Question

The value of $\underset{n\to \infty }{lim}\frac{1}{n^2}\sum _{r=1}^n\left[rx\right]$, is
(where [.] denotes the greatest integer function)

• A. $\frac{1}{2}$
• B. $x$
• C. $\frac{x}{4}$
• D. $\frac{x}{2}$

Answer 1

The correct option is D $\frac{x}{2}$

$L=\underset{n\to \infty }{lim}\frac{\left[x\right]+\left[2x\right]+...+\left[nx\right]}{n^2}nx-1<\left[nx\right]\le nx$
Putting $n=1,2,3,..,n$ and adding them
$x\sum n-n<\sum \left[nx\right]\le x\sum n\Rightarrow \frac{x\sum n}{n^2}-\frac{1}{n}<\frac{\sum \left[nx\right]}{n^2}\le \frac{x\sum n}{n^2}$

$\underset{n\to \infty }{lim}(\frac{x\sum n}{n^2}-\frac{1}{n})=\underset{n\to \infty }{lim}(\frac{nx(n+1)}{2n^2}-\frac{1}{n})=x\times \frac{1}{2}=\frac{x}{2}$

$\underset{n\to \infty }{lim}(\frac{x\sum n}{n^2})=x\underset{n\to \infty }{lim}\frac{n(n+1)}{2n^2}=\frac{x}{2}$

Therefore, using Sandwich theorem, we have,
$\underset{n\to \infty }{lim}\frac{1}{n^2}\sum _{r=1}^n\left[rx\right]=\frac{x}{2}$