Given any sequence $(\alpha_n) \in l^2$, the series $\sum_{k=1}^{\infty} \alpha_k e_k$ converges to some element $u \in H$ such that $(u, e_k) = \alpha_k$, for all $k$, and $|u|^2 = \sum_{k=1}^{\infty} \alpha_k^2$.
Comments: This is the corollary 5.10 of book "Functional Analysis, Sobolev Spaces and Partial Differential Equations" of Brezis. The first part says that:
Let $(e_n)$ be an orthonormal basis. Then for every $u \in H$ (Hilbert space), we have:
$u = \sum_{k=1}^{\infty} (u,e_k)e_k $ and $|u|^2 = \sum_{k=1}^{\infty} |(u,e_k)|^2$.
Thanks for your help.
Let $u_m = \sum_{k=1}^m \alpha_k e_k$ for each $m \ge 1$. Then each $u_m \in H$, and whenever $n > m$ you get $$||u_n - u_m||^2 = \sum_{k=m+1}^n |\alpha_k|^2$$ owing to the fact that the $\{e_k\}$ are orthonormal in $H$. Since $\sum_k |\alpha_k|^2 < \infty$, the right-hand-side of the equality tends to $0$ as $m,n \to \infty$. This means $\{u_m\}$ is Cauchy in $H$, and has a limit $u$.
For any element $v \in H$ strong convergence implies $(u_m,v) \to (u,v)$. In particular for each $k$ it happens that $(u_m,e_k) \to (u,e_k)$. However, the sequence $(u_m,e_k)$ is eventually constant: if $m > k$ then $(u_m,e_k) = \alpha_k$. Thus $(u,e_k) = \alpha_k$.
The rest follows immediately from the part you state above.