Prove that $0< \sum_{k=1}^{n}g(k)/k-2n/3 < 2/3$

How to prove that for all positive integers $n$ $$0< \sum_{k=1}^{n}\frac{g(k)}{k}-\frac{2n}{3}<\frac{2}{3}$$ where $g(k)$ denotes the greatest odd divisor of $k$

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Just a note (not a big deal, since it's easily corrected and I'm not sure whether it's written down in the rules somewhere): I think it's considered bad form to have "display style" ($$...$$ or $...$) math in titles, since it takes up a lot of room on the front page. –  Dylan Moreland Jun 27 '12 at 17:34
Have you made any attempt at the problem yourself? –  Ben Millwood Jun 27 '12 at 17:39

Hint: $k/g(k)$ is the greatest power of $2$ dividing $k$.
Add up the contributions of $g(k)/k=1, 1/2, 1/4, \ldots$, which are at most $(n+1)/2, (n+2)/8, (n+4)/16, \ldots$. What do you get? –  Robert Israel Jun 27 '12 at 20:37