Let $\{ a_n \}_{n=0}^\infty$ be the sequence given by $a_0 = 3$ and $$a_{n+1} = a_n – \frac{1}{a_n}, \quad n\ge 0.$$(you can easily check that the sequence is well defined).
Question: Is this sequence bounded or not?
I was only able to show that there are infinitely many $n \geq 0$ such that $a_n > 0$ and infinitely many $n \geq 0$ such that $a_n < 0$ in this way:
suppose for example that we have eventually $a_n > 0$. But then $a_n$ is eventually strictly decreasing and so, using the fact that eventually $a_n > 0$ together with the monotone convergence theorem for sequences we get that $a_n$ is convergent, which is absurd because this would imply, together with the fact that $a_{n+1} a_n = a_n^2 – 1$ for all $n$, that it's limit $L \in \mathbb{R}$ satisfies the equation $L^2 = L^2 – 1$. Using the same reasoning you can show that $a_n < 0$ eventually leads to a contradiction.
I think the boundedness question is much harder to understand because this sequence behaves in a very chaotic way.
Best Answer
$\color{brown}{\textbf{Infinity points.}}$
Easily to check that the functions $$f_n(x)=\,\underbrace {f(f(f(\dots(f(x)))))}_n,\quad\text{where}\quad f(x)=x-\frac1x=2\sinh \ln x,\quad f_0(x)=x,\tag1$$ map $\;\mathbb Q\mapsto \mathbb Q.\;$
On the other hand, there are exactly two functions $$g_\pm(x)= \dfrac{x\pm\sqrt{4+x^2}}2=\dfrac2{x\mp\sqrt{4+x^2}},\quad\text{such as}\quad f(g_\pm(x))=x,\quad\text{wherein}$$ $$ g_\pm(+\infty)=\dbinom{-0}{+\infty},\quad g_\pm(-\infty)=\dbinom{+0}{-\infty},\quad g_\pm(\pm0)=\dbinom{1}{-1},\quad g_\pm(\pm1)=\frac{\pm\sqrt5\pm1}2.$$ If $\;a_n=\pm\infty,\;$ then $$a_{n-2}\in \left(\pm\infty \bigcup \frac{\pm\sqrt5\pm1}2\right),\quad a_{n-k}=\frac{\pm\sqrt5\pm1}2\not\in\mathbb Q.$$ Therefore, $\;\forall (N)\, \forall (n\le N)\; a_n\not=\pm\infty.\;$
I.e. the given sequence does not contain infinity as a value.
$\color{brown}{\textbf{Periodic sequences.}}$
Let us define periodic sequences via the equation $\;f_T(\tilde x)=\tilde x,\;$ where $\,\tilde x\,$ is a base and $\,T\,$ i a period.
For example, $\;\dbinom {\tilde x}T=\dbinom {\sqrt2^{\,-1}}2.\;$
Rewriting the equation in the form of $\;f_{k-1}(x)=g_\pm(x)\;$ and taking in account, that $\;g_\pm(3)=\dfrac{3\pm\sqrt{13}}2\not\in\mathbb Q,\;$ easily to prove that the given sequence is not periodic.
At the same time, $$f'(x)=1+\dfrac1{x^2},\quad f'_k(x)=\prod\limits_{j=1}^{k-1} f'(f_j(x)) =\prod\limits_{j=1}^k\left(1+\dfrac1{f^2_{k-1}(x)}\right)>f'_{k-1}>1,\quad (k\in\mathbb N),$$ so should be a pole between the neighbor bases with the given period.
Therefore, $\,f_k(x)\,$ has negative infinity gaps in the poles and increasing pieces with $\,f'_k(x)>1\,$ between the poles.
If to consider the equation $\;x_{n+1}=f_k(x_n)\;$ as the simple iteration, then there are not such neighborhood of $\,x=3\,$ which can be compressed via the iteration. I.e. sufficient condition of the simple iteration convergency is not satisfied in the neighborhoods of $\;x=3.$
$\color{brown}{\textbf{Heuristic model of the process.}}$
Let us consider the model with a pair of repeated steps:
Modeling the first step via the ODE $$\dfrac{\text dx}{\text dt}=-\dfrac1x,\quad x(0)=M_k,\tag2$$ one can get $\;x=\sqrt{M^2_k-2t},\quad t_k =\dfrac12(M_k^2-1)\;\text{iterations},\;$ and with 2 additonal iterations per step we have $$N_s(M)=\dfrac12(M^2+3).\tag3$$
The average quantity of iterations in the step with $\;M_k<M,\;m_k>|g_-(M)|=\frac1{g_+(M)}\;$ equals to $$\bar N_s=\dfrac {g_+(M)}{2(g_+(M)+1)}\int\limits_{\frac1{g_+(M)}}^1 \left(\left(\dfrac1m-m\right)^2+3\right)\,\text dm =\dfrac {g_+(M)}{2(g_+(M)+1)}\left(-\dfrac1m+m+\dfrac13m^3\right)\bigg|_{\frac1{g_+(M)}}^1$$ $$ =\dfrac {g_+(M)\,m}{6(g_+(M)+1)}\left(-\dfrac4{m^2}+1+\left(\dfrac1m-m\right)^2\right)\bigg|_{\frac1{g_+(M)}}^1 =\dfrac {4g^2_+(M)-3g_+(M)-1-M^2}{6(g_+(M)+1)},$$ $$\bar N_s(M) =\dfrac{2M^2-3M+6+(4M-3)\sqrt{4+M^2}}{12(g_+(M)+1)}. \tag4$$ Assuming the average quantity of the required steps before $M_k>M$ to $\;g_+(M)+1\;$ gives $$\bar N_\Sigma(M)\approx\dfrac{2M^2-3M+6+(4M-3)\sqrt{4+M^2}}{12}\tag5$$ iterations totally.
$\color{brown}{\mathbf{Tables\;a_{0\dots49},\;a_{50\dots99},\;a_{100\dots149},\;a_{150\dots189}.}}$
$\color{brown}{\textbf{Conclusions.}}$
There are the reasons for the next hypotheses: