if $\lim \limits_{n \to \infty}a_n a_{n+1}=1$ and $0

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

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. — Martin R, Apr 08 '21 at 11:17
If $1 < a_n$ for all $n$ then it is not possible that $\lim \limits_{j \to \infty}a_{n_j}=L < 1$. — Martin R, Apr 08 '21 at 11:19
My apologies it should be $a_n > 0 $ , I would love some help regardless — Loai Nakhly, Apr 08 '21 at 11:38
+1 for showing your work. New users need to take note of this. — Paramanand Singh, Apr 08 '21 at 12:15
What are you trying to prove? You obviously can't prove $\lim a_{n_j}$ exists, since you're not given any information about $n_j$. My guess is you want to prove that there exists a subsequence $a{n_j}$_ such that etc. (If so you should say so! Really - you need to at least state the problem correctly...) — David C. Ullrich, Apr 08 '21 at 12:37

Martin R · Answer 1 · 2021-04-08T16:44:10.437

4

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 $(a_n)$.

edited Apr 08 '21 at 16:44

answered Apr 08 '21 at 12:00

Martin R

113,040

Great Answer , May I ask why $a_n$ is not bounded? – Loai Nakhly Apr 08 '21 at 12:04
@LoaiNakhly: I did not say that it is unbounded. I said that it may be unbounded. – Martin R Apr 08 '21 at 12:05
But I proved that $a_{n+1} -> \infty$ would lead to $a_n -> 0$, which would be wrong – Loai Nakhly Apr 08 '21 at 12:12
I was hinting something similar in comments but then saw your answer. +1 – Paramanand Singh Apr 08 '21 at 12:13
1

@LoaiNakhly: $(a_n)$ can have a subsequence converging to infinity. I have added an example. – Martin R Apr 08 '21 at 12:13
Still , wouldn’t my answer work even for $n -> \infty$ as ε>0 ? – Loai Nakhly Apr 08 '21 at 12:20
@LoaiNakhly: Sorry, I don't get what you mean. It is correct that $\lim_{n \to \infty} a_n = \infty$ is not possible. But it can be that $\lim_{k \to \infty} a_{n_k} = \infty$ for some subsequence, as the example shows. – Martin R Apr 08 '21 at 12:29
Your example doesn’t work , your sequence should have two different subsequence limits, 0 is not positive. – Loai Nakhly Apr 08 '21 at 12:53
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@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. – Martin R Apr 08 '21 at 12:56
But the limit becomes undetermined $\lim \limits_{n \to \infty}a_n a_{n+1}$, it could mean both n/n and 3/2 or 4/3 .. etc – Loai Nakhly Apr 08 '21 at 13:12
@LoaiNakhly: You are right, the example does not work yet ... – Martin R Apr 08 '21 at 13:15
So $a_n$ is bounded ? – Loai Nakhly Apr 08 '21 at 13:59
@LoaiNakhly: I have updated the example. Hopefully it is correct now. – Martin R Apr 08 '21 at 16:44
Appreciate the response but it's still wrong, even ignoring the $0*\frac{1}{n}=0\neq 1$ , Also $\Rightarrow \lim_{n \to \infty}a_n a_{n+1}= \lim_{n \to \infty}\frac{n^2-1}{n}\cdot n\neq 1$ – Loai Nakhly Apr 08 '21 at 19:32
@LoaiNakhly Did you mix up $x_n$ and $a_n$ in the example? – Martin R Apr 08 '21 at 19:42

score 0 · Answer 2 · answered Apr 08 '21 at 12:52

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 $

if $\lim \limits_{n \to \infty}a_n a_{n+1}=1$ and $0

2 Answers2