Consider the Bergman space $A^2(\mathbb{D})$, where $\mathbb{D}$ is the open unit disk in $\mathbb{C}$. Show that if $\sum_{n=0}^\infty |a_n|^2$ converges, then $\sum_{n=0}^\infty (n+1)^{1/2} a_n z^n$ is holomorphic in $\mathbb{D}$.
I want to show $f(z)=\sum_{n=0}^\infty (n+1)^{1/2} a_n z^n$ is also in $A^2(\mathbb{D})$. I know that $\{e_n=\sqrt{\frac{n+1}{\pi}}z^n\}_{n\geq0}$ is an orthonormal basis of $A^2(\mathbb{D})$. Therefore I want to show that
$f =\sum_{n=0}^\infty \langle f, e_n\rangle e_n$
$\langle f,e_n \rangle=\int_0^1 f \overline{e_n}dz = \int_0^1 \overline{\sqrt{\frac{n+1}{\pi}}z^n}\sum_{k=0}^\infty (k+1)^{1/2} a_n z^k dz$
Is this approach correct? I feel like I am complicating the problem. Why do we need $\sum |a_n|^2$ converges?
To consider $\langle f, e_n\rangle$, that expression needs to make sense. A priori, considering the inner product requires $f$ to belong to $A^2(\mathbb{D})$, which is just what is to be shown. Since the functions $e_n$ are bounded, we can relax the condition, $f$ being integrable would suffice to make sense of the integral. However, we don't initially know that $f$ is integrable either.
Recall the definition of $A^2(\mathbb{D})$. It's the space of holomorphic square integrable functions on the unit disk. So let's look when a holomorphic function on the unit disk is square integrable. By continuity, every holomorphic function on the unit disk is square integrable over all disks $D_r(0)$ for $0 < r < 1$. Fix an arbitrary $r\in (0,1)$, and consider $g \in \mathscr{O}(\mathbb{D})$. Let
$$g(z) = \sum_{n = 0}^{\infty} b_n z^n.$$
Since the power series converges absolutely and uniformly on $D_r(0)$, we have
\begin{align} \int_{\lvert z\rvert < r} \lvert g(z)\rvert^2\,d\lambda &= \int_0^{2\pi} \int_0^r \lvert g(\rho e^{i\varphi})\rvert^2\rho\,d\rho\,d\varphi \\ &= \int_0^{2\pi} \int_0^r \Biggl(\sum_{n = 0}^{\infty} b_n \rho^n e^{in\varphi}\Biggr)\Biggl(\sum_{k = 0}^{\infty} \overline{b_k} \rho^k e^{-ik\varphi}\Biggr)\rho\,d\rho\,d\varphi \\ &= \sum_{k,n} \int_0^{2\pi} \int_0^r b_n\overline{b_k} \rho^{n+k+1} e^{i(n-k)\varphi}\,d\rho\,d\varphi \\ &= \sum_{n,k} b_n\overline{b_k}\int_0^{2\pi} \frac{r^{n+k+2}}{n+k+2}e^{i(n-k)\varphi}\,d\varphi \\ &= \sum_{n,k} \frac{b_n\overline{b_k}r^{n+k+2}}{n+k+2} 2\pi \delta_{n,k} \\ &= \sum_{n = 0}^{\infty} \pi \frac{\lvert b_n\rvert^2r^{2(n+1)}}{n+1}. \end{align}
Taking the limit $r \to 1$, we see that
$$\int_{\mathbb{D}} \lvert g(z)\rvert^2\,d\lambda = \pi\sum_{n = 0}^{\infty} \frac{\lvert b_n\rvert^2}{n+1},$$
i.e. $g$ is square integrable over the unit disk if and only if
$$\sum_{n = 0}^{\infty} \frac{\lvert b_n\rvert^2}{n+1} < +\infty.$$
Now apply this with
$$b_n = \sqrt{n+1} a_n.$$
We find
$$\sum_{n = 0}^{\infty} \frac{\lvert b_n\rvert^2}{n+1} = \sum_{n = 0}^{\infty} \frac{(\sqrt{n+1})^2\lvert a_n\rvert^2}{n+1} = \sum_{n = 0}^{\infty} \lvert a_n\rvert^2.$$
Thus $\sum \lvert a_n\rvert^2 < +\infty$ is precisely the condition for
$$\sum_{n = 0}^{\infty} \sqrt{n+1} a_n z^n$$
to be in $A^2(\mathbb{D})$.