The value of _ n → ∞ [r] + [2r] + . . . + [ nr ] n ^ 2 , where r is...

We know that $r \le \left[r\right] < r + 1$

$2r \le \left[2r\right] < 2r + 1$

$3r \le \left[3r\right] < 3r + 1$

$⋮ ⋮ ⋮$

$nr \le \left[nr\right] < nr + 1$

On adding all the above inequalities, we get

$r + 2r + \ldots + nr \le \left[r\right] + \left[2r\right] + . . . + \left[nr\right] < r + 2r + . . . nr + n$

Using $1 + 2 + 3 + . . . + n = \dfrac{n\left(n + 1\right)}{2}, n \in N,$

$\dfrac{n\left(n + 1\right)}{2} \cdot r \le \left[r\right] + \left[2r\right] + . . . + \left[nr\right] < \dfrac{n\left(n + 1\right)}{2} \cdot r + n$

$⇒ \dfrac{\dfrac{n\left(n + 1\right)}{2} \cdot r}{{n}^{2}} \le \dfrac{\left[r\right] + \left[2r\right] + . . . + \left[nr\right]}{{n}^{2}} < \dfrac{\dfrac{n\left(n + 1\right)}{2} \cdot r + n}{{n}^{2}}$

Now, $\lim_{n \to \infty}\dfrac{n ( n + 1 ) \cdot r}{2 \cdot {n}^{2}} = \lim_{n \to \infty}\dfrac{{n}^{2}\left(1 + \dfrac{1}{n}\right) \cdot r}{2 \cdot {n}^{2}} = \dfrac{r}{2}$

and $\lim_{n \to \infty}\dfrac{\dfrac{n\left(n + 1\right)}{2} \cdot r + n}{{n}^{2}} = \lim_{n \to \infty}\dfrac{{n}^{2}\left\{\left(1 + \dfrac{1}{n}\right) \cdot r + \dfrac{1}{n}\right\}}{2 \cdot {n}^{2}} = \dfrac{r}{2}$

So, by Sandwich Theorem, we can conclude that

$\lim_{n \to \infty}\dfrac{\left[r\right] + \left[2r\right] + . . . + \left[nr\right]}{{n}^{2}} = \dfrac{r}{2} .$