Question

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

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

Answer 1

Answer: (C)

Let $L= \displaystyle\lim _{n \rightarrow \infty} \frac{1}{n^{3}}\left(\displaystyle\sum_{k=1}^{n}\left[k^{2} x\right]\right)$
Since $k^{2} x-1 \leq\left[k^{2} x\right]< k^{2} x$
$\Rightarrow \displaystyle\sum_{k=1}^{n}\left(k^{2} x-1\right) \leq \displaystyle\sum_{k=1}^{n}\left[k^{2} x\right]< \displaystyle\sum_{k=1}^{n} k^{2} x a$
$\Rightarrow x\left(\displaystyle\sum_{k=1}^{n} k^{2}\right)-\sum_{k=1}^{n}(1) \leq \sum_{k=1}^{n}\left[k^{2} x\right]< x\left(\displaystyle\sum_{k=1}^{n} k^{2}\right)$
$\Rightarrow \frac{x n(n+1)(2 n+1)}{6}-n \leq \displaystyle\sum_{k=1}^{n}\left[k^{2} x\right]<\frac{x n(n+1)(2 n+1)}{6}$
Dividing throughout by $n^{3}$, we have
$\frac{x n(n+1)(2 n+1)}{6 n^{3}}-\frac{1}{n^{2}} \leq \displaystyle\sum_{k=1}^{n} \frac{\left[k^{2} x\right]}{n^{3}}< \frac{x n(n+1)(2 n+1)}{6 n^{3}} $
$\Rightarrow \frac{x}{6}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)-\frac{1}{n^{2}} \leq \displaystyle\sum_{k=1}^{n} \frac{\left[k^{2} x\right]}{n^{3}}< \frac{x}{6}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)$
Taking limits as $n \rightarrow \infty$, we get
$\displaystyle\lim _{n \rightarrow \infty}\left\{\frac{x}{6}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)-\frac{1}{n^{2}}\right\} \leq L< \displaystyle\lim _{n \rightarrow \infty} \frac{x}{6}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)$
Since, as $n \rightarrow \infty$, we have $\frac{1}{n} \rightarrow 0$
$\Rightarrow \frac{x}{3} \leq L<\frac{x}{3}$
According to Squeeze Principle or Sandwich Theorem, we have
$L=\frac{x}{3}$