TIFR2021-Maths-A: 1

Question

For each positive integer $n$, let

$$s_n=\frac{1}{\sqrt{4n^2-1^2}}+\frac{1}{\sqrt{4n^2-2^2}}+\dots+\frac{1}{\sqrt{4n^2-n^2}}$$

Then the $\displaystyle \lim_{n\rightarrow \infty}s_n$ equals

• A. $\pi/2$
• B. $\pi/6$
• C. $1/2$
• D. $\infty$

Comments

B option

Answer 1

Answer: B

$\large{\lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{1}{\sqrt{4n^2 – 1^2}} + \frac{1}{\sqrt{4n^2 – 2^2}} + \dots + \frac{1}{\sqrt{4n^2 – n^2}}}$

                       $= \large{\lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{\sqrt{4n^2 – r^2}}}$            -------- ($\text{take } n^2 \text{ common from root}$)

                       $= \large{\lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n}.\frac{1}{\sqrt{4 - (\frac{r}{n}})^2}}$

$[\text{Converting limit of sum as definite integral}]$

$\text{Let }\large{ \frac{r}{n} = x, \frac{1}{n} = dx}$

$\text{When r = 1, }\large{x =  \lim_{n \to \infty} \frac{1}{n} = 0}$

$\text{When r = n, }\large{x = \lim_{n \to \infty} \frac{n}{n} = 1}$

 

$\large{\int_{0}^{1} \frac{dx}{\sqrt{4 – x^2}} = {\sin^{-1}\frac{x}{2}} \Big|_0^1 = \frac{\pi}{6}}$