The exewrcise is:
Let $\mathcal{H}$ be an infinite-dimensional Hilbert space, with an orthonormal basis $\{e_n\}$ and let $T \in \mathcal{B}(\mathcal{H})$. Show that if T is compact then $\lim_{n\rightarrow \infty}\|Te_n\|=0$.
Here is the solution given:
But I am wondering how they get the inequality given. It seems there is some sort of the triangle inequality. We have that for every n(r): $\|y\| \le \|y-Te_{n(r)}\|+\|Te_{n(r)}\|$. So $\|\frac{1}{r}Te_{n(r)}\|\ge \frac{1}{r}\|y\|-\frac{1}{r^2}$. Together this gives that $\sum\limits_{k=1}^m\|\frac{1}{r}Te_{n(r)}\|\ge \sum\limits_{k=1}^m\frac{1}{r}\|y\|-\sum\limits_{k=1}^m\frac{1}{r^2}$. But this is not quite what they have, they have the norm outside the sum in the first part: $\|\sum\limits_{k=1}^m\frac{1}{r}Te_{n(r)}\|\ge \sum\limits_{k=1}^m\frac{1}{r}\|y\|-\sum\limits_{k=1}^m\frac{1}{r^2}$, how can they have gotten this form, where the norm is outside the sum?