Problem #WSP-000131

Problem

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 $n\ge 1$, $$\sum _{k=1}^{n}s(k)=s(1)+s(2)+...+s(n)\le \frac{{\pi }^2}{12}{n}^2+\frac{n\log n}{2}+\frac{n}{2}.$$