Question

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

• A. $\log_e \, 2 $
• B. $\frac{\pi}{2}$
• C. $\frac{4}{\pi}$
• D. $e$

Answer 1

The Correct Option is C

We have, $\displaystyle\lim _{n \rightarrow \infty} \frac{1}{n}\left\{\sec ^{2} \frac{\pi}{4 n}+\sec ^{2} \frac{2 \pi}{4 n}+\ldots+\sec ^{2} \frac{n \pi}{4 n}\right\}$
$=\displaystyle\lim _{n \rightarrow \infty} \displaystyle\sum_{r=1}^{n} \frac{1}{n} \sec ^{2}\left(\frac{r \pi}{4 n}\right)=\int\limits_{0}^{1} \sec ^{2}\left(\frac{\pi x}{4}\right)$
$=\frac{4}{\pi}\left[\tan \left(\frac{\pi x}{4}\right)\right]_{0}^{1}$
$=\frac{4}{\pi} \times 1=\frac{4}{\pi}$