Let an=∫0π/2(1-sint)nsin2t dt then limn→∞∑1nann is equal to

an=∫0π/2(1-sint)nsin2tdt Let 1-sint=u⇒-costdt=du=2∫01un(1-u)du=2∫01undu-∫01un+1du=21n+1-1n+2 Hence, ann=21n(n+1)-1n(n+2)limn→∞∑1nann=2∑1n-1n+1-12∑1n-1n+2=2∑1n1n-1n+1-∑1n1n-1n+2=2(1)-1-13+12-14+13-15+….=2-32=12