How to calculate the sum of floors

Question

The problem is to calculate the value

$$ \sum_{i = 1}^{1000} \left\lfloor \frac{1}{\sqrt{i} - \lfloor \sqrt{i} \rfloor} \right\rfloor $$

Where $ i $ is not perfect square number.

My thought is, if I can calculate the sum

$$ \sum_{m = 1}^{M} \sum_{k = 1}^{2m} \left\lfloor \frac{\sqrt{m^2+k} + m}{k} \right\rfloor $$

then I can solve the original problem.

And I also know that $ \left\lfloor \frac{x + m}{n} \right\rfloor = \left\lfloor \frac{\lfloor x \rfloor + m}{n} \right\rfloor $

So I just need to calculate

$$ \sum_{m = 1}^{M} \sum_{k = 1}^{2m} \left\lfloor \frac{2m}{k} \right\rfloor $$

Then, it's difficult for me.

I have tried to exchange the sum-order of $ m $ and $ k $, that is

$$ \sum_{k = 1}^{2M} \sum_{m = 1}^{M} \left\lfloor \frac{2m}{k} \right\rfloor $$

While I could not calculate the sum of $ \frac{2m}{k} $, in fact it is easier if it is $ \frac{m}{k} $

So I ask some advices for you : )

Your sum should start from $i=2$; the summand is infinite at $i=1$. — Ron Gordon, Mar 16 '13 at 13:22
@Ron Gordon Yes you're right! i is not perfect square number. — easymath3, Mar 16 '13 at 15:01
Call that final sum $S_M$. Then $M(M+1) < S_M \le H_{2M} M(M+1)$, where $H_n$ is the $n^\text{th}$ harmonic number. — Snowball, Mar 16 '13 at 21:10
In fact, I think that lower bound can be changed to $M(M+1)\log(M)$, but I haven't proven it. — Snowball, Mar 16 '13 at 21:26
see the same problem:http://math.stackexchange.com/questions/326283/finding-the-sum-sum-limits-n-1-n-neq-m21000-left-frac1-sqrtn — lsr314, Mar 19 '13 at 09:34

0 Answers0