Question

If [.] denotes the greatest integer function then $\underset{n\to \infty }{lim}\frac{\left[x\right]+\left[2x\right]+...+\left[nx\right]}{n^2}$ is

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

Answer 1

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

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

As $kx-1<\left[kx\right]<kx+1$

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

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

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

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

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

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