JEE MAIN - Mathematics (2021 - 17th March Evening Shift - No. 11)
The value of $$\mathop {\lim }\limits_{n \to \infty } {{[r] + [2r] + ... + [nr]} \over {{n^2}}}$$, where r is a non-zero real number and [r] denotes the greatest integer less than or equal to r, is equal to :
r
$${r \over 2}$$
0
2r
Explanation
We know,
(x $$-$$ 1) $$ \le $$ [x] < x
$$ \therefore $$ (r $$-$$ 1) $$ \le $$ [r] < r
(2r $$-$$ 1) $$ \le $$ [2r] < 2r
.
.
.
(nr $$-$$ 1) $$ \le $$ [nr] < nr
Adding
$${{n(n + 1)} \over 2}r - n \le [r] + [2r] + .......[nr] < {{n(n + 1)} \over 2}r$$
$${{{{n\left( {n + 1} \right)} \over 2}r - n} \over {{n^2}}} \le {{\left[ r \right] + \left[ {2r} \right] + .... + \left[ {nr} \right]} \over {{n^2}}} \le {{{{n\left( {n + 1} \right)} \over 2}r} \over {{n^2}}}$$
$$\mathop {\lim }\limits_{n \to \infty } \left( {{{{{n(n + 1)} \over 2}r - n} \over {{n^2}}}} \right) \le L < \mathop {\lim }\limits_{n \to \infty } {{n(n + 1)} \over 2}r$$
$$ \Rightarrow {r \over 2} \le L < {r \over 2}$$
$$ \Rightarrow L = {r \over 2}$$
(x $$-$$ 1) $$ \le $$ [x] < x
$$ \therefore $$ (r $$-$$ 1) $$ \le $$ [r] < r
(2r $$-$$ 1) $$ \le $$ [2r] < 2r
.
.
.
(nr $$-$$ 1) $$ \le $$ [nr] < nr
Adding
$${{n(n + 1)} \over 2}r - n \le [r] + [2r] + .......[nr] < {{n(n + 1)} \over 2}r$$
$${{{{n\left( {n + 1} \right)} \over 2}r - n} \over {{n^2}}} \le {{\left[ r \right] + \left[ {2r} \right] + .... + \left[ {nr} \right]} \over {{n^2}}} \le {{{{n\left( {n + 1} \right)} \over 2}r} \over {{n^2}}}$$
$$\mathop {\lim }\limits_{n \to \infty } \left( {{{{{n(n + 1)} \over 2}r - n} \over {{n^2}}}} \right) \le L < \mathop {\lim }\limits_{n \to \infty } {{n(n + 1)} \over 2}r$$
$$ \Rightarrow {r \over 2} \le L < {r \over 2}$$
$$ \Rightarrow L = {r \over 2}$$
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