convergence in operator norm Let $H$ be Hilbert space, $\{a_n\}_{n=1},$ be ONS and $K$ be a compact operator. Suppose $K_nx:=(Kx,a_n)a_n$, $\sum _{n=1}^{N} K_n$ convergent as $N \to \infty$.
So,$||K_nx||=|(Kx,a_n)|=|(x,K^* a_n)| \leq ||x|| ||K^* a_n||$ and so $\lim ||K_n||\leq \lim ||K^* a_n||=0$ since $K^*$ is compact and $\{a_n\}$ weakly convergent.

Let $x$ be an element of the closed unit ball. Then by orthogonality of $\left(b_n\right)$, 
 $$ \left\lVert \sum_{n=M}^N K_nx\right\rVert^2=\sum_{n=M}^N \left(Kx,a_n\right)^2. $$
Since the set $\left\{Kx,x\in H,\left\lVert x\right\rVert\leqslant 1\right\}$ has a compact closure, it suffices to prove that for each subset $C$ of $H$ having a compact closure,
$$
\lim_{M,N\to +\infty}\sup_{y\in C}\sum_{n=M}^N \left(y,a_n\right)^2=0.
$$
This is clear when $C$ is finite; use precompactness to generalize.
