Question

Evaluate :
$\underset{n\to \infty }{lim}\frac{\left[1.x\right]+\left[2.x\right]+\left[3.x\right]+......+[n.x]}{n^2}$, where $[.]$ denotes the greatest integer function.

Answer 1

We have

${lim}_{n\to \infty }\frac{\left[1.x\right]+\left[2.x\right]+\left[3.x\right]......[n.x]}{n^2}$

Given that,

$[.]$ denotes the greatest integer function.

Then, $\left[x\right]$ can be written as $X+\left\{x\right\}\to \left\{x\right\}$ is dfractional part, between $0$ and $1$.

Therefore,

$={lim}_{n\to \infty }\frac{\left[1.x\right]+\left[2.x\right]+\left[3.x\right]......[n.x]}{n^2}={lim}_{n\to \infty }\frac{x+2x+3x+.......+nx+\left\{x\right\}+\left\{2x\right\}+\left\{3x\right\}+......+\left\{nx\right\}}{n^2}$

$={lim}_{n\to \infty }\frac{x(1+2+3+......+n)+\left\{x\right\}+.......+\left\{nx\right\}}{n^2}$

$={lim}_{n\to \infty }\frac{x(1+2+3+......+n)+\left\{x\right\}+.......+\left\{nx\right\}}{n^2}$

$={lim}_{n\to \infty }\frac{x(1+2+3+......+n)}{n^2}+{lim}_{n\to \infty }\frac{\left\{x\right\}+.......+\left\{nx\right\}}{n^2}$

$={lim}_{n\to \infty }\frac{xn(n+1)}{2n^2}+{lim}_{n\to \infty }\frac{\left\{x\right\}+.......+\left\{nx\right\}}{n^2}$

Since $\left\{\right\}$ is only between $0$ and $1$,

On solving the first part, we get,

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

$=\frac{x}{2}\left(\text{taking}\text{limit}\right)$

An easy way to limits when greatest integer function is there, is to take out the common element which does not change in this case so $x$ remains same.

Hence, this is the answer.