The value of lim _{n rightarrow infty} frac{1 | vedclass

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(C) Given the limit: $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \sec ^{2} \left( \frac{r \pi}{4 n} \right)$.By the definition of a defin

Vedclass · Accepted Answer

$\frac{4}{\pi}$

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(C) Given the limit: $\lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \sec ^{2} \left( \frac{r \pi}{4 n} ight)$.By the definition of a definite integral as the limit of a sum, $\lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f \left( \frac{r}{n} ight) = \int_{0}^{1} f(x) dx$.Here, $f(x) = \sec ^{2} \left( \frac{\pi x}{4} ight)$.So, the integral becomes $\int_{0}^{1} \sec ^{2} \left( \frac{\pi x}{4} ight) dx$.Integrating $\sec ^{2} \left( \frac{\pi x}{4} ight)$, we get $\frac{4}{\pi} 	an \left( \frac{\pi x}{4} ight)$.Evaluating the definite integral from $0$ to $1$: $\left[ \frac{4}{\pi} 	an \left( \frac{\pi x}{4} ight) ight]_{0}^{1} = \frac{4}{\pi} \left( 	an \frac{\pi}{4} - 	an 0 ight)$.Since $	an \frac{\pi}{4} = 1$ and $	an 0 = 0$, the value is $\frac{4}{\pi} (1 - 0) = \frac{4}{\pi}$.

The value of $\lim _{n \rightarrow \infty} \frac{1}{n} \left\{ \sec ^{2} \frac{\pi}{4 n} + \sec ^{2} \frac{2 \pi}{4 n} + \ldots + \sec ^{2} \frac{n \pi}{4 n} \right\}$ is

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