JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Limits

The value of

lim_{n \to \infty} \frac{[ r ] + [ 2 r ] + \dots + [ n r ]}{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:

Answer: (A)

Solution

Write

[k r] = k r - {k r}

, where the fractional part satisfies

0 \leq {k r} < 1

. Summing from

k = 1

n

k = 1 \sum n [k r] = r \frac{n ( n + 1 )}{2} - k = 1 \sum n {k r}, 0 \leq k = 1 \sum n {k r} < n

Dividing by

n^{2}

\frac{1}{n ^{2}} k = 1 \sum n [k r] = \frac{r}{2} \cdot \frac{n + 1}{n} - \frac{1}{n ^{2}} k = 1 \sum n {k r}

The first part tends to

\frac{r}{2}

, and the second is trapped between

0

and

\frac{1}{n}

, so it tends to

0

. The limit is

\frac{r}{2}

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option

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