Prove or disprove existence of a sequence converging weakly to $0$ in an infinite dim Hilbert space This is a problem on an old analysis qual, the prompt is: 
"Prove or give a counter example: if $H$ is an infinite dimensional Hilbert space and $0$ is the zero vector in $H$, then there exists a sequence $\{x_n\}$ in $H$ so that $||x_n|| \ge 1$ and $\{x_n\}$ converges weakly to the zero vector $0$ in $H$."
I know that the unit ball is not necessarily weakly compact in an infinite dimensional space if it is not reflexive. But this is specifying the existence of a single sequence, which doesn't say anything about every sequence having a convergent subsequence etc.
Since it is a Hilbert space I know it is equivalent to $(x_n,y) \rightarrow 0$ for all $y \in H$ for such a space. I was tempted to assume a countable orthonormal  use Parseval's Identity to show $||x_n||^2$ could be made less than 1, but this would seem to require $(x_n,e_k)$ to converge uniformly (ie independently of $k$ where the $e_k$'s are the orthonormal set).
Anyway, I am stuck. Any suggestions? Thanks!

Okay, based on David Mitra's comment we can construct an orthonormal sequence. For ease of showing the required result let's call this $\{x_n\}$. Then Bessel's inequality gives for all $y \in H$ that:
$\sum_{n=1} ^\infty |(x_n,y)|^2 \le ||y||^2$. Assume $||y||^2$ is not infinite, which seems reasonable then this is a convergent sequence which implies the terms much approach zero, hence $\lim_{n \rightarrow \infty} (x_n,y) = 0$, thus by definition we have a sequence that converges to the 0 vector (and since it is orthonormal $||x_n|| = 1$ so it satisfies $||x_n|| \ge 1$.
I wonder why the question specified the norm being greater than or equal to 1. Just to throw to make it more confusing?
