Question Number 55907 by gunawan last updated on 06/Mar/19
$$\mathrm{for}\:\mathrm{every}\:{n}\:\in\:\mathbb{N}\:,\:{f}_{{n}} \left({x}\right)={nx}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{n}} , \\ $$$$\mathrm{for}\:\mathrm{every}\:{x},\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1} \\ $$$$\mathrm{and}\:{a}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {f}_{{n}} \left({x}\right)\:{dx}. \\ $$$$\mathrm{If}\:\mathrm{S}_{\mathrm{n}} =\mathrm{sin}\:\left(\pi{a}_{{n}} \right),\:\mathrm{for}\:\mathrm{every} \\ $$$$\mathrm{n}\in\:\mathbb{N},\:\mathrm{then}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{s}_{\mathrm{n}} =… \\ $$
Commented by maxmathsup by imad last updated on 06/Mar/19
![we have a_n =∫_0 ^1 nx(1−x^2 )^n dx ⇒πa_n =nπ ∫_0 ^1 x(1−x^2 )^n dx but ∫_0 ^1 x(1−x^2 )^n dx =−(1/(2(n+1)))[(1−x^2 )^(n+1) ]_0 ^1 =(1/(2n+2)) ⇒πa_n =((nπ)/(2n+2)) =((nπ)/(2n(1+(1/n)))) =(π/(2(1+(1/n)))) ∼(π/2)(1−(1/n)) (n→+∞ ⇒sin(πa_n )∼sin((π/2)−(π/(2n))) ⇒lim_(n→+∞) S_n =sin((π/2)) =1 .](https://www.tinkutara.com/question/Q55953.png)
$${we}\:{have}\:{a}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {nx}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{n}} {dx}\:\Rightarrow\pi{a}_{{n}} ={n}\pi\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{n}} {dx}\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{n}} \:{dx}\:=−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\left[\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}}\:\Rightarrow\pi{a}_{{n}} =\frac{{n}\pi}{\mathrm{2}{n}+\mathrm{2}} \\ $$$$=\frac{{n}\pi}{\mathrm{2}{n}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)}\:=\frac{\pi}{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)}\:\sim\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\:\left({n}\rightarrow+\infty\:\Rightarrow{sin}\left(\pi{a}_{{n}} \right)\sim{sin}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}{n}}\right)\right. \\ $$$$\Rightarrow{lim}_{{n}\rightarrow+\infty} {S}_{{n}} ={sin}\left(\frac{\pi}{\mathrm{2}}\right)\:=\mathrm{1}\:. \\ $$$$ \\ $$
Answered by 121194 last updated on 06/Mar/19
![a_n =∫_0 ^1 nx(1−x^2 )^n dx y=1−x^2 ⇒dy=−2xdx⇒xdx=−(dy/2) x=0⇒y=1 x=1⇒y=0 a_n =n∫_0 ^1 (1−x^2 )^n xdx=−(n/2)∫_1 ^0 y^n dy=(n/2)∫_0 ^1 y^n dy =(n/(2(n+1))) lim_(n→∞) sin [((πn)/(2(n+1)))]=sin ((π/2)lim_(n→∞) (n/(n+1)))=sin (π/2)=1](https://www.tinkutara.com/question/Q55935.png)
$${a}_{{n}} =\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{nx}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{n}} {dx} \\ $$$${y}=\mathrm{1}−{x}^{\mathrm{2}} \Rightarrow{dy}=−\mathrm{2}{xdx}\Rightarrow{xdx}=−\frac{{dy}}{\mathrm{2}} \\ $$$${x}=\mathrm{0}\Rightarrow{y}=\mathrm{1} \\ $$$${x}=\mathrm{1}\Rightarrow{y}=\mathrm{0} \\ $$$${a}_{{n}} ={n}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{n}} {xdx}=−\frac{{n}}{\mathrm{2}}\underset{\mathrm{1}} {\overset{\mathrm{0}} {\int}}{y}^{{n}} {dy}=\frac{{n}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{y}^{{n}} {dy} \\ $$$$=\frac{{n}}{\mathrm{2}\left({n}+\mathrm{1}\right)} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}sin}\:\left[\frac{\pi{n}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\right]=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}}{{n}+\mathrm{1}}\right)=\mathrm{sin}\:\frac{\pi}{\mathrm{2}}=\mathrm{1} \\ $$