Space generated by orthonormal sets is closed

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Let $H$ be a Hilbert space and $\left\{ e_{n}:n\in\mathbb{N}\right\}$ be orthonormal. $$V=\left\{ x\in H:\text{ there exists }d_{n}=\sum_{j=1}^{n}a_{j}e_{j},a_{j}\in\mathbb{R},\text{ such that }\left\Vert d_{n}-x\right\Vert \rightarrow0\right\}$$

How can we show that $V$ is closed in $H$? ($d_n$ can only be represented as sum of the first $n$ terms)

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Let $M$ be the closed subspace spanned by $(e_n)$. Then $(e_n)$ is an orthonormal basis for $M$ and every element $x$ of $M$ is of the type $\sum \langle x, e_j \rangle e_j$. It follows that $x \in V$. On the other it is also obvious that $x \in V$ implies $x \in M$. Hence $V$ is the closed subspace spanned by $(e_n)$.