I have been unable to find anywhere a sharp bounding for the partial sum of the first $n$ square-free integers $\sum_{k=1}^{n} k$, where $k$ runs over the square-free integers. I guess it should be around $\frac{\pi^2}{6} \left(\frac{n^2+n}{2}\right)$, but I found no reference to check and I do not have a rigorous way to prove my guess, just the heuristic following from the partial sum of the first positive integers and the distribution of square-free numbers.
Any reference or proof of a sharp bound would be welcomed.
Thanks!
EDIT
After some trial an error experimentation, it seems that $$\sum_{k=1}^{n} k = \frac{\pi^2}{6}\left(\frac{n^2+n}{2}\right)-\frac{6}{\pi^2}n\frac{\log(n)}{\log\log(n)}+O\left(\frac{n}{\sqrt{\log\log(n)}}\right)$$ Here the graphical representation of the diferences between $\sum_{k=1}^{n} k$ and $\frac{\pi^2}{6}\left(\frac{n^2+n}{2}\right)-\frac{6}{\pi^2}n\frac{\log(n)}{\log\log(n)}$:
There is still a clear pattern in the differences function, so I believe that the theoretical formula can be improved even further.
Is there any theoretical explanation supporting this formula? It seems much better than the one obtained using abelian partial summation.
By Abelian partial summation,
$$ \sum_{n=1}^x n 1_{n \text{ square-free}} = \frac{6}{π^2} x^2 + O(x^{3/2}) - \int_1^x 1 \frac{6t}{π^2} dt + O(x^{3/2}) $$
Thus, your asymptotic is correct, and the error term has the order $x^{3/2}$.
EDIT: If you want to sum the first $m$ (changing letters here to avoid notational overload) square-free integers, then you may choose $x = π^2/6 m + O(\sqrt{m})$, so that there are $m$ square-free integers up to $x$, and insert this into the above asymptotic.