Question

If $n\to \infty$ then the limit of series in $n$ can be evaluated by following the rule : $\underset{n\to \infty }{lim}\sum _{r=an+b}^{cn+d}\frac{1}{n}f\left(\frac{r}{n}\right)={\int }_a^{c}f\left(x\right)dx,$ where in $LHS$, $\frac{r}{n}$ is replaced by $x$, $\frac{1}{n}$ by $dx$ and the lower and upper limits are ${lim}_{n\to \infty }\frac{an+b}{n}and{lim}_{n\to \infty }\frac{cn+d}{n}$ respectively. Then answer the following question.

${lim}_{n\to \infty }\{\frac{1}{\sqrt{4n-1^2}}+\frac{1}{\sqrt{8n-2^2}}+...+\frac{1}{\sqrt{3}n}\}$ equals?

• A. $\frac{\pi }{6}$
• B. $\frac{4\pi }{6}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (C)

$\frac{\pi }{3}$

The given series $=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{\sqrt{4rn-r^2}}$

$=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n}\frac{1}{\sqrt{\frac{4r}{n}-{\left(\frac{r}{n}\right)}^2}}$

$={\int }_0^1\frac{dx}{\sqrt{4x-x^2}}$

$($Lower limits $=\underset{n\to \infty }{lim}\frac{1}{n}=0$ and upper limits $=\underset{n\to \infty }{lim}\frac{n}{n}=1)$

$={\int }_0^1\frac{dx}{\sqrt{4-(x-2{)}^2}}$

$={\left\{{\sin }^{-1}\left(\frac{x-2}{2}\right)\right\}}_0^1$

$={\sin }^{-1}(-\frac{1}{2})-{\sin }^{-1}(-1)=-\frac{\pi }{6}+\frac{\pi }{2}=\frac{\pi }{3}$