5
$\begingroup$

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?

  • 0
    right of course. tha$n$ks guys!2012-10-24

1 Answers 1

3

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.