I searched a bit and examined this post:
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?
Hint
Since $$u-1<\lfloor u\rfloor\le u$$we can write$${\sum_{i=1}^n(i\sqrt{2007}-1)\over \sum_{i=1}^ni\sqrt{2008}}\le{\sum_{i=1}^n\lfloor i\sqrt{2007}\rfloor\over \sum_{i=1}^n\lfloor i\sqrt{2008}\rfloor} \le {\sum_{i=1}^ni\sqrt{2007}\over \sum_{i=1}^n(i\sqrt{2008}-1)}$$