Can an element of the closure of the span of an orthonormal sequence in a Hilbert space be represented by a Fourier series? A problem I'm struggling with is this: If $(e_k)$ is an orthonormal sequence in a Hilbert space $H$, and we denote $M=\operatorname{span}(e_k)$, then for all $x\in \bar M$ we have that $x$ can be represented as 
$$\sum_{k=1}^{\infty}\langle x,e_k\rangle e_k$$
Now as $x\in \bar M \implies x\in H$ quite obviously it follows that the above series converges. However, I am struggling to prove that this series converges to $x$. I'm pretty sure I need to use the fact that $x$ is in the closure of $M$ but I'm not quite sure how to bring this in. Any hints as to how to start the proof would be appreciated.
PROGRESS:
If we let $x'$ denote the limit of the series I'm able to show using Daniel's hint that $\langle x',e_j\rangle=\langle x,e_j\rangle \ \  \forall j\in \mathbb{N}$. As this is true for every $j$ I'd like to be able to conclude that $x'=x$, but I'm not really sure if I have any justification for such a conclusion.
 A: If $x \in \overline{M}$, then, for every $\epsilon > 0$, there exists $m = \sum_{k=1}^{N}\alpha_k e_k$ such that $\|x-\sum_{k=1}^{N}\alpha_ke_k\| < \epsilon$. For every $n \ge N$, 
$$
      \left(x-\sum_{k=1}^{n}\langle x,e_k\rangle e_k\right)\perp \left(\sum_{k=1}^{n}\langle x,e_k\rangle e_k-m\right)
$$
because the difference on the left is orthogonal to $e_1,e_2,\cdots,e_n$, and because the vector on the right is a linear combination of such $e_k$. Therefore, by the Pythagorean Theorem,
\begin{align}
     \epsilon^2 > \|x-m\|^2
   & =\|x-\sum_{k=1}^{n}\langle x,e_k\rangle e_k\|^2+\|\sum_{k=1}^{n}\langle x,e_k\rangle e_k-m\|^2 \\
   & \ge \|x-\sum_{k=1}^{n}\langle x,e_k\rangle e_k\|^2,\;\;\; n \ge N.
\end{align}
By the definition of vector limit, $x=\sum_{k=1}^{\infty}\langle x,e_k\rangle e_k$, because, for every $\epsilon > 0$, there exists $N$ such that $\|x-\sum_{k=1}^{n}\langle x,e_k\rangle e_k\| < \epsilon$ whenever $n \ge N$.
Conversely, if $x=\sum_{k=1}^{\infty}\langle x,e_k\rangle e_k$, then $x \in \overline{M}$ because the partial sums are in $M$.
