Analytic function on $\mathbb{D}$ belongs to Hardy space $H^2$ iff $\int_{\mathbb{D}}|f'(z)|^2(1-|z|^2)dA(z)<\infty$ where $dA$ is the area measure.
By definition, $f\in H^2$ if $$\|f\|^2_{H^2}=:\sup_{0<r<1}\frac{1}{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^2d\theta<\infty$$ Moreover we know that $\|f\|^2_{H^2}=\sum_{k=0}^{\infty}|a_k|^2$ where $\{a_k\}\in \ell^2$.
Then, we have, $$\begin{align*}\int_{\mathbb{D}}|f'(z)|^2(1-|z|^2)dA(z)&=\int_0^{2\pi}\int_0^1|f'(re^{i\theta})|^2(1-r^2)rdrd\theta\\ &=\int_0^{2\pi}|f'(re^{i\theta})|^2d\theta\int_0^1r(1-r^2)dr\\ &=\frac{1}{4}\sum k^2|a_k|^2r^{2n-2}\end{align*}$$
I think the calculation is a little messed up, and I'm not sure how I should go about proving the statement.
\begin{align*}I=\int_{\mathbb{D}}|f'(z)|^2(1-|z|^2)dA(z)&=\int_0^1\int_0^{2\pi}|f'(re^{i\theta})|^2(1-r^2)rd\theta dr\\ &=2\pi\sum_{k \ge 1} k^2|a_k|^2\int_0^1r^{2k-2}r(1-r^2)dr \end{align*}
(by applying orthogonality in $\theta$ and switching sum and integral by absolute convergence)
Integrating in $r$ one gets that:
$I=2\pi\sum_{k \ge 1} k^2|a_k|^2 (1/(2k)-1/(2k+2))=\pi\sum_{k \ge 1} \frac{k}{k+1}|a_k|^2$ so obviously $I$ coneverges or diverges precisely when $\sum_{k \ge 1}|a_k|^2$ does so hence we have the required if and only if ($f(0)$ has no bearing on the convergence/divergence of $\sum |a_k|^2$)