Question

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