compute norm and weak limit in Hilbert space of $u^N = \frac{1}{\sqrt{N}} \sum_{k=1}^N e_k$ Let $H$ be a Hilbert space with ONB $\{e_k\}_{k=1}^\infty$.
Define for every $N \in \mathbb{Z}_+$ the vector $$u^N = \frac{1}{\sqrt{N}} \sum_{k=1}^N e_k$$
The task is now:

Compute the norm $||u^N||$ for every $N$ and show that $u^N$ converges weakly and compute the weak limit.

To compute the norm, I thought about applying Parsevals formula, wich states that for every $h \in H$
$$ ||h||^2 = \sum_{k=1}^\infty |(h,e_k)|^2,$$
hence I need to compute 
$$ ||u^N||^2 = \sum_{k=1}^\infty |(u^N,e_k)|^2 .$$
I get $$ (u^N,e_k) = \left(\frac{1}{\sqrt{N}} \sum_{i=1}^N e_i,e_k\right) 
= \frac{1}{\sqrt{N}} \sum_{i=1}^N(e_i,e_k) = \frac{1}{\sqrt{N}} \sum_{i=1}^N \delta_{ik} = \frac{1}{\sqrt{N}}$$ since the $e_k$'s are an ONB and hence $(e_i,e_k)=0%$ for $i \neq k$.
Moreover I have $(u^N,e_k) = 0 $ for $k>N$ because of the same reason.
All in all I get
$$ ||u^N||^2 = \sum_{k=1}^\infty |(u^N,e_k)|^2 = \sum_{k=1}^N|\frac{1}{\sqrt{N}}|^2 = \sum_{k=1}^N \frac{1}{N} = N*\frac{1}{N} = 1$$
Is this right so far?
Now to the second part with weak convergence.
Weak convergence from $x_n$ to $x$ means that it holds 
$$ \xi(x_n) \rightarrow \xi(x) \;\; \operatorname{for all} \xi \in H^*.$$
Since I have a Hilbert space, I know from the Riesz representation Theorem, that every $\xi \in H^*$ can be written as an inner product, which means that there is an $y \in H$ such that $\xi(x)=(y,x)$ for all $x \in H$. Now I know that the $e_k$ are an ONB and I can write $y=\sum_{k=1}^\infty y_k e_k$ with coefficients $y_k \in \mathbb{R}$. I can plug it into the defintion of weak convergence and I need to find an $u\in H$ such that for every $y \in H$ it holds that 
$$(y,u^N) \rightarrow (y,u) $$
Now I don't know how to go on. How can I compute the $u$ which satisfies this convergence?
I could expand 
$$ (y,u^N) = \left(\sum_{i=1}^\infty y_i e_i, \frac{1}{\sqrt{N}} \sum_{k=1}^N e_k \right) = \frac{1}{\sqrt{N}} \sum_{i=1}^\infty y_i \left(e_i, \sum_{k=1}^N e_k\right) = \frac{1}{\sqrt{N}} \sum_{i=1}^N y_i,$$
but I cant interpret this. 
Some hints or solutions would be really helpful.
 A: Computations with norms are correct. As for second part $u^N$ weakly converges to zero.

$\mathit{Lemma}$. Assume that $v^n$ is a bounded sequence in $H$ and $S \subset H$ is a dense subset. Also assume that for all $x \in S$ $\lim\limits_{n \rightarrow \infty} (v^n,x) = 0$. Then $v^n$ weakly converges to $0$.

It is easy to see that $(u^N,e_k) \rightarrow 0$ for all $k \in \mathbb{N}$ (since this sequence coincides with $\frac{1}{\sqrt{N}}$ for big enough $N$). Therefore $(u^N,x) \rightarrow 0$ for all $x \in S = span\{e_k,\; k \in \mathbb{N}\}$. This set $S$ is dense in $H$ and sequence $u^N$ is bounded (since $||u^N|| = 1$). Therefore $u^N$ weakly converges to $0$ by foregoing lemma.
$\mathit{Proof\; of\; lemma}.$ Let $||v^n|| \le C$. Consider arbitrary $x \in H$ and fix $\varepsilon > 0$. Then there exists $x_0 \in S$ s.t. $||x - x_0|| \le \frac{\varepsilon}{2C}$. Therefore $|(v^n,x)| \le |(v^n,x_0)| + |(v^n,x - x_0)| \le |(v^n,x_0)| + \frac{\varepsilon}{2}$. But sequence $|(v^n,x_0)|$ converges to zero and therefore it is less than $\frac{\varepsilon}{2}$ for big enough $n$. Therefore $|(v^n,x)| \le \varepsilon$ for big enough $n$. Hence $(v^n,x)$ converges to $0$ for all $x \in H$.
