∀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


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

$\forall n ⩾ 2, u_{n} = \underset{k = 2}{\prod^{n}} \cos (\frac{π}{2^{k}}) e t v_{n} = u_{n} \sin (\frac{π}{2^{n}})$ $c o n v e r g e n c e, n a t u r e, s e n s o f v a r i a t i o n s a n d a d j a n t e s ?$ $u_{n} a n d v_{n}$ $h e l p m e p l e a s e$
	Answered by Olaf_Thorendsen last updated on 15/Jul/21

	$\sin 2 θ = 2 \sin θ \cos θ$ $\cos θ = \frac{\sin 2 θ}{2 \sin θ}$ $\Rightarrow \cos (\frac{π}{2^{k}}) = \frac{\sin (\frac{π}{2^{k - 1}})}{2 \sin (\frac{π}{2^{k}})}$ $u_{n} = \underset{k = 2}{\prod^{n}} \cos (\frac{π}{2^{k}}) = \underset{k = 2}{\prod^{n}} \frac{\sin (\frac{π}{2^{k - 1}})}{2 \sin (\frac{π}{2^{k}})}$ $(telescopic product)$ $u_{n} = \frac{\sin (\frac{π}{2})}{2^{n - 1} \sin (\frac{π}{2^{n}})} = \frac{1}{2^{n - 1} \sin (\frac{π}{2^{n}})}$ $u_{n} \sim \frac{1}{2^{n - 1} \times \frac{π}{2^{n}}} = \frac{2}{π}$ $v_{n} = u_{n} \sin (\frac{π}{2^{n}}) = \frac{1}{2^{n - 1}} \underset{\infty}{\to} 0$ $\frac{u_{n + 1}}{u_{n}} = \frac{\sin (\frac{π}{2^{n}})}{2 \sin (\frac{π}{2^{n + 1}})} = \frac{\sin (\frac{π}{2^{n}})}{4 \sin (\frac{π}{2^{n}}) \cos (\frac{π}{2^{n}})}$ $\frac{u_{n + 1}}{u_{n}} = \frac{1}{4 \cos (\frac{π}{2^{n}})}$ $n ⩾ 2 \Leftrightarrow \frac{π}{2^{n}} ⩽ \frac{π}{4}$ $\cos (\frac{π}{2^{n}}) ⩾ \frac{1}{\sqrt{2}}$ $\frac{1}{4 \cos (\frac{π}{2^{n}})} ⩽ \frac{1}{\frac{4}{\sqrt{2}}} = \frac{\sqrt{2}}{4} < 1 \Rightarrow (u_{n}) ↘$ $\frac{v_{n + 1}}{v_{n}} = \frac{2^{n - 1}}{2^{n}} = \frac{1}{2} < 1 \Rightarrow (v_{n}) ↘$
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