Question

The value of $\underset{n\to \infty }{lim}[\frac{1}{n^2}{\sec }^2\frac{\pi }{3n^2}+\frac{2}{n^2}{\sec }^2\frac{4\pi }{3n^2}+\frac{3}{n^2}{\sec }^2\frac{9\pi }{3n^2}+\cdots +\frac{1}{n}{\sec }^2\frac{\pi }{3}]$ is

• A. $\frac{3}{2\pi }$
• B. $\frac{\sqrt{3}}{2\pi }$
• C. $\frac{3\sqrt{3}}{2\pi }$
• D. $\frac{4}{\sqrt{3}\pi }$

Answer 1

The correct option is C $\frac{3\sqrt{3}}{2\pi }$

Let
$L=\underset{n\to \infty }{lim}[\frac{1}{n^2}{\sec }^2\frac{\pi }{3n^2}+\frac{2}{n^2}{\sec }^2\frac{4\pi }{3n^2}+\frac{3}{n^2}{\sec }^2\frac{9\pi }{3n^2}+\cdots +\frac{1}{n}{\sec }^2\frac{\pi }{3}]=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n}\cdot \frac{r}{n}{\sec }^2(\frac{\pi }{3}\cdot \frac{r^2}{n^2})=\underset{0}{\overset{1}{\int }}x{\sec }^2\left(\frac{\pi }{3}{x}^2\right)dx$
put $\left(\frac{\pi }{3}{x}^2\right)=t\Rightarrow \frac{2\pi }{3}xdx=dt$
$=\underset{0}{\overset{1}{\int }}\frac{3}{2\pi }{\sec }^2\left(t\right)dt$
$=\frac{3}{2\pi }[\tan \left(\frac{\pi }{3}{x}^2\right){]}_0^1=\frac{3\sqrt{3}}{2\pi }$