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This was in one of the calculus textbook exercise problem. Now, I am pretty sure that there was a typo in the problem.

Problem

Let $a_n$ be a sequence of real numbers satisfying the following: $$ \lim_{n\rightarrow\infty} (2-a_n)a_{n+1} = 1. $$ Prove that $\lim\limits_{n\rightarrow\infty} a_n= 1$.

My tries

  1. The conclusion is true under assumption of $0

  2. The conclusion is false without any further assumption. So, the problem is incorrect as it is stated.

Question

Is the conclusion true under an additional assumption of boundedness of $a_n$?

Remarks

I was able to obtain a sequence $a_n$ that the conclusion is false. However, under the assumption of boundedness, I am wondering if we can prove the conclusion or give a counterexample.

Proof of My try 1

Let $\overline{a} = \limsup a_n$ and $\underline{a} = \liminf a_n$. Let $\{n_k\}_{k=1}^{\infty}$ be a subsequence of natural numbers such that $a_{n_k}\rightarrow \overline{a}$. Then we have $$ (2-a_{n_k}) a_{n_k +1} \rightarrow 1. $$ Taking a convergent subsequence of $\{a_{n_k +1}\}_{k=1}^{\infty}$ which converges to $b$, we have $$(2-\overline{a}) b = 1.$$ This gives $b=1/(2-\overline{a})$. Since $\overline{a}$ is the limsup, we have $$ \frac1{2-\overline{a}}\leq \overline{a}. $$ By assumption $00$. Thus, $$1\leq (2-\overline{a})\overline{a}.$$ Then it follows that $(\overline{a}-1)^2 \leq 0$, which yields $\overline{a}=1$.

Similar argument for $\underline{a}$ gives $\underline{a}=1$. Therefore, $a_n\rightarrow 1$.

Proof of My try 2

Since $a_n=0$ or $a_n=2$ do not happen infinitely often, we may assume that $a_n\neq 0$, $a_n\neq 2$ for all $n$.

Consider $\{1/k \}_{k=1}^{\infty}$, and the sequence given by the recurrence: $$ a_{n+1} = \frac{1+\frac1 1}{2-a_n}. \ \ \textrm{(Round 1)} $$ Then there exists $n$ such that $a_n>2$. We will put the first time that it happens as $n=n_1$, and we will say that we exit the Round 1. We have $a_{n_1+1}<0$ by the recurrence.

Once we exit the Round 1, we enter into the Round 2 which starts with $n=n_1+1$ and the recurrence: $$ a_{n+1} = \frac{1+\frac12}{2-a_n}. \ \ \textrm{(Round 2)} $$ Then there exists $n$ such that $a_n>2$. We put the first time $n\geq n_1+1$ that it happens as $n=n_2$, and we exit the Round 2. Then $a_{n_2+1}<0$.

Continue this process with using the recurrence $$ a_{n+1} = \frac{1+\frac1k}{2-a_n}. \ \ \textrm{(Round $k$)} $$

Then we have the sequcne $a_n$ such that $a_n>2$ infinitely many often and $(2-a_n)a_{n+1}\rightarrow 1$.

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    Would be nice if you could tell us how did you obtain the sequence, and what sequence was it for which it's false. – 2017-02-17
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    Not sure, but Isn't $a_n \sim a_{n+1}$ so that $(2-a_n)(a_n) \sim 1$ and thus $a_n \sim 1$. – 2017-02-17
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    $a_n=(-1)^n+1$. – 2017-02-17
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    @MyGlasses That sequence $a_n$ is $2$ infinitely many often. So, it does not satisfy the condition $(2-a_n)a_{n+1} \rightarrow 1$. – 2017-02-17
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    @kingW3 I will upload the proofs of my tries 1 and 2 soon. – 2017-02-17
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    @AhmedS.Attaalla That's what I thought too when I first saw the problem. After trying few small numbers, you will see that we do not have $a_n\sim a_{n+1}$ without further assumptions. – 2017-02-17
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    What textbook is this from? – 2017-02-17
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    That was a textbook I read 16 years ago, and it was a translated text. Sorry for not having the exact reference. – 2017-02-17
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    You can construct a bounded counterexample similar to what you already have: let's denote $f_c(x)=c/(2-x)$, then for every positive integer $k$ there is $c_k>1$ such that $-0.5$c_k$ form a decreasing sequence with limit $1$. Let Round $k$ consist of $k+2$ terms: $-0.5, f_{c_k}(-0.5), f_{c_k}(f_{c_k}(-0.5)), ..., f_{c_k}^{(k)}(-0.5) = 1.75, 4$. Then, if $a_n$ belongs to the Round $k$, then $(2-a_n)a_{n+1}$ is either $c_k$ or 1, so the limit of $(2-a_n)a_{n+1}$ is 1, and $a_n$ is bounded by -0.5 below and 4 above. – 2017-02-17
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    @Litho thank you. Would you put your comment as an answer so that I can accept it? – 2017-02-17

1 Answers 1

1

It's possible to construct a bounded counterexample in a way similar to what you have done: let's denote $f_c(x)=c/(2-x)$, then for every positive integer $k$ there is $c_k>1$ such that $-0.5

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    One question: How can we make sure that we arrive at 1.75 in exactly $k$ steps? – 2017-02-17
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    Oh, I got it now. Thanks! This is very good. – 2017-02-17