Question

The value of $\underset{n\to \infty }{lim}\frac{\left[r\right]+\left[2r\right]+\cdots +\left[nr\right]}{n^2},$ where $r$ is a non zero real number and $\left[r\right]$ denotes the greatest integer less than or equal to $r$, is equal to

• A. $0$
• B. $r$
• C. $\frac{r}{2}$
• D. $2r$

Answer 1

The correct option is C $\frac{r}{2}$

Let $L=\underset{n\to \infty }{lim}\frac{\left[r\right]+\left[2r\right]+\cdots +\left[nr\right]}{n^2}$
We know that
$r-1<\left[r\right]\le r2r-1<\left[2r\right]\le 2r3r-1<\left[3r\right]\le 3r\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot nr-1<\left[nr\right]\le nr\Rightarrow \frac{n(n+1)r-n}{2}<\left[r\right]+\left[2r\right]+\cdots +\left[nr\right]\le \frac{n(n+1)r}{2}\Rightarrow \underset{n\to \infty }{lim}\frac{n(n+1)r-n}{2n^2}<L\le \underset{n\to \infty }{lim}\frac{n(n+1)r}{2n^2}\Rightarrow \frac{r}{2}<L\le \frac{r}{2}\therefore L=\frac{r}{2}\left(\text{By sandwich theorem}\right)$