Show $\{u_n\}$ orthonormal, A compact implies $\|Au_n\| \to 0$

I'm having a bit trouble with this homework exercise.

Let $\mathcal{H}$ be a Hilbert space and $\{u_n\}_{n=1}^\infty$ an orthonormal sequence in $\mathcal{H}$. Let $A$ be a compact operator on $\mathcal{H}$. Show that $\|Au_n \| \to 0$ as $n\to \infty$.

My book defines a compact operator as an operator $A$ such that whenever $f_n$ is bounded, then $Af_n$ has a convergent subsequence (equivalently, the image of $A$ is relatively compact).

It seems I must somehow combine the fact that $Au_n$ has a convergent subsequence with the fact that $\{u_n\}$ is orthonormal. This is where I get stuck. Maybe I can somehow use the fact that $\|u_n-u_m \| = \sqrt{2}\,$ for $m \neq n$.

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If you know $u_n$ converges weakly, and find what its weak limit is, that may help you. – GEdgar Oct 16 '11 at 17:01

I suppose it is clear for you that a compact operator is clearly bounded so continuous. Suppose that the $(Au_n)$ has a convergent subsequence towards $v$ not zero. For the sake of simplicity, let us note also $(Au_n)$ this subsequence. $(u_n)$ is then also an orthonormal sequence in $\mathcal{H}$. Set $v_n=\frac{1}{n}\sum_{k=n}^{2n}u_k$. It is clear that the sequence $(v_n)$ converges towards $0$. But $(Av_n)$ converge towards $v$ which is not zero. QEA.
Thanks for the answer. However, doesn't this just show that the convergent subsequence of $Au_n$ converges to zero? It doesn't show that $Au_n$ itself converges. (I guess it now suffices to show, for example, that $Au_n$ is Cauchy) – Fredrik Meyer Oct 16 '11 at 17:13
@FredrikMeyer it shows actually that ANY convergent subsequence of $(Au_n)$ converges towards $0$. Therefore $(Au_n)$ converges as a whole towards $0$ (if not, being relatively compact as a set, you can easily extract a subsequence not converging towards $0$ ...) – brunoh Oct 16 '11 at 17:27