Let n be a positive integer. We denote by s(n) the sum of the divisors of n. For example, the divisors of n=6 are 1, 2, 3 and 6, so s(6)=1+2+3+6=12. Prove that, for all n≥1, ∑k=1ns(k)=s(1)+s(2)+...+s(n)≤π212n2+nlogn2+n2.