Question

If $\left[x\right]$ denotes the integral part of $x$, then

${lim}_{x\to \infty }\frac{1}{n^3}\left({\sum }_{k=1}^n\left[k^{2}x\right]\right)$

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

Answer 1

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

Let $L=\underset{x\to \infty }{lim}\frac{1}{n^3}\left(\sum _{k=1}^n\left[k^{2}x\right]\right)$

Since $k^{2}x-1\le \left[k^{2}x\right]<k^{2}x$

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

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

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

Dividing throughout by $n^3,$ we have

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

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

Taking limits as $n\to \infty ,$ we get

$\underset{n\to \infty }{lim}\{\frac{x}{6}(1+\frac{1}{n})(2+\frac{1}{n})-\frac{1}{n^2}\}\le L<\underset{n\to \infty }{lim}\frac{x}{6}(1+\frac{1}{n})(2+\frac{1}{n})$

Since, as $n\to \infty ,$ we have $\frac{1}{n}\to 0$

$\Rightarrow \frac{x}{3}\le L<\frac{x}{3}$

According to Squeeze principle or Sanhich Theorem,we have

$L=\frac{x}{3}.$