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Question Number 146793 by KONE last updated on 15/Jul/21

∀n≥2, u_n =Π_(k=2) ^n cos ((π/2^k )) et v_n =u_n sin ((π/2^n )) convergence, nature, sens of variations and adjantes? u_n and v_n help me please

$$\forall{n}\geqslant\mathrm{2},\:{u}_{{n}} =\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}^{{k}} }\right)\:{et}\:{v}_{{n}} ={u}_{{n}} \mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}^{{n}} }\right) \\ $$$${convergence},\:{nature},\:{sens}\:{of}\:{variations}\:{and}\:{adjantes}? \\ $$$${u}_{{n}} \:{and}\:{v}_{{n}} \\ $$$${help}\:{me}\:{please} \\ $$

Answered by Olaf_Thorendsen last updated on 15/Jul/21

sin2θ = 2sinθcosθ cosθ = ((sin2θ)/(2sinθ)) ⇒ cos((π/2^k )) = ((sin((π/2^(k−1) )))/(2sin((π/2^k )))) u_n = Π_(k=2) ^n cos((π/2^k )) =Π_(k=2) ^n ((sin((π/2^(k−1) )))/(2sin((π/2^k )))) (telescopic product) u_n =((sin((π/2)))/(2^(n−1) sin((π/2^n )))) = (1/(2^(n−1) sin((π/2^n )))) u_n ∼ (1/(2^(n−1) ×(π/2^n ))) = (2/π) v_n = u_n sin((π/2^n )) = (1/2^(n−1) ) →_∞ 0 (u_(n+1) /u_n ) = ((sin((π/2^n )))/(2sin((π/2^(n+1) )))) = ((sin((π/2^n )))/(4sin((π/2^n ))cos((π/2^n )))) (u_(n+1) /u_n ) = (1/(4cos((π/2^n )))) n ≥ 2 ⇔ (π/2^n ) ≤ (π/4) cos((π/2^n )) ≥ (1/( (√2))) (1/(4cos((π/2^n )))) ≤ (1/(4/( (√2)))) = ((√2)/4) < 1 ⇒ (u_n ) ↘ (v_(n+1) /v_n ) = (2^(n−1) /2^n ) = (1/2) < 1 ⇒ (v_n ) ↘

$$\mathrm{sin2}\theta\:=\:\mathrm{2sin}\theta\mathrm{cos}\theta \\ $$$$\mathrm{cos}\theta\:=\:\frac{\mathrm{sin2}\theta}{\mathrm{2sin}\theta} \\ $$$$\Rightarrow\:\mathrm{cos}\left(\frac{\pi}{\mathrm{2}^{{k}} }\right)\:=\:\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}^{{k}−\mathrm{1}} }\right)}{\mathrm{2sin}\left(\frac{\pi}{\mathrm{2}^{{k}} }\right)} \\ $$$${u}_{{n}} \:=\:\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\:\mathrm{cos}\left(\frac{\pi}{\mathrm{2}^{{k}} }\right)\:=\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\:\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}^{{k}−\mathrm{1}} }\right)}{\mathrm{2sin}\left(\frac{\pi}{\mathrm{2}^{{k}} }\right)} \\ $$$$\left(\mathrm{telescopic}\:\mathrm{product}\right) \\ $$$${u}_{{n}} \:=\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}\right)}{\mathrm{2}^{{n}−\mathrm{1}} \mathrm{sin}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} \mathrm{sin}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)} \\ $$$${u}_{{n}} \:\sim\:\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} ×\frac{\pi}{\mathrm{2}^{{n}} }}\:=\:\frac{\mathrm{2}}{\pi} \\ $$$${v}_{{n}} \:=\:{u}_{{n}} \mathrm{sin}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\:\underset{\infty} {\rightarrow}\:\mathrm{0} \\ $$$$\frac{{u}_{{n}+\mathrm{1}} }{{u}_{{n}} }\:=\:\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)}{\mathrm{2sin}\left(\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} }\right)}\:=\:\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)}{\mathrm{4sin}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)\mathrm{cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)} \\ $$$$\frac{{u}_{{n}+\mathrm{1}} }{{u}_{{n}} }\:=\:\frac{\mathrm{1}}{\mathrm{4cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)} \\ $$$${n}\:\geqslant\:\mathrm{2}\:\Leftrightarrow\:\frac{\pi}{\mathrm{2}^{{n}} }\:\leqslant\:\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)\:\geqslant\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)}\:\leqslant\:\frac{\mathrm{1}}{\frac{\mathrm{4}}{\:\sqrt{\mathrm{2}}}}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\:<\:\mathrm{1}\:\Rightarrow\:\left({u}_{{n}} \right)\:\searrow \\ $$$$\frac{{v}_{{n}+\mathrm{1}} }{{v}_{{n}} }\:=\:\frac{\mathrm{2}^{{n}−\mathrm{1}} }{\mathrm{2}^{{n}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:<\:\mathrm{1}\:\Rightarrow\:\left({v}_{{n}} \right)\:\searrow \\ $$

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