Question Number 146793 by KONE last updated on 15/Jul/21
$$\forall{n}\geqslant\mathrm{2},\:{u}_{{n}} =\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}^{{k}} }\right)\:{et}\:{v}_{{n}} ={u}_{{n}} \mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}^{{n}} }\right) \\ $$$${convergence},\:{nature},\:{sens}\:{of}\:{variations}\:{and}\:{adjantes}? \\ $$$${u}_{{n}} \:{and}\:{v}_{{n}} \\ $$$${help}\:{me}\:{please} \\ $$
Answered by Olaf_Thorendsen last updated on 15/Jul/21
$$\mathrm{sin2}\theta\:=\:\mathrm{2sin}\theta\mathrm{cos}\theta \\ $$$$\mathrm{cos}\theta\:=\:\frac{\mathrm{sin2}\theta}{\mathrm{2sin}\theta} \\ $$$$\Rightarrow\:\mathrm{cos}\left(\frac{\pi}{\mathrm{2}^{{k}} }\right)\:=\:\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}^{{k}−\mathrm{1}} }\right)}{\mathrm{2sin}\left(\frac{\pi}{\mathrm{2}^{{k}} }\right)} \\ $$$${u}_{{n}} \:=\:\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\:\mathrm{cos}\left(\frac{\pi}{\mathrm{2}^{{k}} }\right)\:=\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\:\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}^{{k}−\mathrm{1}} }\right)}{\mathrm{2sin}\left(\frac{\pi}{\mathrm{2}^{{k}} }\right)} \\ $$$$\left(\mathrm{telescopic}\:\mathrm{product}\right) \\ $$$${u}_{{n}} \:=\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}\right)}{\mathrm{2}^{{n}−\mathrm{1}} \mathrm{sin}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} \mathrm{sin}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)} \\ $$$${u}_{{n}} \:\sim\:\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} ×\frac{\pi}{\mathrm{2}^{{n}} }}\:=\:\frac{\mathrm{2}}{\pi} \\ $$$${v}_{{n}} \:=\:{u}_{{n}} \mathrm{sin}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\:\underset{\infty} {\rightarrow}\:\mathrm{0} \\ $$$$\frac{{u}_{{n}+\mathrm{1}} }{{u}_{{n}} }\:=\:\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)}{\mathrm{2sin}\left(\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} }\right)}\:=\:\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)}{\mathrm{4sin}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)\mathrm{cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)} \\ $$$$\frac{{u}_{{n}+\mathrm{1}} }{{u}_{{n}} }\:=\:\frac{\mathrm{1}}{\mathrm{4cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)} \\ $$$${n}\:\geqslant\:\mathrm{2}\:\Leftrightarrow\:\frac{\pi}{\mathrm{2}^{{n}} }\:\leqslant\:\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)\:\geqslant\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)}\:\leqslant\:\frac{\mathrm{1}}{\frac{\mathrm{4}}{\:\sqrt{\mathrm{2}}}}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\:<\:\mathrm{1}\:\Rightarrow\:\left({u}_{{n}} \right)\:\searrow \\ $$$$\frac{{v}_{{n}+\mathrm{1}} }{{v}_{{n}} }\:=\:\frac{\mathrm{2}^{{n}−\mathrm{1}} }{\mathrm{2}^{{n}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:<\:\mathrm{1}\:\Rightarrow\:\left({v}_{{n}} \right)\:\searrow \\ $$