Question

If $\left[x\right]$ denotes the greatest integer less than or equal to $x$, then $\underset{x\to \infty }{lim}\frac{\left[x\right]+\left[2x\right]+\left[3x\right]+\cdots +\left[nx\right]}{n^2}=$

• A. $\frac{x}{2}$
• B. $\frac{x}{3}$
• C. $x$
• D. $0$

Answer 1

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

$\underset{n\to \infty }{lim}\frac{\left[x\right]+\left[2x\right]+...+\left[nx\right]}{n^2}=\underset{n\to \infty }{lim}\frac{1}{n^2}\sum _{k=1}^n\left[kx\right]$
As $kx-1<\left[kx\right]$

$\Rightarrow \sum _{k=1}^n(kx-1)<\sum _{k=1}^n\left[kx\right]<\sum _{k=1}^n(kx+1)$

$\Rightarrow \frac{xn(n+1)}{2}-n<\sum _{k=1}^n\left[kx\right]<\frac{xn(n+1)}{2}+n\Rightarrow \frac{x}{2}(1+\frac{1}{n})-\frac{1}{n}<\frac{1}{n^2}\sum _{k=1}^n\left[kx\right]<\frac{x}{2}(1+\frac{1}{n})+\frac{1}{n}$

$\Rightarrow \underset{n\to \infty }{lim}(\frac{x}{2}(1+\frac{1}{n})-\frac{1}{n})<\underset{n\to \infty }{lim}\frac{1}{n^2}\sum _{k=1}^n\left[kx\right]<\underset{n\to \infty }{lim}(\frac{x}{2}(1+\frac{1}{n})+\frac{1}{n})$

$\Rightarrow \frac{x}{2}<\underset{n\to \infty }{lim}\frac{1}{n^2}\sum _{k=1}^n\left[kx\right]<\frac{x}{2}$
$\Rightarrow \underset{n\to \infty }{lim}\frac{1}{n^2}\sum _{k=1}^n\left[kx\right]=\frac{x}{2}$
Hence, option 'A' is correct.