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Also Given

Suppose $x_1=tan^{-1}2>x_2>x_3>......$ are positive real numbers

I cannot understand as to how I should approach but I have a hunch which is a bit hand-wavy way to prove it but tried to show that

$lim_{n \to \infty}x_n =\dfrac{\pi}{4}$ (For the proof see my edits in @hypernova's answer)

But can anybody prove it and then use it to find $cot \space x_n$ please?

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Note that $$ \sin\left(x_{n+1}-x_n\right)=\sin x_{n+1}\cos x_n-\cos x_{n+1}\sin x_n. $$ The iterative scheme thus reads $$ \sin x_{n+1}\cos x_n-\cos x_{n+1}\sin x_n+2^{-n-1}\sin x_{n+1}\sin x_n=0. $$ Thanks to the given condition that $x_j\in\left(0,\arctan 2\right]\subseteq\left(0,\pi\right)$, we have $\sin x_j\ne 0$. Hence the above scheme is equivalent to $$ \cot x_n-\cot x_{n+1}+2^{-n-1}=0, $$ or in a more friendly way, $$ \cot x_{n+1}-\cot x_n=\frac{1}{2^{n+1}}. $$ Therefore, \begin{align} \cot x_n&=\cot x_1+\sum_{j=2}^n\left(\cot x_j-\cot x_{j-1}\right)\\ &=\cot x_1+\sum_{j=2}^n\frac{1}{2^j}\\ &=\cot x_1+\frac{2^n-2}{2^{n+1}}\\ &=\cot\left(\arctan 2\right)+\frac{2^n-2}{2^{n+1}}\\ &=\frac{1}{2}+\frac{2^n-2}{2^{n+1}}\\ &=1-\frac{1}{2^n}. \end{align}

For the proof of $lim_{n\to \infty}(x_n)=\dfrac{\pi}{4}$ which is infact a piece of cake now,

Using the above result we get,

$tan(x_n)=\dfrac{1}{1-\dfrac{1}{2^n}}=\dfrac{2^n}{2^n-1}$

$x_n= tan^{-1}\bigg(\dfrac{2^n}{2^n-1}\bigg)$

Now $lim_{n\to \infty}(x_n)=lim_{n\to \infty}\bigg(tan^{-1}\bigg(\dfrac{2^n}{2^n-1}\bigg)\bigg)$

$\implies \space lim_{n\to \infty}(x_n)=tan^{-1}(1)=\dfrac{\pi}{4}$ (Proved)

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  • $\begingroup$ so easy yet it slipped my eyes. Why????I was complicating it as usual. $\endgroup$
    – Saradamani
    Apr 27 '18 at 8:16
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    $\begingroup$ @Saradamani: I was scared at the first glimpse as well. Then tried with what I know and happened to find it not as complicated as imagined... $\endgroup$
    – hypernova
    Apr 27 '18 at 8:18
  • $\begingroup$ I wanted to give you a thousand upvotes @hypernova but there is provision for 1 only. But consider from my end a thousand upvotes. These are the cutest and prettiest solutions to those mind boggling problem which people fear the most. $\endgroup$
    – Saradamani
    Apr 27 '18 at 8:25
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    $\begingroup$ Wow, thank you @Saradamani. It is so beyond the generous of you! I have absolutely no doubt that, for both of us, and for all of us, it is that we share with each other what we know and what we have doubt about that bring us here. We benefit from this, and we enjoy this :-) $\endgroup$
    – hypernova
    Apr 27 '18 at 8:34
  • $\begingroup$ @Saradamani: Thank you very much for your suggestion! This last part is crucial :-) $\endgroup$
    – hypernova
    Apr 27 '18 at 8:41

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