Question Number 11089 by suci last updated on 11/Mar/17
$${for}\:{each}\:{n}\in\mathbb{N},\:{f}_{{n}} \left({x}\right)={nx}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{n}} \\ $$$${for}\:{each}\:{x},\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:{and}\:{a}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {f}_{{n}} \left({x}\right){dx} \\ $$$${if}\:{s}_{{n}} ={sin}\left(\pi{a}_{{n}} \right),\:{for}\:{each}\:{n}\in\mathbb{N},\:{then} \\ $$$${l}\underset{{n}\rightarrow\sim} {{i}m}\:{s}_{{n}} =….??? \\ $$
Commented by FilupS last updated on 11/Mar/17
![n∈N f_n (x)=nx(1−x^2 )^n 0≤x≤1 a_n =∫_0 ^( 1) f_n (x)dx s_n =sin(πa_n ) lim_(n→∞) s_n =L L=? a_n =∫_0 ^( 1) f_n (x)dx a_n =∫_0 ^( 1) nx(1−x^2 )^n dx =n∫_0 ^( 1) x(1−x^2 )^n dx u=1−x^2 du=−2xdx =−n(1/2)∫_0 ^( 1) u^n du =−n(1/2)((1/(n+1))u^(n+1) )_0 ^1 =−(n/(2(n+1)))(u^(n+1) )_0 ^1 =−(n/(2(n+1)))((1−x^2 )^(n+1) )_0 ^1 =−(n/(2(n+1)))(0^(n+1) −1^(n+1) ) =(n/(2(n+1))) ∴a_n =(n/(2(n+1))) s_n =sin(πa_n ) s_n =sin(π(n/(2(n+1)))) s_n =sin((1/2)π(n/((n+1)))) lim_(n→∞) s_n =lim_(n→∞) sin((1/2)π(n/((n+1)))) =sin((1/2)π[lim_(n→∞) (n/((n+1)))]) L′Ho^ pital′s Law =sin((1/2)π lim_(n→∞) (1/1)) =sin((1/2)π) ∴ L = 1 lim_(n→∞) s_n =1](https://www.tinkutara.com/question/Q11090.png)
$${n}\in\mathbb{N} \\ $$$${f}_{{n}} \left({x}\right)={nx}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{n}} \:\:\:\:\:\:\:\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1} \\ $$$${a}_{{n}} =\int_{\mathrm{0}} ^{\:\mathrm{1}} {f}_{{n}} \left({x}\right){dx} \\ $$$${s}_{{n}} =\mathrm{sin}\left(\pi{a}_{{n}} \right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{s}_{{n}} ={L} \\ $$$${L}=? \\ $$$$\: \\ $$$${a}_{{n}} =\int_{\mathrm{0}} ^{\:\mathrm{1}} {f}_{{n}} \left({x}\right){dx} \\ $$$${a}_{{n}} =\int_{\mathrm{0}} ^{\:\mathrm{1}} {nx}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{n}} {dx} \\ $$$$={n}\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{n}} {dx} \\ $$$${u}=\mathrm{1}−{x}^{\mathrm{2}} \\ $$$${du}=−\mathrm{2}{xdx} \\ $$$$=−{n}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {u}^{{n}} {du} \\ $$$$=−{n}\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}{u}^{{n}+\mathrm{1}} \right)_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=−\frac{{n}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\left({u}^{{n}+\mathrm{1}} \right)_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=−\frac{{n}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\left(\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} \right)_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=−\frac{{n}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\left(\mathrm{0}^{{n}+\mathrm{1}} −\mathrm{1}^{{n}+\mathrm{1}} \right) \\ $$$$=\frac{{n}}{\mathrm{2}\left({n}+\mathrm{1}\right)} \\ $$$$\therefore{a}_{{n}} =\frac{{n}}{\mathrm{2}\left({n}+\mathrm{1}\right)} \\ $$$$\: \\ $$$${s}_{{n}} =\mathrm{sin}\left(\pi{a}_{{n}} \right) \\ $$$${s}_{{n}} =\mathrm{sin}\left(\pi\frac{{n}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\right) \\ $$$${s}_{{n}} =\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\pi\frac{{n}}{\left({n}+\mathrm{1}\right)}\right) \\ $$$$\: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{s}_{{n}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\pi\frac{{n}}{\left({n}+\mathrm{1}\right)}\right) \\ $$$$=\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\pi\left[\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}}{\left({n}+\mathrm{1}\right)}\right]\right) \\ $$$$\mathrm{L}'\mathrm{H}\hat {\mathrm{o}pital}'\mathrm{s}\:\mathrm{Law} \\ $$$$=\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\pi\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{1}}\right) \\ $$$$=\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\pi\right) \\ $$$$\therefore\:{L}\:=\:\mathrm{1} \\ $$$$\: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{s}_{{n}} =\mathrm{1} \\ $$