Can someone help me the summation of the given series.
$$\sum_{i=1}^n\left\lfloor\frac{n}i\right\rfloor$$
Negative of the above summation looks similar to the expansion of the $\log(1-x)$ without the floor.
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According to a recent answer, it is $n\log n + (2\gamma-1)n + O(n^{1/2})$. – Thomas Andrews Sep 06 '13 at 19:35
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As the floor is being used can't we get the numerical value.. – user1580096 Sep 06 '13 at 19:37
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OEIS A006218 gives various formulae and references – Henry Sep 06 '13 at 19:38
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Yes, but we can give an approximation, which is what I gave. @user1580096 – Thomas Andrews Sep 06 '13 at 19:39
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@user1580096 and so? for example you can easily prove that $\sum_{r=0}^{n-1}\bigl\lfloor\frac{a+r}n\bigr\rfloor=a$ for any $a,n\in\mathbb Z$ with $n\ne0$. – Matemáticos Chibchas Sep 06 '13 at 20:07
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Strictly speaking, you are right when you call your sum a "series", because you can change the upper limit of summation $n$ by $\infty$ (because $\lfloor n/i\rfloor=0$ for $i>n$)...but I think that it is better you call it simply "a sum". – Matemáticos Chibchas Sep 06 '13 at 20:10
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This is OEIS A$006218$. No closed form is known; the problem of finding a precise asymptotic estimate is the Dirichlet divisor problem. You’ll find references, asymptotic estimates, and other information at the two links and here.
A useful tip: I simply calculated the first half-dozen terms and ran them through The On-Line Encyclopedia of Integer Sequences (OEIS); that’s worth a try any time you want information about an integer sequence.
Brian M. Scott
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