Question

The value of $\underset{n\to \infty }{\lim }\frac{1}{n}\left\{{\sec }^2\frac{\pi }{4n}+{\sec }^2\frac{2\pi }{4n}+.....{\sec }^2\frac{n\pi }{4n}\right\}$ is

• A. ${\log }_{e}2$
• B. $\frac{\pi }{2}$
• C. $\frac{4}{\pi }$
• D. $e$

Answer 1

Answer: (C)

We have, $\underset{n\to \infty }{\lim }\frac{1}{n}\left\{{\sec }^2\frac{\pi }{4n}+{\sec }^2\frac{2\pi }{4n}+\dots +{\sec }^2\frac{n\pi }{4n}\right\}$
$=\underset{n\to \infty }{\lim }\sum _{r=1}^n\frac{1}{n}{\sec }^2\left(\frac{r\pi }{4n}\right)=\underset{0}{\overset{1}{\int }}{\sec }^2\left(\frac{\pi x}{4}\right)$
$=\frac{4}{\pi }{\left[\tan \left(\frac{\pi x}{4}\right)\right]}_0^1$
$=\frac{4}{\pi }\times 1=\frac{4}{\pi }$