Let $H$ be a separable Hilbert space, define a bounded linear operator $T:H \rightarrow H$, show it is compact if $\sum \|Te_n\|_H < \infty$.
My attempt:
We show that $T(B)$ is totally bounded.
For each $h\in B$ the unit ball, we have $$\|h\| = \bigg\|\sum_{n=1}^\infty h_n e_n \bigg\| = \sum_{n=1}^\infty h_n^2 = 1$$ Let $\epsilon >0$ be given, there exists an $N$ such that $$\sum_{n=N}^\infty \|Te_n\| <\epsilon/2.$$
Define $$T_N (h) = \sum_{n=1}^{N-1} h_n Te_n$$ and we see that $$\|T(h) - T_N(h)\| = \bigg\|\sum_{n=N}^\infty h_n Te_n\bigg\| \leq \sum_{n=N}^\infty \|Te_n\|\leq \epsilon/2 .$$ (I am not sure about the above inequality)
Then since $T_N(B)$ is finite dimensional, it is totally bounded, it can be covered with finite many balls with radius $\epsilon/2$. Since we have shown $$\|T(h) - T_N(h)\| \leq \epsilon/2,$$ expand each ball's radius from $\epsilon/2$ to $\epsilon$, the finite collection will cover $T(B)$.
Thank you very much!
Define $T_{N}h = \sum_{n=1}^{N}h_{n}Te_{n}$. Because $\sum_{n}\|Te_{n}\| < \infty$, then there is a constant $M$ such that $\|Te_{n}\| \le M$, or $\|Te_{n}\|^{1/2} \le M^{1/2}$ for all $n \ge 1$. Then you use the Cauchy-Schwarz inequality to get what you want: $$ \begin{align} \|T_{N}h-Th\| & \le \sum_{n=N+1}^{\infty}|h_{n}|\|Te_{n}\| \\ & \le M^{1/2}\sum_{n=N+1}^{\infty}|h_{n}|\|Te_{n}\|^{1/2} \\ & \le M^{1/2}\left(\sum_{n=N+1}^{\infty}|h_{n}|^{2}\right)^{1/2}\left(\sum_{n=N+1}^{\infty}\|Te_{n}\|\right)^{1/2} \\ & \le M^{1/2}\left(\sum_{n=N+1}^{\infty}\|Te_{n}\|\right)^{1/2}\|h\|. \end{align} $$ This inequality gives you a uniform approximation of $T$ by a finite rank operator $T_{N}$, with $$ \|T_{N}-T\|^{2} \le M\sum_{n=N+1}\|Te_{n}\|. $$