decomposing and the summing a sum. 
Prove:
$$\sum_{k\leq n} k \left\{ \frac {n}{k} \right\} = n^2\left(1 - \frac {\pi^2}{12}\right) + O (n \log n) \quad \text{ where } \{x\} = x-\lfloor x \rfloor $$

We have that $$\sum_{k\leq n} k ( \frac {n}{k} - \lfloor \frac {n}{k} \rfloor ) = \sum_{k\leq n}  n -\sum_{k\leq n} k \lfloor \frac {n}{k} \rfloor $$
EDIT:
$$= n^2 - \sum_{k \leq n} \sigma (k)= n^2 - (\frac{\pi^2}{12}n^2 + O(n \log n)) $$
Hence $$\sum_{k\leq n} k \left\{ \frac {n}{k} \right\} = n^2\left(1 - \frac {\pi^2}{12}\right) + O (n \log n) $$

We have that $$\sum_{k\leq n} k ( \frac {n}{k} - \lfloor \frac {n}{k} \rfloor ) = \sum_{k\leq n}  n -\sum_{k\leq n} k \lfloor \frac {n}{k} \rfloor $$
$$= n^2 - \sum_{k \leq n} \sigma (k) $$ 
Since $ \sum_{k \leq n} \sigma (k)= (\frac{\pi^2}{12}n^2 + O(n \log n) )$ 
Hence we have that $$= n^2 - \sum_{k \leq n} \sigma (k)= n^2 - (\frac{\pi^2}{12}n^2 + O(n \log n)) $$ 
Hence $$\sum_{k\leq n} k \left\{ \frac {n}{k} \right\} = n^2\left(1 - \frac {\pi^2}{12}\right) + O (n \log n) $$
As desired.
