If limit of $f(n)$ is zero then the operator is compact

Question

I want to prove the following:

Suppose $\mathfrak{H}$ is the Hilbertspace $l^2(\Bbb{N})$ and $T_f$ the multiplication operator on $\mathfrak{H}$, thus $T_f\psi=f\psi$ for $f:\Bbb{N}\rightarrow\Bbb{C}$. Suppose $\lim_{n\rightarrow\infty}{|f(n)|}=0$. Then $T_f$ is compact.

Can one prove this with finite rank operators or just with Bolzano-Weierstrass? Can someone help me? Thanks.