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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})