Q. The value of $\displaystyle\lim _{n \rightarrow \infty} \frac{[ r ]+[2 r ]+\ldots . .+[ nr ]}{ n ^{2}}$, where $r$ is non-zero real number and $[ r ]$ denotes the greatest integer less than or equal to $r$, is equal to :

JEE MainJEE Main 2021Limits and Derivatives Report Error

A

$\frac{ r }{2}$

58%

B

$r$

21%

C

$2 r$

8%

D

0

13%

Solution:

We know that
$r \le [r] < r + 1$
and $2r \le [2r] < 2r + 1$
$3r \le [3r] < 3r + 1$
$⋮ \,\,\,\,⋮\,\,\,⋮$
$nr \le [nr] < nr + 1$
$r + 2r + .... + nr$
$\le [r] + [2r] + ... + [nr] < (r + 2r + .... + nr ) + n$
$\frac{\frac{n(n + 1)}{2} \cdot r}{n^2} \le \frac{[r] + [2r] + ...[nr]}{n^2} < \frac{\frac{n(n + 1)}{2} r + n}{n^2}$
Now,
$\displaystyle\lim _{n \rightarrow \infty} \frac{n(n+1) \cdot r}{2 \cdot n^{2}}=\frac{r}{2}$
and
$\displaystyle\lim _{n \rightarrow \infty} \frac{\frac{n(n+1) r}{2}+n}{n^{2}}=\frac{r}{2}$
So, by Sandwich Theorem, we can conclude that
$\displaystyle\lim _{n \rightarrow \infty} \frac{[ r ]+[2 r ]+\ldots \ldots+[ nr ]}{ n ^{2}}=\frac{ r }{2}$

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