Question

If $[x]$ denotes the integral part of $x$, then $\underset{n\to \infty }{\lim }\frac{1}{n^3}\left(\sum _{k=1}^n\left[k^{2}x\right]\right)=$

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

Answer 1

Answer: (C)

Let $L=\underset{n\to \infty }{\lim }\frac{1}{n^3}\left(\sum _{k=1}^n\left[k^{2}x\right]\right)$
Since $k^{2}x-1\le \left[k^{2}x\right]<k^{2}x$
$\Rightarrow \sum _{k=1}^n\left(k^{2}x-1\right)\le \sum _{k=1}^n\left[k^{2}x\right]<\sum _{k=1}^n{k}^{2}xa$
$\Rightarrow x\left(\sum _{k=1}^n{k}^2\right)-{\sum }_{k=1}^n(1)\le {\sum }_{k=1}^n\left[k^{2}x\right]<x\left(\sum _{k=1}^n{k}^2\right)$
$\Rightarrow \frac{xn(n+1)(2n+1)}{6}-n\le \sum _{k=1}^n\left[k^{2}x\right]<\frac{xn(n+1)(2n+1)}{6}$
Dividing throughout by $n^3$, we have
$\frac{xn(n+1)(2n+1)}{6n^3}-\frac{1}{n^2}\le \sum _{k=1}^n\frac{\left[k^{2}x\right]}{n^3}<\frac{xn(n+1)(2n+1)}{6n^3}$
$\Rightarrow \frac{x}{6}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)-\frac{1}{n^2}\le \sum _{k=1}^n\frac{\left[k^{2}x\right]}{n^3}<\frac{x}{6}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)$
Taking limits as $n\to \infty$, we get
$\underset{n\to \infty }{\lim }\left\{\frac{x}{6}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)-\frac{1}{n^2}\right\}\le L<\underset{n\to \infty }{\lim }\frac{x}{6}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)$
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 Sandwich Theorem, we have
$L=\frac{x}{3}$