Let \(a_n=\displaystyle\int_0^{\pi/2}(1-\sin t)^n \sin(2t)\,dt\), then \(\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\frac{a_k}{k}\) is equal to
1/2
Write \(\sin(2t)=2\sin t\cos t\) and substitute \(u=1-\sin t\), so \(du=-\cos t\,dt\) and \(\sin t=1-u\); as \(t:0\to\pi/2\), \(u:1\to 0\).
Then \(a_n=\displaystyle\int_0^{\pi/2}(1-\sin t)^n\cdot 2\sin t\,(\cos t\,dt)=2\int_0^1 u^n(1-u)\,du=2\left[\dfrac{1}{n+1}-\dfrac{1}{n+2}\right]=\dfrac{2}{(n+1)(n+2)}\).
So \(\dfrac{a_n}{n}=\dfrac{2}{n(n+1)(n+2)}\), and using the identity \(\dfrac{1}{n(n+1)(n+2)}=\dfrac12\left[\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}\right]\), this equals \(\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}\).
Summing from \(n=1\) to \(\infty\), this telescopes: \(\displaystyle\sum_{n=1}^{\infty}\left[\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}\right]=\dfrac{1}{1\cdot2}-\lim_{n\to\infty}\dfrac{1}{(n+1)(n+2)}\).
Since the last term tends to 0, the sum equals \(\dfrac12\).
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