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n-2-u-n-k-2-n-cos-pi-2-k-et-v-n-u-n-sin-pi-2-n-convergence-nature-sens-of-variations-and-adjantes-u-n-and-v-n-help-me-please-




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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