How can I approximate the sum$\sum_{k=1}^n \left(\frac{2k}{n} \left\lceil \frac{n}{k} \right\rceil \left\{ \frac{n}{k} \right\}-1\right)$ where $\{x\}$ is the fractional part function, and $\lceil x\rceil$ is the ceiling function.
I know that if I divide by $n$ and let $n\to\infty$, it's equal to $0$. At first I thought the sum might be of the order $n^a$, but now I think it could be logarithmic. The partial sums are really weird. I would appreciate any help on giving an approximate value to the sum.