from basic number theory we know
$$ \sum_{n\le x}\sigma _{0}(n)= \sum_{n=1}^{\infty}[x/n] $$
where $ [x] $ is the floor function
then for the divison function of any order
can we evaluate exactly the sums $ \sum_{n\le x}\sigma _{k}(n) $
for an positive value of 'x' integer ? in terms of the floor function or another arithmetical functions
The main idea is switching the order of summation with a substitution $n=dm$: $$ \begin{align} \sum_{n\leq x}\sigma_k(n)&=\sum_{n\leq x}\sum_{d|n}d^k =\sum_{d\leq x}\sum_{m\leq \frac xd} d^k\\ &=\sum_{d\leq x} d^k \sum_{m\leq \frac xd} 1 =\sum_{d\leq x} d^k \left\lfloor \frac xd \right\rfloor\\ &=\sum_{d=1}^{\infty} d^k \left\lfloor \frac xd\right \rfloor. \end{align} $$
