If [.] denotes the greatest integer function, then find the value of $lim n \to \infty \frac{[x] + [2 x] + . . . . . . . . . . . . + [n x]}{n^{2}} .$

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Solution

We have for $x \in R$ ,
$x - 1 < [x] \leq x$
or, $n x - 1 < [n x] \leq n x$ [For $n \in N]$ .
Then,
$(x - 1) + (2 x - 1) + . . . . + (n x - 1) < [x] + [2 x] + . . . . + [n x] \leq x + 2 x + . . . . . + n x$
$\Rightarrow \frac{n (n + 1)}{2} x - n < [x] + [2 x] + . . . . + [n x] \leq \frac{n (n + 1)}{2} x$
$\Rightarrow (\frac{1}{2} + \frac{1}{2 n}) x - \frac{1}{n} < \frac{[x] + [2 x] + . . . . + [n x]}{n^{2}} \leq (\frac{1}{2 n} + \frac{1}{2 n}) x$
$\Rightarrow lim n \to \infty (\frac{1}{2} + \frac{1}{2 n}) x - \frac{1}{n} \leq lim n \to \infty \frac{[x] + [2 x] + . . . . + [n x]}{n^{2}} \leq lim n \to \infty (\frac{1}{2} + \frac{1}{2 n}) x$ [Using limit property]
$\Rightarrow \frac{x}{2} \leq lim n \to \infty \frac{[x] + [2 x] + . . . . + [n x]}{n^{2}} \leq \frac{x}{2}$
Using Sandwich property we get,
$lim n \to \infty \frac{[x] + [2 x] + . . . . + [n x]}{n^{2}} = \frac{x}{2}$

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