _{n \in \mathbb{Z}} \overline{a_n} (r + s(R-r))^n e^{-2 \pi i t n} \right )} \text {dt}$
$\displaystyle{=\sum _{n \in \mathbb{Z}} |a_n|^2 (r + s(R-r))^{2n}}$.
So, now the original integral (*) becomes,
$$\displaystyle{2 \pi (R-r) \int_{[0,1]}\sum_{n \in \mathbb{Z}} |a_n|^2 (r + s(R-r))^{2n+1} \text{ds}}$$
Uniform convergence of the power series inside the annulus allows us to switch sum and integration, which gives
$$\displaystyle{2 \pi (R-r) \sum_{n \in \mathbb{Z}} \int_{[0,1]}|a_n|^2 (r + s(R-r))^{2n+1} \text{ds}}$$.
This is easy to integrate and gives you the required answer, i.e $$ \pi \sum _{n \neq -1} \frac{R^{2n+2} - r^{2n+2}}{n+1}|a_n|^2 + 2 \pi \log \frac{R}{r}|a_{-1}|^2$$