Find $\displaystyle \lim_{n\to\infty}2^n\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\dots+\sqrt2}}}}_{n \textrm{ square roots}}$.
By geometry method, I know that this is $\pi$. But is there algebraic method to find this ?
Thank you.
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2What's the geometry method? Circumference approximated by a $2^n-$gon? – Calvin Lin Jun 24 '13 at 05:44
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You're right. ${}$ – JSCB Jun 24 '13 at 05:45
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1That would suggest that you can use $\lim \frac{\sin x } {x} = 1 = \lim \frac{\tan x} { x} $, with the same working as the circumference method. – Calvin Lin Jun 24 '13 at 05:47
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4You can check the expression is in fact $2^{n+1} \sin \frac{\pi}{2^{n+1}}$ (by induction using trigonometric formulae) and compute the limit that way. Would that count as an algebraic method ? – Joel Cohen Jun 24 '13 at 05:56
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2@JoelCohen Post that as an answer? – Steven Stadnicki Jun 24 '13 at 06:45
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2Possible duplicate of Why is this series of square root of twos equal $\pi$? – chx Apr 04 '17 at 13:20
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Let $2=4\cos^{2}{\theta}$ we get $$\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\dots+\sqrt2}}}}_{n \textrm{ square roots}}=4\left|\sin{\frac{\theta}{2^n}}\right|$$
And as $n \longrightarrow \infty$ ,$\frac{\theta}{2^n} \longrightarrow 0$. Hence $\displaystyle \lim_{n \to \infty}2^n\sin{\frac{\theta}{2^n}}=\theta$
$\text{Limit}=|4\theta|$
As $\cos{\theta}=\pm \frac{1}{\sqrt{2}}$ we have $\theta=\pm \frac{\pi}{4}+n\pi$ $\quad ,n \in \mathrm{I}$
Here we take $\theta =\pm \frac{\pi}{4}$ (To know why please refer the note at the bottom) .
Hence the value of limit is $\pi$
Note: the reason why $\theta=\pm \frac{\pi}{4}$ only.
We know that by the definition of square-root function $\sqrt{x^2}=|x|$ using the same reasoning. We have $\sqrt{2+2\cos{x}}=2\cos{\frac{x}{2}}$ iff $\cos{\frac{x}{2}}\ge 0 \Rightarrow \frac{x}{2}\in \left[2n\pi,\frac{\left(4n+1\right)\pi}{2}\right] \cup \left[\frac{\left(4n-1\right)\pi}{2},2n\pi\right] ,n\in \mathbb{I}$.
As we applied the process for $n$ times for every whole number $j \le n$ we need to have the following inequality $\cos{\frac{x}{2^j}}\ge 0 \Rightarrow \frac{x}{2^j}\in \left[2N\pi,\frac{\left(4N+1\right)\pi}{2}\right] \cup \left[\frac{\left(4N-1\right)\pi}{2},2N\pi\right] ,N\in \mathbb{I}$.
Now it is easy to reason out that our $\theta$ must belong to the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$
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@Kaster. Sorry I finally reasoned out why should omit the rest of the values other than $\frac{\pi}{4}$. I will post it in detail now. – hrkrshnn Jun 24 '13 at 08:09
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1In fact, you could've taken $\theta = \frac \pi 4$ explicitly. All your calculations would hold anyway. – Kaster Jun 24 '13 at 09:06
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@Kaster, yes you are right. But to some extend that raises a doubt in a reader (including me) why I tried $\frac{\pi}{4}$ and why not $\frac{9\pi}{4}$. – hrkrshnn Jun 24 '13 at 11:18