Question

The value of $\underset{n \rightarrow \infty}{\text{Lim}} \left(\frac{1-\cos \frac{2}{n}}{1+\cos \frac{2}{n}}\right)\left(\displaystyle\sum_{r=1}^{2 n} \frac{n}{\sqrt{8 n^2-r^2}}\right)\left(\displaystyle\sum_{r=1}^{3 n} \frac{n}{\sqrt{18 n^2-r^2}}\right)$ is equal to

• A. $\frac{\pi^2}{4}$
• B. $\frac{\pi^2}{8}$
• C. $\frac{\pi^2}{16}$
• D. $\frac{3 \pi^2}{16}$

Answer 1

Answer: (C)

$\underset{n \rightarrow \infty}{\text{Lim}} \left(\frac{1-\cos \frac{2}{n}}{1+\cos \frac{2}{n}}\right)\left(\displaystyle\sum_{r=1}^{2 n} \frac{n}{\sqrt{8 n^2-r^2}}\right)\left(\displaystyle\sum_{r=1}^{3 n} \frac{n}{\sqrt{18 n^2-r^2}}\right) $
$=\underset{n \rightarrow \infty}{\text{Lim}} \frac{1}{2} \cdot\left(\frac{1-\cos \frac{2}{n}}{\frac{4}{n^2}}\right) \cdot 4 \cdot\left(\displaystyle\sum_{r=1}^{2 n} \frac{n}{n \cdot n \sqrt{8-\frac{r^2}{n^2}}}\right)\left(\displaystyle\sum_{r=1}^{3 n} \frac{n}{n \cdot n \sqrt{18-\frac{r^2}{n^2}}}\right) $
$=\frac{1}{2} \times \frac{1}{2} \times 4\left(\int\limits_0^2 \frac{d x}{\sqrt{8-x^2}}\right)\left(\int\limits_0^3 \frac{d x}{\sqrt{18-x^2}}\right)=\left(\sin ^{-1} \frac{x}{2 \sqrt{2}}\right)_0^2\left(\sin ^{-1} \frac{x}{3 \sqrt{2}}\right)_0^3=\frac{\pi}{4} \cdot \frac{\pi}{4}=\frac{\pi^2}{16}$