Let an=∫0π2(1−sint)nsin2t dt then limn→∞∑k=1nakk is equal to

1−sint=u⇒costdt=−duan=∫10un2(1−u)(−du)=2∫01un(1−u)du = 2∫01(un−un+1)du=2[1n+1−1n+2]ann=2[1n(n+1)−1n(n+2)]∑n=1nann=2[11.2+12.3+−−−+1n(n+1)]−2[11.3+12.4+−−−+1n(n+2)]=2[(1−12)+(12−13)+−−−−+(1n−1n+1)]− 22[(1−13)+(12−14)+(13−19)+−−−−+(1n−1n+2)]limn→∞∑k=1nakk=2−32=12

Let an=∫0π2(1−sint)nsin2t dt then limn→∞∑k=1nakk is equal to

1−sint=u⇒costdt=−duan=∫10un2(1−u)(−du)=2∫01un(1−u)du = 2∫01(un−un+1)du=2[1n+1−1n+2]ann=2[1n(n+1)−1n(n+2)]∑n=1nann=2[11.2+12.3+−−−+1n(n+1)]−2[11.3+12.4+−−−+1n(n+2)]=2[(1−12)+(12−13)+−−−−+(1n−1n+1)]− 22[(1−13)+(12−14)+(13−19)+−−−−+(1n−1n+2)]limn→∞∑k=1nakk=2−32=12

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Let $a_{n} = \int_{0}^{\frac{π}{2}} {(1 - \sin t)}^{n} \sin 2 t d t$ then $\begin{matrix} \lim \\ n \to \infty \end{matrix} \sum_{k = 1}^{n} \frac{a_{k}}{k}$ is equal to

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$\frac{1}{2}$

$\frac{4}{3}$

$\frac{3}{2}$

answer is A.

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Detailed Solution

$1 - \sin t = u \Rightarrow \cos t d t = - d u$

$\begin{array}{l} a_{n} = \int_{1}^{0} u^{n} 2 (1 - u) (- d u) = 2 \int_{0}^{1} u^{n} (1 - u) d u \\ = 2 \int_{0}^{1} (u^{n} - u^{n + 1}) d u = 2 [\frac{1}{n + 1} - \frac{1}{n + 2}] \end{array}$

$\frac{a_{n}}{n} = 2 [\frac{1}{n (n + 1)} - \frac{1}{n (n + 2)}]$

$\sum_{n = 1}^{n} \frac{a_{n}}{n} = 2 [\frac{1}{1.2} + \frac{1}{2.3} + - - - + \frac{1}{n (n + 1)}] - 2 [\frac{1}{1.3} + \frac{1}{2.4} + - - - + \frac{1}{n (n + 2)}]$

$= 2 [(1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + - - - - + (\frac{1}{n} - \frac{1}{n + 1})] - \frac{2}{2} [(1 - \frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{9}) + - - - - + (\frac{1}{n} - \frac{1}{n + 2})]$