 Floor-sum rearrangement

For an arithmetic function $a$, changing the order of the finite summation gives ${\sum }_{d=1}^{n}a(d)\lfloor n/d\rfloor ={\sum }_{t=1}^n{\sum }_{d\le \lfloor n/t\rfloor }a(d)$. Both sides count pairs of positive integers $(d,t)$ with $dt\le n$, assigning weight $a(d)$ to each pair. This expresses a weighted floor sum in terms of a summatory function.