An orthonormal set in a separable Hilbert space is complete (is a basis) if its distance to another orthonormal basis is bounded

Question

Let $H$ be a separable Hilbert space, $\{a_n\}$ be an orthonormal set in $H$ and $\{b_n\}$ be an orthonormal basis in $H$. Show that $\{a_n\}$ is also a basis if $\sum_{n} ||a_n - b_n ||^2 < 1$.

I guess I should show that $||x - \sum_{n} \langle a_n, x \rangle a_n || \rightarrow 0$ for all $x \in H$ but stopped at \begin{equation} ||x - \sum_{n} \langle a_n, x \rangle a_n || = \lim_{M,N \rightarrow \infty} || \sum_{n=N+1}^{M} (\langle b_n,x \rangle b_n - \langle a_n,x \rangle a_n) || \end{equation} And I don't know where to use the condition $H$ is separable. Any hint?

Thank you in advance.

Answer 1

Hint: Write $\|\langle b_n,x\rangle b_n-\langle a_n,x\rangle a_n\|=\|\langle b_n,x\rangle b_n-\langle b_n,x\rangle a_n+\langle b_n,x\rangle a_n-\langle a_n,x\rangle a_n\|$

$\leq \|\langle b_n,x\rangle(b_n-a_n)\|+\|\langle b_n-a_n,x\rangle a_n\|\leq (|\langle b_n,x\rangle|+\|a_n\|\|x\|)\|b_n-a_n\|$.