Showing that the operator is compact

Question

Let $H$ separable Hilbert space and $(e_n)_n$ a orthogonal basis of $H$. Let $T$ a Hilbert-Schimidt operator and $a_n = T^*(e_n)$ where $T^* $ is Hermitian adjoint. I proved that $T(x)= \sum\langle x,a_n\rangle e_n$ as well as $(\lVert a_n\rVert)_n \in \ell_2$. The exercise says that after proving this it is only to conclude that $T$ is compact, but I am not able to show it.

Answer 1

Write $T_n(x)=\sum_{i=1}^{n}\langle a_i,x\rangle e_i$.

$\|T(x)-T_n(x)\|^2=\sum_{i>n}|\langle x,a_i\rangle|^2 \leq \sum_{i>n}\|x\|^2\|a_i\|^2$ $=\|x\|^2\sum_{i>n}\|a_i\|^2$.

Since $(\|a_n\|)\in l_2$, for every $c>0,$ there exists an integer $N_c$ such that $n>N_c$ implies that $\sum_{i>n}\|a_i\|^2<c^2$. We deduce that for $n>N_c$, $\|T(x)-T_n(x)\|<c\|x\|$. This implies that for $n>N_c$, $\|T-T_n\|<c$ and $\lim_nT_n=T$. This shows that $T$ is compact since it is the limit of the sequence $(T_n)$ and each $T_n$ has a finite rank.