# The sequence $(x_n)$ : $x_{n+2}=\frac{x_{n+1}\sqrt{x_n^2+1}+x_{n}\sqrt{x_{n+1}^2+1}-x_n-x_{n+1}}{x_{n+1}x_n-(\sqrt{x_{n+1}^2+1}-1)(\sqrt{x_n^2+1}-1)}$

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.

• Where does the problem come form? Homework, a contest, ...? – I would guess that the substitution $x_n = \sinh(y_n)$ simplifies things ... Aug 30 '21 at 7:20
• 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. Aug 30 '21 at 7:21
• @Jochen: I know that, but how to prove $(x_n)$ is converges. That's I don't know.
– MrCR
Aug 30 '21 at 7:23
• 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. Aug 30 '21 at 7:34
• 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 $1+ \sinh^2{y} = \cosh^2{y}$. Aug 30 '21 at 7:37

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 $$\cot\frac{7\pi}{36}$$.