Q:

Show that $0$ belongs to the weak closure of $K$

I'm working on the following problem from my introductory course on functional analysis:
Problem: Let $H$ be a separable Hilbert space with orthonormal basis $(e_n)_{n \in \mathbb{N}}$ and consider its weak topology (we defined the weak topology on $H$ as follows. You have a family $\mathcal{P}$ of seminorms $\mathcal{P} = \left\{ p_y \mid y \in H \right\}$ with $p_y(x) = | \langle y, x \rangle |$ for all $x \in H$. The resulting semi-norm topology on $H$ is called the weak topology).
Let $K := \left\{ \sqrt{n} e_n \mid n \geq 1 \right\}$. Prove that the point $0$ belongs to the weak closure of $K$.
Attempt: I'm given an explicit hint: proof by contradiction. (Recall the definition of closure: for a pseudo metric space $(X, \mathcal{D})$, a point $x \in X$ lies in the closure of a subset $Z \subset X$ if and only if for all $d_1, \ldots, d_n \in \mathcal{D}$ and all $\epsilon > 0$ there exists a $z \in Z$ satisfying $d_i (x, z) < \epsilon$ for all $i=1, \ldots n$.)
So suppose that $0$ does not belong to the weak closure of $K$.Then by definition there exist $x_1, \ldots, x_n \in H$ and a $\delta > 0$ such that for all $n \in \mathbb{N}$ there exists an $i \in \left\{ 1, \ldots, k \right\}$ such that $$ | \langle \sqrt{n} e_n, x_i \rangle| \geq \delta. $$
However, I already proved that $e_n \to 0$ weakly when $(e_n)$ is an orthonormal sequence in $H$. I think I have to use this. Since $e_n \to 0$ weakly, we know that for every $x \in H$, we have $d_x (e_n, 0) = | \langle x, e_n \rangle | \to 0$. Since this holds for all $x \in H$ and the index $i \in \left\{ 1, \ldots, n \right\}$ is fixed (as above), I could define $x_n := \sqrt{n} x_i$? Then this would tell me that $$ | \langle \sqrt{n} x_i, e_n \rangle | \to 0. $$ However, this seems to me be a contradiction with my assumption above? So that would conclude the proof?
Is my reasoning correct? Thank you in advance.

A:

As was commented by egorovik, note that 
$$ | \langle \sqrt{n} e_n, x_i \rangle| \geq \delta. $$
implies that for all $N$
$$\sum_{n=1}^N|\langle x_i,e_n\rangle|^2\geq\delta\sum_{n=1}^N\frac{1}{n}$$
contradicting the fact that the series on the l.h.s converges.

