Question-129825

Question Number 129825 by 0731619177 last updated on 19/Jan/21

Answered by Olaf last updated on 20/Jan/21

Let u_{0} = 0 and u_{n} = \sqrt{2 + \sqrt{2 + \sqrt{2 \dots}}}, n ⩾ 1

u_{n + 1} = \sqrt{2 + u_{n}}

By induction :

0 ⩽ u_{n} ⩽ 2 \Rightarrow 0 ⩽ u_{n + 1} ⩽ 2

So let u_{n} = 2 \cos θ_{n} (u_{0} = 0 \Rightarrow θ_{0} = \frac{π}{2})

u_{n + 1} = \sqrt{2 + 2 \cos θ_{n}} = 2 \sqrt{\frac{1 + \cos θ_{n}}{2}}

u_{n + 1} = 2 \sqrt{\cos^{2} \frac{θ_{n}}{2}} = 2 \cos \frac{θ_{n}}{2} = 2 \cos θ_{n + 1}

\Rightarrow θ_{n + 1} = \frac{θ_{n}}{2}

Finally, θ_{n} = \frac{θ_{0}}{2^{n}} = \frac{π}{2^{n + 1}}, u_{n} = 2 \cos \frac{π}{2^{n + 1}}

\sin 2 α = 2 \sin α \cos α

\cos α = \frac{\sin 2 α}{2 \sin α}

\Rightarrow u_{n} = 2 \cos θ_{n} = \frac{\sin 2 θ_{n}}{\sin θ_{n}} = \frac{\sin \frac{π}{2^{n}}}{\sin \frac{π}{2^{n + 1}}}

\underset{n = 1}{\prod^{N}} \frac{2}{u_{n}} = 2^{N} \underset{n = 1}{\prod^{N}} \frac{\sin \frac{π}{2^{n + 1}}}{\sin \frac{π}{2^{n}}} = 2^{N} \frac{\sin \frac{π}{2^{N + 1}}}{\sin \frac{π}{2}}

(telescopic product)

\underset{n = 1}{\prod^{N}} \frac{2}{u_{n}} = 2^{N} \sin \frac{π}{2^{N + 1}} \underset{\infty}{\sim} 2^{N} \frac{π}{2^{N + 1}} = \frac{π}{2} (1)

3!! = 6! = 720 (2)

\sqrt{Γ (2)} = \sqrt{1!} = 1 (3)

With (1), (2) and (3) the result is :

{\ln {[- (\frac{π}{2})]}^{720}}^{1} = 720 . \ln \frac{π}{2}

Commented by mr W last updated on 20/Jan/21

3!! \neq (3!)!

3!! = 3 \times 1 = 3

Commented by Olaf last updated on 20/Jan/21

Sir, why 3!! = 3 \times 1 ?

{It}^{'} s the first time I see this notation .

Commented by Olaf last updated on 20/Jan/21

Dear mr W, my own calculator

gives : 3!! = (3!)! = 720

Commented by Olaf last updated on 20/Jan/21

Commented by mr W last updated on 20/Jan/21

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