Question

The value of $\underset{n\to \infty }{lim}\frac{1}{n^3}\left(\right[1^{2}x]+[2^{2}x]+\cdots +[n^{2}x\left]\right),x\in R$ is equal to $($Here $[.]$ denotes the greatest integer function$)$

• A. $\frac{x}{6}$
• B. $\frac{x}{3}$
• C. $\frac{x}{2}$
• D. $0$

Answer 1

The correct option is B $\frac{x}{3}$

We know $k-1<\left[k\right]\le k,\forall k\in R$
$\Rightarrow k^{2}x-1<\left[k^{2}x\right]\le k^{2}x$

$\therefore \sum _{k=1}^n(k^{2}x-1)<\sum _{k=1}^n\left[k^{2}x\right]\le \sum _{k=1}^n{k}^{2}x$

$\Rightarrow \frac{n(n+1)(2n+1)}{6}\cdot x-n<\sum _{k=1}^n\left[k^{2}x\right]\le \frac{n(n+1)(2n+1)}{6}\cdot x$

$\Rightarrow \frac{n(n+1)(2n+1)x-6n}{6n^3}<\frac{1}{n^3}\sum _{k=1}^n\left[k^{2}x\right]\le \frac{n(n+1)(2n+1)}{6n^3}\cdot x$

Since $\underset{n\to \infty }{lim}\frac{n(n+1)(2n+1)x-6n}{6n^3}=\underset{n\to \infty }{lim}\frac{n(n+1)(2n+1)\cdot x}{6n^3}=\frac{x}{3}$
By Sandwich theorem of limits,
$\underset{n\to \infty }{lim}\frac{1}{n^3}\sum _{k=1}^n\left[k^{2}x\right]=\frac{x}{3}$