First slide

Introduction to limits

Question

$lim_{n \to \infty} \frac{[x] + [2 x] + [3 x] + \dots + [nx]}{n^{2}}$ where [.] denotes the greatest integer function, is equal to

Moderate

A

$\frac{x}{2}$
B

$x$
C

$2 x$
D

None of these

Solution

We know that $x - 1 < [x] \leq x$
$\Rightarrow x + 2 x + \dots + nx - n < \sum_{r = 1}^{n} [rx] \leq x + 2 x + \dots + nx$
$\begin{array}{l} \Rightarrow \frac{x \cdot n (n + 1)}{2} - n < \sum_{r = 1}^{n} [rx] \leq \frac{x \cdot n (n + 1)}{2} \\ \Rightarrow \frac{x}{2} (1 + \frac{1}{n}) - \frac{1}{n} < \frac{1}{n^{2}} \sum_{r = 1}^{n} [rx] \leq \frac{x}{2} (1 + \frac{1}{n}) \end{array}$
now,
$\begin{array}{l} lim_{n \to \infty} \frac{x}{2} (1 + \frac{1}{n}) = \frac{x}{2} \\ lim_{n \to \infty} \frac{x}{2} (1 + \frac{1}{n}) - \frac{1}{n} = \frac{x}{2} \end{array}$
Using Sandwich theorem, we find that
$lim_{x \to \infty} \frac{[x + [2 x] + \dots + [nx]]}{n^{2}} = \frac{x}{2}$

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