Calculating limit-sequences So i have this limit to calculate:
$\lim_{n\to\infty}\frac{[x] + [2x]+ ... +[nx]}{n^2}$
And i tried to make some boundaries and got this two limits:
$\lim_{n\to\infty}\frac{[x] + [x]+ ... +[x]}{n^2}$
$\lim_{n\to\infty}\frac{[nx] + [nx]+ ... +[nx]}{n^2}$
But i am not sure if it's correct,while the first limit is 0,and the second one is not.
And also those [  ] are floor functions.
Any help would be appreciated.
 A: Using $$x-1 \le [x] \le x,$$ you can safely remove the floor function signs without changing the limit. From there you should be good.
A: $$x-1\leqslant \left \lfloor x \right \rfloor< x\\2x-1\leqslant \left \lfloor 2x \right \rfloor< 2x\\3x-1\leqslant \left \lfloor 3x \right \rfloor< 3x\\...\\\\nx-1\leqslant \left \lfloor nx \right \rfloor< nx$$ 
now sum of them 
$$(x+2x+3x+..+nx)-(1+1+1..1)\leq \left \lfloor x \right \rfloor+\left \lfloor 2x \right \rfloor+...+\left \lfloor nx \right \rfloor< (x+2x+3x+..+nx)\\ \frac{n(n+1)}{2}x -n\leq  \left \lfloor x \right \rfloor+\left \lfloor 2x \right \rfloor+...+\left \lfloor nx \right \rfloor< \frac{n(n+1)}{2}x $$ 
now divide by $n^2$ 
$$\frac{n(n+1)}{2n^2}x -\frac{n}{n^2}\leq  \frac{(\left \lfloor x \right \rfloor+\left \lfloor 2x \right \rfloor+...+\left \lfloor nx \right \rfloor)}{n^2}< \frac{n(n+1)}{2n^2}x$$ 
now apply limit ,by squeeze theorem
$$\lim_{n \to \infty} \frac{n(n+1)}{2n^2}x -\frac{n}{n^2}\leq  \lim_{n \to \infty}\frac{(\left \lfloor x \right \rfloor+\left \lfloor 2x \right \rfloor+...+\left \lfloor nx \right \rfloor)}{n^2}< \lim_{n \to \infty}\frac{n(n+1)}{2n^2}x\\ \frac{1}{2}x  \leq  \lim_{n \to \infty}\frac{(\left \lfloor x \right \rfloor+\left \lfloor 2x \right \rfloor+...+\left \lfloor nx \right \rfloor)}{n^2} <\frac{1}{2} x$$ so the limit =$\frac{1}{2}x$
