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Question Number 179657 by mathlove last updated on 31/Oct/22
prove that lim_(n→∞) ((√((1/2)×(√((1/2)+(1/2)(√(1/2))))))×(√((1/2)+(1/2)(√((1/2)+(1/2)(√(1/2))))))×∙∙∙∙∙(√((1/2)+(1/2)(√((1/2)+(1/2)(√((1/(2+)).....+(1/2)(√(1/2))))))))_(n term) )=(2/π)
$${prove}\:{that} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt{\frac{\mathrm{1}}{\mathrm{2}}×\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}}×\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}}×\centerdot\centerdot\centerdot\centerdot\centerdot\underset{{n}\:{term}} {\underbrace{\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}+}…..+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}}}}}\right)=\frac{\mathrm{2}}{\pi} \\ $$
Commented by mr W last updated on 01/Nov/22
the question should be: lim_(n→∞) ((√(1/2))×(√((1/2)+(1/2)(√(1/2))))×(√((1/2)+(1/2)(√((1/2)+(1/2)(√(1/2))))))×∙∙∙∙∙(√((1/2)+(1/2)(√((1/2)+(1/2)(√((1/(2+)).....+(1/2)(√(1/2))))))))_(n term) )=(2/π)
$${the}\:{question}\:{should}\:{be}: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}×\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}×\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}}×\centerdot\centerdot\centerdot\centerdot\centerdot\underset{{n}\:{term}} {\underbrace{\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}+}…..+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}}}}}\right)=\frac{\mathrm{2}}{\pi} \\ $$
Commented by mathlove last updated on 02/Nov/22
yes sir
$${yes}\:{sir} \\ $$
Answered by mr W last updated on 01/Nov/22
a_1 =(√(1/2)) a_2 =(√((1/2)+(1/2)(√(1/2)))) ... a_n =(√((1/2)+(1/2)a_(n−1) )) 2a_n ^2 −1=a_(n−1) say a_n =cos θ_n 2cos^2 θ_n −1=cos θ_(n−1) cos 2θ_n =cos θ_(n−1) 2θ_n =θ_(n−1) θ_n =(θ_(n−1) /2)=(θ_(n−2) /2^2 )=...=(θ_1 /2^(n−1) ) a_1 =cos θ_1 =(√(1/2)) ⇒θ_1 =(π/4)=(π/2^2 ) θ_n =(θ_1 /2^(n−1) )=(π/2^(n+1) ) a_n =cos θ_n =cos (π/2^(n+1) ) a_1 a_2 ...a_n =cos (π/2^2 )cos (π/2^3 )...cos (π/2^(n+1) ) 2sin (π/2^(n+1) )(a_1 a_2 ...a_n )=cos (π/2^2 )cos (π/2^3 )...sin (π/2^n ) 2^2 sin (π/2^(n+1) )(a_1 a_2 ...a_n )=cos (π/2^2 )cos (π/2^3 )...sin (π/2^(n−1) ) ...... 2^n sin (π/2^(n+1) )(a_1 a_2 ...a_n )=sin (π/2)=1 a_1 a_2 ...a_n =(1/( 2^n sin (π/2^(n+1) ))) a_1 a_2 ...a_n =(2/π)×((π/2^(n+1) )/( sin (π/2^(n+1) ))) LHS=lim_(n→∞) (a_1 a_2 ...a_n )=(2/( π))
$${a}_{\mathrm{1}} =\sqrt{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${a}_{\mathrm{2}} =\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}} \\ $$$$… \\ $$$${a}_{{n}} =\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{a}_{{n}−\mathrm{1}} } \\ $$$$\mathrm{2}{a}_{{n}} ^{\mathrm{2}} −\mathrm{1}={a}_{{n}−\mathrm{1}} \\ $$$$ \\ $$$${say}\:{a}_{{n}} =\mathrm{cos}\:\theta_{{n}} \\ $$$$\mathrm{2cos}^{\mathrm{2}} \:\theta_{{n}} −\mathrm{1}=\mathrm{cos}\:\theta_{{n}−\mathrm{1}} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta_{{n}} =\mathrm{cos}\:\theta_{{n}−\mathrm{1}} \\ $$$$\mathrm{2}\theta_{{n}} =\theta_{{n}−\mathrm{1}} \\ $$$$\theta_{{n}} =\frac{\theta_{{n}−\mathrm{1}} }{\mathrm{2}}=\frac{\theta_{{n}−\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }=…=\frac{\theta_{\mathrm{1}} }{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$$${a}_{\mathrm{1}} =\mathrm{cos}\:\theta_{\mathrm{1}} =\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}\:\Rightarrow\theta_{\mathrm{1}} =\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{2}^{\mathrm{2}} } \\ $$$$\theta_{{n}} =\frac{\theta_{\mathrm{1}} }{\mathrm{2}^{{n}−\mathrm{1}} }=\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} } \\ $$$${a}_{{n}} =\mathrm{cos}\:\theta_{{n}} =\mathrm{cos}\:\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} } \\ $$$${a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}} =\mathrm{cos}\:\frac{\pi}{\mathrm{2}^{\mathrm{2}} }\mathrm{cos}\:\frac{\pi}{\mathrm{2}^{\mathrm{3}} }…\mathrm{cos}\:\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} } \\ $$$$\mathrm{2sin}\:\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} }\left({a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}} \right)=\mathrm{cos}\:\frac{\pi}{\mathrm{2}^{\mathrm{2}} }\mathrm{cos}\:\frac{\pi}{\mathrm{2}^{\mathrm{3}} }…\mathrm{sin}\:\frac{\pi}{\mathrm{2}^{{n}} } \\ $$$$\mathrm{2}^{\mathrm{2}} \mathrm{sin}\:\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} }\left({a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}} \right)=\mathrm{cos}\:\frac{\pi}{\mathrm{2}^{\mathrm{2}} }\mathrm{cos}\:\frac{\pi}{\mathrm{2}^{\mathrm{3}} }…\mathrm{sin}\:\frac{\pi}{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$$$…… \\ $$$$\mathrm{2}^{{n}} \mathrm{sin}\:\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} }\left({a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}} \right)=\mathrm{sin}\:\frac{\pi}{\mathrm{2}}=\mathrm{1} \\ $$$${a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}} =\frac{\mathrm{1}}{\:\mathrm{2}^{{n}} \:\mathrm{sin}\:\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} }} \\ $$$${a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}} =\frac{\mathrm{2}}{\pi}×\frac{\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} }}{\:\mathrm{sin}\:\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} }} \\ $$$${LHS}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left({a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}} \right)=\frac{\mathrm{2}}{\:\pi} \\ $$
Commented by mathlove last updated on 01/Nov/22
thanks
$${thanks} \\ $$

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