The value of

Question:

The value of $\lim _{n \rightarrow \infty} \frac{[\mathrm{r}]+[2 \mathrm{r}]+\ldots \ldots+[\mathrm{nr}]}{\mathrm{n}^{2}}$, where $\mathrm{r}$ is non-zero real number and

$[r]$ denotes the greatest integer less than or equal to $r$, is equal to :

  1. (1) $\frac{\mathrm{r}}{2}$

  2. (2) $\mathrm{r}$

  3. (3) $2 \mathrm{r}$

  4. (4) 0


Correct Option: 1

Solution:

We know that

$\mathrm{r} \leq[\mathrm{r}]<\mathrm{r}+1$

and

$2 r \leq[2 r]<2 r+1$

$3 r \leq[3 r]<3 r+1$

Now,

$\lim _{n \rightarrow \infty} \frac{n(n+1) \cdot r}{2 \cdot n^{2}}=\frac{r}{2}$

and

$\lim _{n \rightarrow \infty} \frac{\frac{n(n+1) r}{2}+n}{n^{2}}=\frac{r}{2}$

So, by Sandwich Theorem, we can conclude that

$\lim _{n \rightarrow \infty} \frac{[r]+[2 r]+\ldots \ldots+[n r]}{n^{2}}=\frac{r}{2}$

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