0
$\begingroup$

I searched a bit and examined this post:

Floor function properties: $[2x] = [x] + [ x + \frac12 ]$ and $[nx] = \sum_{k = 0}^{n - 1} [ x + \frac{k}{n} ] $ .

I thought I could use that in the following problem:

\begin{equation} \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} \end{equation} (source: undergraduate student contest, Faculty of Science, Mathematics Department, Zagreb, 2007.)

My attempt:

if\begin{equation}\lfloor nx\rfloor=\sum_{k=0}^{n-1}\left\lfloor x+\frac{k}{n}\right\rfloor\end{equation} then:\begin{equation}\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}\end{equation} and: \begin{equation}\lfloor\sqrt{2007}\rfloor=\lfloor\sqrt{2008}\rfloor=\lfloor\sqrt{2009}\rfloor=\alpha\end{equation} because: $44^2<2007<45^2\\44^2<2008<45^2\\44^2<2009<45^2$,

thus:\begin{equation}\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\end{equation}

Is this correct?

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ Jan 29, 2020 at 21:34
  • 1
    $\begingroup$ @InterstellarProbe, may I ask for your advice on how to solve this problem? $\endgroup$
    – PinkyWay
    Jan 29, 2020 at 21:46
  • 1
    $\begingroup$ 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 $$\dfrac{44}{45+1} = \dfrac{44\dbinom{n+1}{2}}{45\dbinom{n+1}{2}+\dbinom{n+1}{2}} \le \dfrac{T_n}{T_n+D_n} = S_n \le 1$$ $\endgroup$ Jan 29, 2020 at 22:02
  • $\begingroup$ @InterstellarProbe, I appreciate your responce! Thank you! $\endgroup$
    – PinkyWay
    Jan 29, 2020 at 22:03

1 Answer 1

2
$\begingroup$

Hint

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

$\endgroup$
2
  • 1
    $\begingroup$ Very nice solution! $\endgroup$ Jan 29, 2020 at 22:18
  • 1
    $\begingroup$ Thank you...... $\endgroup$ Jan 29, 2020 at 22:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .