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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.