# Limit involving floor function summation

I searched a bit and examined this post:

I thought I could use that in the following problem:

$$$$\lim _{n \to \infty} \frac{\lfloor\sqrt{2007}\rfloor+\lfloor 2 \sqrt{2007}\rfloor+\cdots+\lfloor n \sqrt{2007}\rfloor}{\lfloor\sqrt{2008}\rfloor+\lfloor 2 \sqrt{2008}\rfloor+\cdots+\lfloor n \sqrt{2008}\rfloor}$$$$ (source: undergraduate student contest, Faculty of Science, Mathematics Department, Zagreb, 2007.)

My attempt:

if$$$$\lfloor nx\rfloor=\sum_{k=0}^{n-1}\left\lfloor x+\frac{k}{n}\right\rfloor$$$$ then:$$$$\lim_{n\to\infty}\frac{\displaystyle\sum_{i=0}^{n-1}(n-i)\left\lfloor\sqrt{2007+\frac{i}{n}}\right\rfloor}{\displaystyle\sum_{i=0}^{n-1}(n-i)\left\lfloor\sqrt{2008+\frac{i}{n}}\right\rfloor}$$$$ and: $$$$\lfloor\sqrt{2007}\rfloor=\lfloor\sqrt{2008}\rfloor=\lfloor\sqrt{2009}\rfloor=\alpha$$$$ because: $$44^2<2007<45^2\\44^2<2008<45^2\\44^2<2009<45^2$$,

thus:$$$$\lim_{n\to\infty}\frac{\displaystyle\sum_{i=0}^{n-1}(n-i)\left\lfloor\sqrt{2007+\frac{i}{n}}\right\rfloor}{\displaystyle\sum_{i=0}^{n-1}(n-i)\left\lfloor\sqrt{2008+\frac{i}{n}}\right\rfloor}=\frac{\frac{n(n+1)}{2}\alpha}{\frac{n(n+1)}{2}\alpha}=1$$$$

Is this correct?

• Let's consider partial sums to see if what you propose makes sense. $$S_n = \dfrac{\displaystyle \sum_{k=1}^n \left\lfloor k\sqrt{2007} \right\rfloor }{\displaystyle \sum_{k = 1}^n \left\lfloor k\sqrt{2008} \right\rfloor}$$ and $$S_{10} = \dfrac{2458}{2460}$$ But, if I used the sums you proposed, I would get $$S_{10} = \dfrac{1540}{1540}$$ These two are not even remotely similar, so your formula is off. – InterstellarProbe Jan 29 at 21:34
• @InterstellarProbe, may I ask for your advice on how to solve this problem? – Invisible Jan 29 at 21:46
• I'm not sure. The best I can come up off the top of my head would show that the limit is between $$\dfrac{44}{45+1} \le \lim_{n\to \infty} S_n \le 1$$ You can do that by taking $$44\dbinom{n+1}{2} \le T_n = \sum_{k=1}^n \lfloor k\sqrt{2007} \rfloor \le 45\dbinom{n+1}{2}$$ and $$0\le D_n = \sum_{k=1}^n(\lfloor k\sqrt{2008} \rfloor - \lfloor k\sqrt{2007} \rfloor) \le \dbinom{n+1}{2}$$ Then you have $$\dfrac{44}{45+1} = \dfrac{44\dbinom{n+1}{2}}{45\dbinom{n+1}{2}+\dbinom{n+1}{2}} \le \dfrac{T_n}{T_n+D_n} = S_n \le 1$$ – InterstellarProbe Jan 29 at 22:02
• @InterstellarProbe, I appreciate your responce! Thank you! – Invisible Jan 29 at 22:03

Hint

Since $$u-1<\lfloor u\rfloor\le u$$we can write$${\sum_{i=1}^n(i\sqrt{2007}-1)\over \sum_{i=1}^ni\sqrt{2008}}\le{\sum_{i=1}^n\lfloor i\sqrt{2007}\rfloor\over \sum_{i=1}^n\lfloor i\sqrt{2008}\rfloor} \le {\sum_{i=1}^ni\sqrt{2007}\over \sum_{i=1}^n(i\sqrt{2008}-1)}$$

• Very nice solution! – InterstellarProbe Jan 29 at 22:18
• Thank you...... – Mostafa Ayaz Jan 29 at 22:22