lf $[x]$ denotes the greatest integer less than or equal to $x$ then $lim n \to \infty \frac{[x] + [2 x] + \dots . + [n x]}{n^{2}} =$

\frac{x}{2}

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\frac{x}{3}

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x

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Solution

The correct option is C $\frac{x}{2}$
${lim}_{n \to \infty} \frac{[x] + [2 x] + . . . + [n x]}{n^{2}} = {lim}_{n \to \infty} \frac{1}{n^{2}} \sum_{k = 1}^{n} [k x]$
As $k x - 1 < [k x]$ $\Rightarrow \sum_{k = 1}^{n} (k x - 1) < \sum_{k = 1}^{n} [k x] < \sum_{k = 1}^{n} (k x + 1) \Rightarrow \frac{x n (n + 1)}{2} - n < \sum_{k = 1}^{n} [k x] < \frac{x n (n + 1)}{2} + n \Rightarrow \frac{x}{2} (1 + \frac{1}{n}) - \frac{1}{n} < \frac{1}{n^{2}} \sum_{k = 1}^{n} [k x] < \frac{x}{2} (1 + \frac{1}{n}) + \frac{1}{n} \Rightarrow {lim}_{n \to \infty} (\frac{x}{2} (1 + \frac{1}{n}) - \frac{1}{n}) < {lim}_{n \to \infty} \frac{1}{n^{2}} \sum_{k = 1}^{n} [k x] < {lim}_{n \to \infty} (\frac{x}{2} (1 + \frac{1}{n}) + \frac{1}{n}) \Rightarrow \frac{x}{2} < {lim}_{n \to \infty} \frac{1}{n^{2}} \sum_{k = 1}^{n} [k x] < \frac{x}{2}$
$\Rightarrow {lim}_{n \to \infty} \frac{1}{n^{2}} \sum_{k = 1}^{n} [k x] = \frac{x}{2}$
Hence, option 'A' is correct.

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