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

My try , which I got 0/15 on from the exam :

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
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    $\begingroup$ Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title. $\endgroup$
    – Martin R
    Apr 8 at 11:17
  • $\begingroup$ If $1 < a_n$ for all $n$ then it is not possible that $\lim \limits_{j \to \infty}a_{n_j}=L < 1$. $\endgroup$
    – Martin R
    Apr 8 at 11:19
  • $\begingroup$ $a_n>0$ and they mean 0<L<1 $\endgroup$ Apr 8 at 11:21
  • $\begingroup$ But is says $1<a_n$ in the title. $\endgroup$
    – saulspatz
    Apr 8 at 11:32
  • $\begingroup$ My apologies it should be $a_n > 0 $ , I would love some help regardless $\endgroup$ Apr 8 at 11:38
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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 $(a_n)$.

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  • $\begingroup$ Great Answer , May I ask why $a_n$ is not bounded? $\endgroup$ Apr 8 at 12:04
  • $\begingroup$ @LoaiNakhly: I did not say that it is unbounded. I said that it may be unbounded. $\endgroup$
    – Martin R
    Apr 8 at 12:05
  • $\begingroup$ But I proved that $a_{n+1} -> \infty$ would lead to $a_n -> 0$, which would be wrong $\endgroup$ Apr 8 at 12:12
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    $\begingroup$ @LoaiNakhly: $(a_n)$ can have a subsequence converging to infinity. I have added an example. $\endgroup$
    – Martin R
    Apr 8 at 12:13
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    $\begingroup$ @LoaiNakhly: My example sequence has $2$ and $1/2$ as positive subsequence limits, and some more. If you meant that the sequence has exactly two different subsequence limits then you should say so in the question. That would make it only simpler. $\endgroup$
    – Martin R
    Apr 8 at 12:56
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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 $

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