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

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Question

(C) The given expression is $S = \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{k}{n^2} \sec^2 \left(\frac{k^2}{n^4}\right)$.This can be rewritten

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

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(C) The given expression is $S = \lim _{n ightarrow \infty} \sum_{k=1}^{n} \frac{k}{n^2} \sec^2 \left(\frac{k^2}{n^4}ight)$.This can be rewritten as $S = \lim _{n ightarrow \infty} \frac{1}{n^2} \sum_{k=1}^{n} k \sec^2 \left(\left(\frac{k}{n^2}ight)^2ight)$.Let $x = \frac{k}{n^2}$ and $dx = \frac{1}{n^2}$.As $k=1$,$x ightarrow 0$ and as $k=n$,$x ightarrow \frac{n}{n^2} = \frac{1}{n} ightarrow 0$.Wait,let us re-evaluate the limit as a Riemann sum.Let $S = \lim _{n ightarrow \infty} \sum_{k=1}^{n} \frac{k}{n^2} \sec^2 \left(\frac{k^2}{n^4}ight)$.Let $f(x) = x \sec^2(x^2)$.This is equivalent to $\int_{0}^{1} x \sec^2(x^2) dx$.Let $u = x^2$,then $du = 2x dx$,so $x dx = \frac{1}{2} du$.When $x=0, u=0$ and when $x=1, u=1$.Thus,$S = \int_{0}^{1} \frac{1}{2} \sec^2(u) du = \frac{1}{2} [	an(u)]_{0}^{1} = \frac{1}{2} 	an(1)$.

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

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