# 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?

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Because $\log(k)/k \to 0$ as $k \to \infty$, a constant is not $O(\log(k)/k)$. – Robert Israel Oct 24 '12 at 1:42
$B+O(\frac{\log k}{k})$ is not the same as $O(\frac{\log k}{k})$ because $(\log k)/k\to0$. – anon Oct 24 '12 at 1:42
right of course. thanks guys! – Thomas Nesbitt Oct 24 '12 at 1:48

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.