Weak convergence in separable Hilbert spaces Let $H$ be a Hilbert space with ONB $\{b_1, b_2, \cdots \}$.

I want to prove that the sequence $x_n = \frac{1}{n}\sum_{i=1}^{n^2}
 b_i$ converges weakly to $0$.

Notice also that $\|x_n\|=1$.
The claim is easily verified on the basis elements, i.e. for large $n$
$$\langle x_n, b_i \rangle= \frac{1}{n}\to 0$$
but I fail to see why it is true for an arbitrary element $y = \sum_i \langle y, b_i \rangle b_i$. Why does
$$\langle x_n, y\rangle = \frac{1}{n} \sum_{i=1}^{n^2} \langle y, b_i \rangle$$
converge to $0$?
 A: First, I will show that if $v \in \text{span}\{b_{m} \, \mid \, m \in \mathbb{N}\}$, then $\langle x_{n}, v \rangle \to 0$ as $n \to \infty$.  Indeed, any such $v$ can be written $$v = \sum_{j = 1}^{N} c_{j} b_{j}$$
for some $N$ and coefficients $c_{1},c_{2},\dots,c_{N}$ in the base field.  Now
$$\langle x_{n}, v \rangle = n^{-1} \sum_{j = 1}^{N \wedge n^{2}} c_{j},$$
which evidently vanishes as $n \to \infty$.  (Here $N \wedge n^{2} = \min\{N,n^{2}\}$.)  
Now suppose $v \in H$.  Then there is a sequence $(v_{m})_{m \in \mathbb{N}} \subseteq \text{span}\{b_{m} \, \mid \, m \in \mathbb{N}\}$ such that $v_{m} \to v$ as $m \to \infty$.  Fix $\epsilon > 0$.  There is a $M \in \mathbb{N}$ such that $\|v - v_{m}\| < \frac{\epsilon}{2}$ if $m \geq M$.  Moreover, by the previous paragraph, there is a $N \in \mathbb{N}$ such that $|\langle x_{n}, v_{M} \rangle| < \frac{\epsilon}{2}$ if $n \geq N$. 
 Hence
\begin{align*}
|\langle x_{n}, v \rangle| &\leq |\langle x_{n}, v_{M} \rangle| + |\langle x_{n}, v_{M} - v \rangle| \\
&\leq \frac{\epsilon}{2} + \|x_{n}\| \|v_{M} - v\| \\
&< \epsilon.
\end{align*}
where I used the fact that $\|x_{n}\| = 1$.  
Note the trick of proving convergence on a dense set first and then using boundedness is applicable in more general situations.  In particular, if $E$ and $F$ are Banach spaces, $(T_{n})_{n \in \mathbb{N}}$ is a bounded sequence of continuous linear maps from $E$ to $F$, and $\mathcal{D}$ is a dense subset of $E$, then $T_{n} \to 0$ pointwise if and only if 
$$\forall x \in D \quad T_{n}(x) \to 0.$$ 
A: Your sequence $(x_n)_{n=1}^\infty$ indeed converges weakly to $0$.
Lemma:

Let $(e_n)_{n=1}^\infty$ be an orthonormal basis for a separable Hilbert space $H$. Let $(x_n)_{n=1}^\infty$ be a sequence of vectors in $H$.
Then $(x_n)_{n=1}^\infty$ converges weakly to $x \in H$ if and only if it is bounded and converges to $x$ component-wise, i.e. there exists $M > 0$ such that $\|x_n\| \le M$ for all $n \in \mathbb{N}$ and $\langle x_n, e_m\rangle \xrightarrow{n\to\infty} \langle x, e_m\rangle$ for all $m \in \mathbb{N}$.

Proof:
If $x_n \rightharpoonup x$, then $(x_n)_{n=1}^\infty$ is bounded, and $\langle x_n, e_m\rangle \xrightarrow{n\to\infty} \langle x, e_m\rangle$ for all $m \in \mathbb{N}$ by definition of weak convergence and the Riesz representation theorem.
Conversely, assume that there exists $M > 0$ such that $\|x_n\| \le M$ for all $n \in \mathbb{N}$ and $\langle x_n, e_m\rangle \xrightarrow{n\to\infty} \langle x, e_m\rangle$ for all $m \in \mathbb{N}$.
Linearity of the inner product immediately implies that $\langle x_n, y\rangle \xrightarrow{n\to\infty} \langle x, y\rangle$ where $y = \sum_{i=1}^p \langle y, e_i\rangle e_i$ is a finite sum of the basis vectors.
We have to show that $\langle x_n, v\rangle \xrightarrow{n\to\infty} \langle x, v\rangle$ for all $v \in H$.
Let $\varepsilon > 0$ and pick $p \in \mathbb{N}$ such that $\sum_{k=p+1}^\infty \left|\langle v, e_k\rangle\right|^2 < \left(\frac{\varepsilon}{2(M + \|x\|)}\right)^2$.
Since $$\left\langle x_n, \sum_{k=1}^p \langle v, e_k\rangle e_k \right\rangle \xrightarrow{n\to\infty} \left\langle x, \sum_{k=1}^p \langle v, e_k\rangle e_k \right\rangle$$
there exists $n_0 \in \mathbb{N}$ such that $n \ge n_0$ implies $\left|\left\langle x_n, \sum_{k=1}^p \langle v, e_k\rangle e_k \right\rangle - \left\langle x, \sum_{k=1}^p \langle v, e_k\rangle e_k \right\rangle\right| < \frac{\varepsilon}2$.
For every $n \ge n_0$ we have:
\begin{align}
\left|\langle x_n, v\rangle - \langle x, v\rangle\right| &= \left|\left\langle x_n,  \sum_{k=1}^p \langle v, e_k\rangle e_k\right\rangle - \left\langle x,  \sum_{k=1}^p \langle v, e_k\rangle e_k\right\rangle + \left\langle x_n,  \sum_{k=p+1}^\infty \langle v, e_k\rangle e_k\right\rangle - \left\langle x,  \sum_{k=p+1}^\infty \langle v, e_k\rangle e_k\right\rangle\right| \\
&= \left|\left\langle x_n,  \sum_{k=1}^p \langle v, e_k\rangle e_k\right\rangle - \left\langle x,  \sum_{k=1}^p \langle v, e_k\rangle e_k\right\rangle\right| + \left|\left\langle x_n - x,  \sum_{k=p+1}^\infty \langle v, e_k\rangle e_k\right\rangle\right|\\
&< \frac{\varepsilon}2 + \underbrace{\|x_n-x\|}_{<M+ \|x\|}\sqrt{\sum_{k=p+1}^\infty \left|\langle v, e_k\rangle\right|^2}\\
&< \varepsilon
\end{align}
We conclude $x_n \rightharpoonup x$.

Applied to your example:
$$\|x_n\|^2 = \frac{1}{n^2}\left\|\sum_{i=1}^{n^2} b_i\right\|^2 = 1$$
so $(x_n)_{n=1}^\infty$ is bounded. You have already demonstrated that $\langle x_n, b_j\rangle \xrightarrow{n\to\infty} 0 = \langle 0, b_j\rangle$ for all $j \in \mathbb{N}$.
The lemma implies $x_n \rightharpoonup 0$.
