Question

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

• A. $\frac{x}{2}$
• B. $x$
• C. $2x$
• D. None of these

Answer 1

Answer: (A)

We know that, $x-1<\left[x\right]\le x$
$\Rightarrow x+2x+...+nx-n<$ $\sum _{r=1}^n[rx]\le x+2x+...+nx$
$\Rightarrow \frac{x\cdot n\left(n+1\right)}{2}-n<$ $\sum _{r=1}^n$$\left[rx\right]\le \frac{x\cdot n\left(n+1\right)}{2}$
$\Rightarrow \frac{x}{2}\left(1+\frac{1}{n}\right)-\frac{1}{n}<\frac{1}{n^2}$ $\sum _{r=1}^n$$\left[rx\right]\le \frac{x}{2}\left(1+\frac{1}{n}\right)$
Now, $\underset{n\to \infty }{\lim }$$\frac{x}{2}\left(1+\frac{1}{n}\right)=\frac{x}{2}$
and $\underset{n\to \infty }{\lim }$$\frac{x}{2}\left(1+\frac{1}{n}\right)-\frac{1}{n}=\frac{x}{2}$
Using Sandwich theorem, we find that
$\underset{n\to \infty }{\lim }$$\frac{\left[x\right]+\left[2x\right]+...+\left[nx\right]}{n^2}=\frac{x}{2}$