# Finding the summation of the floor of the series.

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 6 '13 at 19:35
As the floor is being used can't we get the numerical value.. – user1580096 Sep 6 '13 at 19:37
OEIS A006218 gives various formulae and references – Henry Sep 6 '13 at 19:38
Yes, but we can give an approximation, which is what I gave. @user1580096 – Thomas Andrews Sep 6 '13 at 19:39
@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 6 '13 at 20:07

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.