lim _{n rightarrow infty}left[frac{1}{n^2} se | vedclass

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Question

(C) दी गई अभिव्यक्ति $S = \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{k}{n^2} \sec^2 \left(\frac{k^2}{n^4}\right)$ है।इसे $S = \lim _{n \rightar

Vedclass · Accepted Answer

$\frac{1}{2} 	an 1$

Vedclass · Answer

(C) दी गई अभिव्यक्ति $S = \lim _{n ightarrow \infty} \sum_{k=1}^{n} \frac{k}{n^2} \sec^2 \left(\frac{k^2}{n^4}ight)$ है।इसे $S = \lim _{n ightarrow \infty} \sum_{k=1}^{n} \frac{k}{n^2} \sec^2 \left(\left(\frac{k}{n^2}ight)^2ight)$ के रूप में लिखा जा सकता है।माना $x = \frac{k}{n^2}$ और $dx = \frac{1}{n^2}$।यह रीमान योग $\int_{0}^{1} x \sec^2(x^2) dx$ के बराबर है।माना $u = x^2$,तो $du = 2x dx$,इसलिए $x dx = \frac{1}{2} du$।जब $x=0, u=0$ और जब $x=1, u=1$।अतः,$S = \int_{0}^{1} \frac{1}{2} \sec^2(u) du = \frac{1}{2} [	an(u)]_{0}^{1} = \frac{1}{2} 	an(1)$।

$\lim _{n \rightarrow \infty}\left[\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\frac{3}{n^2} \sec ^2 \frac{9}{n^2}+\ldots+\frac{n}{n^2} \sec ^2 \frac{n^2}{n^2}\right]=$

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