Let be an infinite-dimensional Hilbert space. Claim: an operator is compact if and only if for every orthonormal sequence .
Proof: Suppose is compact. Given an orthonormal sequence, extend it to an orthonormal basis . Let be the projection onto the span of . We know that for any compact operator. Since is compact as well, we have , hence . Finally, note that , and therefore as .
Conversely, suppose is not compact. Pick an orthonormal basis and let be as above. The non-compactness of implies that there exists such that for all (because has finite rank). Let .
Pick a unit vector such that . Since is a basis, for any vector we have as . Thus, there exists such that . Let be the vector normalized and note that .
Continuing in this way, we form a sequence of unit vectors such that and lies in the range of projection (this is a projection because all commute). These ranges are mutually orthogonal, because for any we have , which implies and therefore for all . Thus, is the desired sequence.