if $\lim \limits_{n \to \infty}a_n a_{n+1}=1$ and $0<a_n$ Prove $\lim \limits_{j \to \infty}a_{n_j}=L < 1$ exists.

Question

It's also given that $a_n$ has two different positive sub limits.

My try :

Given $\forall \varepsilon >0:n>N:N \in \mathbb{N} :|a_n a_{n+1}-1|<\frac{\varepsilon}{2}$

$ \Rightarrow \frac{1}{a_{n+1}}-\frac{\varepsilon}{2}\leq a_n\leq \frac{1}{a_{n+1}}+\frac{\varepsilon}{2}$ $Hence \ a_n \ is \ bounded$

We'll wrongfully assume $\lim \limits_{n \to \infty}a_n=\infty$ then $\lim \limits_{n \to \infty}a_{n+1}=0$ which would mean there's no Two Different Sublimits , Therefore $\lim \limits_{n \to \infty}a_n\neq\infty$

Hence because $\frac{1-{\frac{\varepsilon}{2}}}{a_{n+1}\geq 0 \neq\infty}\leq a_n\leq \frac{1+{\frac{\varepsilon}{2}}}{a_{n+1}\geq 0 \neq\infty}$ , for every $n \in \mathbb{N}$ we'll get a bounded $a_n$ :

$\frac{1}{q}-\frac{\varepsilon}{2} \leq a_n \leq \frac{1}{q}+\frac{\varepsilon}{2} $ ; $q \in \mathbb{R} $

According to bolzano weierstrass theorem , every bounded sequence has a sub-sequence that converge to final limit $L \in \mathbb{R}$ Therefore ,we'll build the sub sequence $a_{n_j}$ as :

$\frac{1}{q}-\frac{\varepsilon}{2} \leq a_{n_j} \leq \frac{1}{q}+\frac{\varepsilon}{2} $ ; $q \in \mathbb{R} $

therefore assuming $n\rightarrow \infty $ would result $q \in \mathbb{R}$

$ \Rightarrow \lim \limits_{n \to \infty}a_{n_j}=\frac{1}{q}<1$

• I know I should've mentioned that if $q \in [0,1]$ then the answer is direct from that
• I also know I should've probably wrote $q \in \mathbb{R} >0 $ , I didn't bother because the question is given where $a_n$ is positive

Answer 1

You have a sequence $(a_n)$ with the following properties:

• $a_n > 0$ for all $n$.
• $\lim_{n \to \infty}a_n a_{n+1}=1$.
• $(a_n)$ has two different positive subsequential limits.

The goal is to show that there is a subsequence $(a_{n_j})$ with $\lim_{j \to \infty}a_{n_j}=L$ and $0 < L < 1$.

Your proof does not work because the given conditions do not imply that $(a_n)$ is bounded.

$(a_n)$ has two different positive subsequential limits, and one of them must be different from $1$. So there is a subsequence $(a_{n_j})$ with $\lim_{j \to \infty}a_{n_j}=L$ and $L > 0$, $L \ne 1$.

If $0 < L < 1$ then you are done. Otherwise $L > 1$, and then $$ a_{n_j + 1} = \frac{a_{n_j} a_{n_j+1}}{a_{n_j}} \to \frac 1 L < 1 $$ does the job.

Addendum: The sequence $$ \begin{align} x_n =\; & 0, 1, \\ & 0, \frac 12, \frac 22, \frac 32, 2, \frac 32, \frac 22, \frac 12, \\ & 0, \frac 13, \frac 23, \ldots, \frac 83, 3, \frac 83, \ldots, \frac 13, \\ & 0, \frac 14, \ldots \, . \end{align} $$ (taken from here and inspired by this) is unbounded with $\lim_{n\to \infty} (x_{n+1} - x_n) = 0$, and every non-negative real number is a subsequential limit of $(x_n)$. Then $$ a_n = e^{x_0},e^{-x_0}, e^{x_1},e^{-x_1}, e^{x_2},e^{-x_2}, e^{x_3},e^{-x_3}, \ldots $$ is an unbounded sequence of non-negative real numbers with $\lim_{n \to \infty}a_n a_{n+1}=1$. Every non-negative real number is a subsequential limit of $(a_n)$.

Answer 2

For all $k \in \Bbb N $ , build a finite sequence $A_k$ :

• $A_k = 1,1,1+\frac{1}{k} , \frac{1}{1+\frac{1}{k}} , \ldots , 1+\frac{i}{k} , \frac{1}{1+\frac{i}{k}} , \ldots , 2,\frac{1}{2} , 2-\frac{1}{k} , \ldots 2-\frac{i}{k} , \frac{1}{2-\frac{i}{k}} , \ldots , 1 , 1$

I assumed here that $i = (1,\cdots , k)$. furthermore lets build a sequence ${a_n}$ with $A_k$ s.t $k= 1,2,\ldots$.

In the sequence ${a_n}$ there exists infinitely many $1 , \frac{1}{2} , 2$ , so there is at least 3 subsequent limits (one of them is $\frac{1}{2}$).

Lets proof that $\lim_{n\to \infty} a_na_{n+1} = 1 $ .

If we consider $a_na_{n+1}$ they will be equal to $i \text{ or } k \text{ or }(\frac{1}{1+\frac{i}{k}})(1+\frac{i+1}{k}) = 1+\frac{1}{k+i}$ , for ($0 \leqslant i\leqslant k $). let $\epsilon >0$ , so for $k\geqslant \lceil \frac{1}{\epsilon} \rceil \text{ and for all } 0 \leqslant i\leqslant k \text{. } \Rightarrow |a_na_{n+1} -1|<\epsilon \Rightarrow \lim_{n\to \infty} a_na_{n+1} = 1 $