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In my analysis class they let us find out if the following sequence converges or not: $$x_1=1,\quad x_2=2,\quad x_{n+2}=\frac{x_{n+1}+x_n}{x_{n+1}-x_n}$$

This sequence appears in the Bartle but the exercise does not ask to prove convergence. The whole class has had problems with proving that the sequence diverges (because according to Wolfram the sequence is divergent). We have tried to do operations of the type $x_{n+1}-x_{n+2}$ and try to prove that it converges to two different limits and thus conclude the divergence, but it doesn't seem to work

$$x_{n+1}=\frac{x_n+x_{n-1}}{x_n-x_{n-1}},\quad x_{n+2}=\frac{x_{n-1}+x_n(1+x_n-x_{n-1})}{x_{n+1}-x_n(-1+x_n-x_{n-2})}$$

where $x_{n+2}\to x$ and $\frac{x_{n-1}+x_n(1+x_n-x_{n-1})}{x_{n+1}-x_n(-1+x_n-x_{n-2})}\to \frac{2x}{2x}=1$

But it seems that attacking the problem from this point is not convincing, does anyone have any suggestions?

EDIT (Progress according to a user's proposal, I think you have good ideas)---------------

The idea is to first assume that $x_n=0$ for some $n\in\mathbb{N}$, and let $N=\min\{ n\in\mathbb{N}\;:\; x_n=0\}$, then

$\displaystyle x_{N+1}=\frac{x_N+x_{N-1}}{x_N-x_{N-1}}=\frac{x_{N-1}}{-x_{N-1}}=-1$

and

$\displaystyle x_{N+2}=\frac{x_{N+1}+x_N}{x_{N+1}-x_N}=\frac{-1}{-1}=1 $

in this way we have the sequence behaves $0, -1, 1, 0, -1, 1,...$ and diverges.

Now assuming that $x_n\neq 0$ for all $n\in\mathbb{N}$ and suppose that $(x_n)$ converges to $L$, then $\displaystyle x_{n+2}(x_{n+1}-x_n)=x_{n+1}+x_n \Rightarrow $ $L(L-L)=L+L$ and $L=0$. Now, let

$a_n=\frac{x_{n+1}}{x_n}$ notice that

$\displaystyle x_{n+2}=\frac{x_{n+1}+x_n}{x_{n+1}-x_n}=1+\frac{2x_n}{x_{n+1}-x_n}=1+\frac{2}{a_n-1}$

therefore $ \lim_{n\to\infty} \frac{2}{a_n-1}=-1$ and $\lim a_n=-1$.

This is what I have, any idea to end the demonstration or some other way?

Luis Alexandher
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    If it converges to a nonzero number then the denominator approaches 0 and the numerator does not. What happens if it approaches 0? – QC_QAOA Jul 03 '23 at 23:56
  • Hint: If the sequence converges, it is Cauchy and for any $\epsilon >0$, $\exists N\in\mathbb{N}$ s.t. $|x_{n+1}-x_n| < \epsilon$ for all $n>N$. – whpowell96 Jul 04 '23 at 00:05
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    I think that this is the way: as QC_QADA said, if it converges then the limit must be $0$. Then this yields that $\frac{x_{n+1}}{x_n}$ must converge to $-1$. This means that after some, the sign alternates. But I think it is impossible for the sign to alternate more than few times in a row. Maybe 3 times but no more. I feel like this can be checked by checking various possible ratios of $x_n$ and $x_{n+1}$. – donaastor Jul 04 '23 at 00:39
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    FYI, using an Approach0 search, there's the fairly closely related is the recursive function $(x_1+x_2)_(x_2-x_1) = x_3$ ever get to a point where it divides by $0$ if the starting conditions are $x_1 = 1, x_2= 2$. – John Omielan Jul 04 '23 at 01:46
  • @donaastor When $x_n=-1$ and $x_{n+1}=1$, we get the three-periodic sequence $(-1,1,0,-1,1,0,-1,1,0,\ldots)$ for which the sign indeed "alternates" if you allow the equality case. If you take a very small perturbation of this situation, you see that the sign can indeed alternate arbitrarily many times in a row. – Ewan Delanoy Jul 06 '23 at 18:23
  • Based on some numerical explorations in Python, I think this sequence will have a lim sup of $+\infty$ though I'm not certain. Every so often, I get a reasonably large number out of it (relative to 0). In any case, it seems to have a lot of frequency over $1$, so you may be able to show that there are infinitely many values larger than $1$. – Cameron Williams Jul 06 '23 at 18:23
  • Why is this problem difficult? There is a sequence with the same recurrence relation $x_{n+2}=\frac{x_{n+1}+x_n}{x_{n+1}-x_n}$ that converge to $0$. This fact implies that any answer that does not use the initial condition $x_1=1,\ x_2=2$ must be incorrect. – Apass.Jack Jul 06 '23 at 21:20
  • To continue the above train of thought: in general, nonlinear recurrence relations are very tricky to analyze as they are often very sensitive to initial conditions. – Cameron Williams Jul 06 '23 at 22:53
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    @Apass.Jack: I'm curious -- what is that sequence, please? – TonyK Jul 06 '23 at 23:07
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    @EwanDelanoy Thank you for replying. Unfortunately, I believe there was a misunderstanding. My concept of alternating sign comes from the ratio $\frac{x_{n+1}}{x_n}$ converging to $-1$. For this to make sense, the only option is not to allow the equality case. Anyway, after giving some thought, I too believe that the number of alternations is not globally bounded, but I still believe that it cannot be infinite. – donaastor Jul 08 '23 at 21:11

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