prove-that-lim-n-1-2-1-2-1-2-1-2-1-2-1-2-1-2-1-2-1-2-1-2-1-2-1-2-1-2-1-2-1-2-1-2-n-term-2-pi-

Question Number 179657 by mathlove last updated on 31/Oct/22

p r o v e t h a t

\lim_{n \to \infty} (\sqrt{\frac{1}{2} \times \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}} \times \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}} \times \cdot \cdot \cdot \cdot \cdot \underset{n t e r m}{\underset{⏟}{\sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2 +} \dots . . + \frac{1}{2} \sqrt{\frac{1}{2}}}}}}}) = \frac{2}{π}

Commented by mr W last updated on 01/Nov/22

t h e q u e s t i o n s h o u l d b e :

\lim_{n \to \infty} (\sqrt{\frac{1}{2}} \times \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}} \times \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}} \times \cdot \cdot \cdot \cdot \cdot \underset{n t e r m}{\underset{⏟}{\sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2 +} \dots . . + \frac{1}{2} \sqrt{\frac{1}{2}}}}}}}) = \frac{2}{π}

Commented by mathlove last updated on 02/Nov/22

y e s s i r

Answered by mr W last updated on 01/Nov/22

a_{1} = \sqrt{\frac{1}{2}}

a_{2} = \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}

\dots

a_{n} = \sqrt{\frac{1}{2} + \frac{1}{2} a_{n - 1}}

2 a_{n}^{2} - 1 = a_{n - 1}

s a y a_{n} = \cos θ_{n}

{2 \cos}^{2} θ_{n} - 1 = \cos θ_{n - 1}

\cos 2 θ_{n} = \cos θ_{n - 1}

2 θ_{n} = θ_{n - 1}

θ_{n} = \frac{θ_{n - 1}}{2} = \frac{θ_{n - 2}}{2^{2}} = \dots = \frac{θ_{1}}{2^{n - 1}}

a_{1} = \cos θ_{1} = \sqrt{\frac{1}{2}} \Rightarrow θ_{1} = \frac{π}{4} = \frac{π}{2^{2}}

θ_{n} = \frac{θ_{1}}{2^{n - 1}} = \frac{π}{2^{n + 1}}

a_{n} = \cos θ_{n} = \cos \frac{π}{2^{n + 1}}

a_{1} a_{2} \dots a_{n} = \cos \frac{π}{2^{2}} \cos \frac{π}{2^{3}} \dots \cos \frac{π}{2^{n + 1}}

2 \sin \frac{π}{2^{n + 1}} (a_{1} a_{2} \dots a_{n}) = \cos \frac{π}{2^{2}} \cos \frac{π}{2^{3}} \dots \sin \frac{π}{2^{n}}

2^{2} \sin \frac{π}{2^{n + 1}} (a_{1} a_{2} \dots a_{n}) = \cos \frac{π}{2^{2}} \cos \frac{π}{2^{3}} \dots \sin \frac{π}{2^{n - 1}}

\dots \dots

2^{n} \sin \frac{π}{2^{n + 1}} (a_{1} a_{2} \dots a_{n}) = \sin \frac{π}{2} = 1

a_{1} a_{2} \dots a_{n} = \frac{1}{2^{n} \sin \frac{π}{2^{n + 1}}}

a_{1} a_{2} \dots a_{n} = \frac{2}{π} \times \frac{\frac{π}{2^{n + 1}}}{\sin \frac{π}{2^{n + 1}}}

L H S = \lim_{n \to \infty} (a_{1} a_{2} \dots a_{n}) = \frac{2}{π}

Commented by mathlove last updated on 01/Nov/22

t h a n k s