Help Me Out!

nice problem.we need to first write the integral as $a_n=2\int_0^{\pi/2} (1-\sin(x))^n \sin(x)\cos(x) \text{dx}$ We make a simple u-sub of $\begin{cases} u=\sin(x)\\ \text{du}=\cos(x)\text{dx}\\\mid_0^{\pi/2}\to \mid_0^1\end{cases}$ so the integral is now $a_n=2\int_0^1 (1-u)^nu \text{du}=2\mathcal{B}(n+1,2)=2\dfrac{\Gamma(n+1)\Gamma(2)}{\Gamma(n+3)}=\dfrac{2}{(n+1)(n+2)}$ where the beta function is used.so the summation is $\sum_{n>0} \dfrac{2}{n(n+1)(n+2)}=\sum_{n>0} \left(\dfrac{1}{n}-\dfrac{1}{n+1}-\dfrac{1}{n+1}+\dfrac{1}{n+2}\right)\\=1-\dfrac{1}{2}=\boxed{\dfrac{1}{2}}$

$\displaystyle \lim _{ m\rightarrow \infty }{ \sum _{ n=1 }^{ m }{ \frac { { a }_{ n } }{ n } } }$

$\displaystyle =\lim _{ m\rightarrow \infty }{ \sum _{ n=1 }^{ m }{ \frac { 1 }{ n } \left( \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( 1-\sin { x } \right) ^{ n }\sin { 2x } dx } \right) } }$

$\displaystyle = \left( \int _{ 0 }^{ \frac { \pi }{ 2 } } \lim _{ m\rightarrow \infty }{ \sum _{ n=1 }^{ m }} { \frac { 1 }{ n }\left( 1-\sin { x } \right) ^{ n }\sin { 2x } dx } \right)$

$\displaystyle = \left( \int _{ 0 }^{ \frac { \pi }{ 2 }} {- \ln { \sin { x } } \sin { 2x } dx } \right)$

$\displaystyle =- \left( \int _{ 0 }^{ \frac { \pi }{ 2 }} { \ln { \sin { x } } d\sin^{2} { x }} \right)$

$\displaystyle =-\left( \sin ^{ 2 }{ x } \ln { \sin { x } } -\frac { \sin ^{ 2 }{ x } }{ 2 } \right )$ from 0 to 1

$\displaystyle =-\left( -\frac { 1 }{ 2 } -0 \right) =\frac{1}{2}$