JEE Main 2021 (Online) 17th March Evening ShiftQuestion 1
The value of $\lim_{n\to\infty}\frac{[r]+[2r]+\ldots+[nr]}{n^{2}}$, where $r$ is a non-zero real number and $[r]$ denotes the greatest integer less than or equal to $r$, is equal to :
Limits & Continuity
Solution
$kr-1<[kr]\le kr$, so $\sum[kr]=r\frac{n(n+1)}2-O(n)$. Dividing by $n^2$: limit $=\frac r2$. Option B.