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Q.
Let $a_n=\int\limits_0^{\pi / 2}(1-\sin t)^n \sin 2 t d t$ then $\underset{n \rightarrow \infty}{\text{Lim}} \displaystyle\sum_{n=1}^n \frac{a_n}{n}$ is equal to
Integrals
Solution:
$\quad a_n=\int\limits_0^{\pi / 2}(1-\sin t)^n \sin 2 t d t$
Let $\quad 1-\sin t = u \Rightarrow-\cos t dt = du$
$a_n=2 \int\limits_0^1 u^n(1-u) d u=2\left(\int\limits_0^1 u^n d u-\int\limits_0^1 u^{n+1} d u\right)=2\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$
hence $\frac{a_n}{n}=2\left(\frac{1}{n(n+1)}-\frac{1}{n(n+2)}\right)$
$\underset{n \rightarrow \infty} {\text{Lim}} \displaystyle\sum_{n=1}^n \frac{a_n}{n} =2\left(\sum\left(\frac{1}{n}-\frac{1}{n+1}\right)-\frac{1}{2} \displaystyle\sum\left(\frac{1}{n}-\frac{1}{n+2}\right)\right)=2 \sum_{n=1}^n\left(\frac{1}{n}-\frac{1}{n+1}\right)-\displaystyle\sum_{n=1}^n\left(\frac{1}{n}-\frac{1}{n+2}\right) $
$=2(1)-\left[\left(1-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\ldots . .\right]=2-\frac{3}{2}=\frac{1}{2} $