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Let $$a_n=\int_{0}^{\pi/2}(1-\sin(t))^{n}\sin(2t)\, dt$$ then $$\lim_{N \rightarrow \infty} \sum_{n=1}^{N}\frac{a_n}{n}$$ is equal to?

I tried to apply L'Hospital's rule initially but it will not work directly as there is a summation involved.

Next I tried to convert the $a_n$ to summation form using first principle and then substitute in second.But that makes things even more complicated with two summations.

Also I tried using King's rule to express $a_n$ differently.But that introduces a cosine term.Doesn't help much.

What else to try?

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Hint: One has $$a_n=2\int_{0}^{\pi/2}(1-\sin t)^n \sin (t)\cos (t) dt$$ $$=2\int_{0}^1(1-x)^n x dx.$$ Denote $$b_n=2\int_{0}^1(1-x)^n dx=\frac{2}{n+1}.$$ Then consider $b_n-a_n$.

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For any $t\in\left(0,\frac{\pi}{2}\right)$ we have $$\sum_{n=1}^{+\infty}\frac{(1-\sin t)^n}{n}=-\log\sin t \tag{1} $$ hence our limit equals $$ -\int_{0}^{\pi/2}\sin(2t)\log\sin t\,dt\stackrel{IBP}{=}\frac{1}{2}\int_{0}^{\pi/2}(1-\cos(2t))\cot(t)\,dt=\color{red}{\frac{1}{2}}.\tag{2} $$

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