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Question Number 55907 by gunawan last updated on 06/Mar/19

$$\mathrm{for}\mathrm{every}n\in \mathbb{N},f_n\left(x\right)=nx{(1-x^2)}^n,$$$$\mathrm{for}\mathrm{every}x,0⩽x⩽1$$$$\mathrm{and}{a}_n={\int }_0^1{f}_n\left(x\right)dx.$$$$\mathrm{If}{\mathrm{S}}_{\mathrm{n}}=\sin \left(\pi a_n\right),\mathrm{for}\mathrm{every}$$$$\mathrm{n}\in \mathbb{N},\mathrm{then}\underset{n\to \mathrm{\infty }}{\lim }{\mathrm{s}}_{\mathrm{n}}=...$$

Commented by maxmathsup by imad last updated on 06/Mar/19

$$wehave{a}_n={\int }_0^{1}nx{(1-x^2)}^{n}dx\Rightarrow \pi a_n=n\pi {\int }_0^{1}x{(1-x^2)}^{n}dxbut$$$${\int }_0^{1}x{(1-x^2)}^{n}dx=-\frac{1}{2(n+1)}{\left[{(1-x^2)}^{n+1}\right]}_0^1=\frac{1}{2n+2}\Rightarrow \pi a_n=\frac{n\pi }{2n+2}$$$$=\frac{n\pi }{2n(1+\frac{1}{n})}=\frac{\pi }{2(1+\frac{1}{n})}\sim \frac{\pi }{2}(1-\frac{1}{n})(n\to +\mathrm{\infty }\Rightarrow sin\left(\pi a_n\right)\sim sin(\frac{\pi }{2}-\frac{\pi }{2n})$$$$\Rightarrow {lim}_{n\to +\mathrm{\infty }}{S}_n=sin\left(\frac{\pi }{2}\right)=1.$$$$$$

Answered by 121194 last updated on 06/Mar/19

$$a_n=\underset{0}{\stackrel{1}{\int }}nx{(1-x^2)}^{n}dx$$$$y=1-x^2\Rightarrow dy=-2xdx\Rightarrow xdx=-\frac{dy}{2}$$$$x=0\Rightarrow y=1$$$$x=1\Rightarrow y=0$$$$a_n=n\underset{0}{\stackrel{1}{\int }}{(1-x^2)}^{n}xdx=-\frac{n}{2}\underset{1}{\stackrel{0}{\int }}{y}^{n}dy=\frac{n}{2}\underset{0}{\stackrel{1}{\int }}{y}^{n}dy$$$$=\frac{n}{2(n+1)}$$$$\underset{n\to \mathrm{\infty }}{\lim sin}\left[\frac{\pi n}{2(n+1)}\right]=\sin \left(\frac{\pi }{2}\underset{n\to \mathrm{\infty }}{\lim }\frac{n}{n+1}\right)=\sin \frac{\pi }{2}=1$$

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