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The sequence $(x_n)$ is defined by the formula: $$\left\{\begin{array}{cc} x_1=1, x_2=\sqrt{3}\\ x_{n+2}=\frac{x_{n+1}\sqrt{x_n^2+1}+x_{n}\sqrt{x_{n+1}^2+1}-x_n-x_{n+1}}{x_{n+1}x_n-\big(\sqrt{x_{n+1}^2+1}-1\big)\big(\sqrt{x_{n}^2+1}-1\big)}, \quad n=1,2,3\dots \end{array} \right.$$ Find $\lim\limits_{n\to \infty}x_n$.

I see: $$\begin{array}{rl} x_{n+2}&=\frac{x_{n+1}\sqrt{x_n^2+1}+x_{n}\sqrt{x_{n+1}^2+1}-x_n-x_{n+1}}{x_{n+1}x_n-\big(\sqrt{x_{n+1}^2+1}-1\big)\big(\sqrt{x_{n}^2+1}-1\big)}\\ &=\frac{x_{n}\big(\sqrt{x_{n+1}^2+1}-1 \big)+x_{n+1}\big(\sqrt{x_{n}^2+1}-1 \big)}{x_{n+1}x_n-\frac{x_n^2x_{n+1}^2}{\big(\sqrt{x_{n+1}^2+1}+1\big)\big(\sqrt{x_{n}^2+1}+1\big)}}\\ &=\frac{x_{n+1}\big(\sqrt{x_{n}^2+1}+1\big)+x_{n}\big(\sqrt{x_{n+1}^2+1}+1\big)}{\big(\sqrt{x_{n+1}^2+1}+1\big)\big(\sqrt{x_{n}^2+1}+1\big)-x_nx_{n+1}} \end{array}$$ That's all I can do. So, I hope hints from you. Thank you.

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  • $\begingroup$ Where does the problem come form? Homework, a contest, ...? – I would guess that the substitution $x_n = \sinh(y_n)$ simplifies things ... $\endgroup$
    – Martin R
    Aug 30 '21 at 7:20
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    $\begingroup$ Hello :) If $(x_n)_n$ converges, let $x$ be its limit. We have $x_{n+2}=f(x_n,x_{n+1})$, where the RHS $f$ is continuous. Hence, $x=f(x,x)$. Maybe, you can solve this equation. $\endgroup$
    – Jochen
    Aug 30 '21 at 7:21
  • $\begingroup$ @Jochen: I know that, but how to prove $(x_n)$ is converges. That's I don't know. $\endgroup$
    – MrCR
    Aug 30 '21 at 7:23
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    $\begingroup$ I found that $x_n = \tan \theta_n$ yields $\tan \theta_{n+2} = \tan\frac{\theta_n + \theta_{n+1}}{2}$. Calculation is somewhat tedious so I'm not sure, but it would solve the problem. $\endgroup$
    – dust05
    Aug 30 '21 at 7:34
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    $\begingroup$ Intuition for this substitution is from terms $\sqrt{1+x_n^2}$; Note that $1+ \tan^2{\theta} = \sec^2\theta$. @MartinR 's comment would be from the similar observation and $1+ \sinh^2{y} = \cosh^2{y}$. $\endgroup$
    – dust05
    Aug 30 '21 at 7:37
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By rationalising the denominator, $x_{n+2}$ simplifies to $$\frac{\sqrt{x_n^2+1}\sqrt{x_{n+1}^2+1}+x_nx_{n+1}-1}{x_n+x_{n+1}}.$$

Letting $$x_n=\frac12\left(p_n-\frac1{p_n}\right),$$ that is (say) $$p_n=x_n+\sqrt{x_n^2+1}$$ and simplifying further, you get $$x_{n+2}=\frac{p_np_{n+1}-1}{p_n+p_{n+1}}$$

So, if $p_n=\cot\theta_n$, then $$x_{n+2}=\cot(\theta_n+\theta_{n+1}).$$

Furthermore, $$x_n=\frac12\left(\cot\theta_n-\frac1{\cot\theta_n}\right)=\cot{2\theta_n}.$$

It follows that $2\theta_{n+2}=\theta_n+\theta_{n+1},$ so that you can solve for $\theta_n$ explicitly from the initial conditions. The solution is $$\frac{2\theta_2+\theta_1}{3}+\frac{4\theta_2-4\theta_1}{3}\left(-\frac12\right)^n,$$ which has limit $$\frac{2\theta_2+\theta_1}{3}$$.

You know that $\cot2\theta_1=1$ and $\cot2\theta_2=\sqrt{3}$, so that $\theta_1=\frac\pi8$ and $\theta_2=\frac\pi{12}$. Therefore, the limit of $\theta_n$ is $\frac{7\pi}{72}$, and so the limit of $x_n$ is $\cot\frac{7\pi}{36}$.

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