Approximation of $\sum_{x \le k} \frac{\log(x)}{x}$ Originally posted as a non-homework question. New to the site, and didn't know asking for homework advice was O.K. Anyways, here's what's going on:
I'm trying to show there exists a constant $B$ such that
$$
\sum_{x \le k} \frac{\log(x)}{x} = \frac{1}{2}\log^2(k) + B + O\left(\frac{\log(k)}{k}\right)
$$
I'm trying via partial summation to establish this. I think some of my trouble lies in understanding the question. If we're using the $O$ notation to bound an error term, and if we just need to show there exists a constant $B$ such that the above holds, why isn't $B$ absorbed into the error term? 

The Euler-Maclaurin Sum Formula gives this immediately because
$$
\int\frac{\log(x)}{x}\,\mathrm{d}x=\frac12\log(x)^2+C
$$
The constant $B$ dominates the error term $O\left(\frac{\log(x)}{x}\right)$, so it is separate.
