How to prove that an operator is compact? Consider $T\colon\ell^2\to\ell^2$ an operator such that 
$Te_k=\lambda_k e_k$ with $\lambda_k\to 0$ as $k \to \infty$ how to prove that it is compact?
 A: Let $T_n$ such that $T_n(e_k)=\begin{cases}\lambda_ke_k&\mbox{ if }k\leq n\\\
0&\mbox{ if }k>n
\end{cases}$. Then $T_n$ is finite ranked hence compact and for $v\in\ell^2$, 
$v=(v_0,v_1,\ldots)$
$$\lVert (T-T_n)v\rVert^2=\sum_{k=0}^{+\infty}|\langle((T-T_n)v)_k\rangle|^2=\sum_{k\geq n+1}|(T-T_n)(v_k)|^2=\sum_{k\geq n+1}|\lambda_k|^2\cdot|v_k|^2\\\leq \sup_{k\geq n+1}|\lambda_k|^2\lVert v\rVert_{\ell^2}^2 $$
so $\lVert T-T_n\rVert\leq \sup_{k\geq n+1}|\lambda_k|$ and we conclude that $T_n\to T$ in norm. A norm-limit of compact operators is compact so $T$ is compact. 
Conversely, if $T$ is compact, then you can extract from $\{Te_n\}$ a converging subsequence so you can extract from $\{\lambda_ne_n\}$ a converging subsequence, say $\{\lambda_{n_k}e_{n_k}\}$. Since $\lVert \lambda_{n_{k+1}}e_{n_k{+1}}-\lambda_{n_k}e_{n_k}\rVert^2=|\lambda_{n_{k+1}}|^2+|\lambda_{n_k}|^2\to 0$, we should have $\lambda_{n_k}\to 0$. We thus can prove that for each subsequence $\{\lambda_{n_j}\}$, we can extract that a further subsequence which converges to $0$, hence the whole sequence converges to $0$.
