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How can I approximate the sum$$\sum_{k=1}^n \left(\frac{2k}{n} \left\lceil \frac{n}{k} \right\rceil \left\{ \frac{n}{k} \right\}-1\right)$$ where $\{x\}$ is the fractional part function, and $\lceil x\rceil$ is the ceiling function.

I know that if I divide by $n$ and let $n\to\infty$, it's equal to $0$. At first I thought the sum might be of the order $n^a$, but now I think it could be logarithmic. The partial sums are really weird. I would appreciate any help on giving an approximate value to the sum.

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    $\begingroup$ I was checking numerically the values for $n\leq 1000$. Looking at the number they seem to be rather spread (both positive and negative and of different magnitude). Given that I am not sure the an approximation in terms of a smooth function does in fact exist. $\endgroup$ – Fabian Dec 26 '12 at 8:23
  • $\begingroup$ Damnet, it would be really helpful if I had one lol $\endgroup$ – Ethan Dec 26 '12 at 8:24
  • $\begingroup$ @Ethan So, did you ever figure out a better approximation? $\endgroup$ – Alexander Gruber Apr 26 '13 at 1:37
  • $\begingroup$ @AlexanderGruber No I sort of gave up on this a while ago, it doesn't really matter to me anymore, but just looking at it now, I would say application of the dirichlet hyperbola method, would probably give me an ok estimate. $\endgroup$ – Ethan Apr 27 '13 at 2:18
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An observation: let $\rm f(n)=\sum_{k=1}^n \left(\frac{2k}{n} \left\lceil \frac{n}{k} \right\rceil \left\{ \frac{n}{k} \right\}-1\right)$ and $\rm \mu(n)$ be the mean of $\rm f(1),\ldots,f(n)$. Then $\rm \mu(n)$ is logarithmic:

log.

as is the standard deviation:

enter image description here

so you could probably come up with a numerical estimate for a bound as a function of $\rm n$ pretty easily, though it appears that $\rm f$ is unbounded as $\rm n\rightarrow\infty$.

EDIT: The upper and lower bounds actually seem to behave like those of the Divisor Summatory function. Maybe you should look at that.

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