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I'm given:

$$\begin{align*} x_1&=\frac32\\\\ x_{n+1}&=\frac3{4-x_n} \end{align*}$$

How do I go about to formally prove the sequence converges and show it?

Thanks in advance.

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  • $\begingroup$ Is $(x_n)_{n\in \Bbb N}$ decreasing? Is $(x_n)_{n\in \Bbb N}$ bounded below? $\endgroup$ – Git Gud Nov 6 '13 at 14:37
  • $\begingroup$ What I wrote is all I am given. $\endgroup$ – GinKin Nov 6 '13 at 14:45
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    $\begingroup$ Let me rephrase: prove that $(x_n)_{n\in \Bbb N}$ is both decreasing and bounded below. Do you know how proving this helps you? $\endgroup$ – Git Gud Nov 6 '13 at 14:46
  • $\begingroup$ If it's bounded below, I can assume the sequence converges right ? And showing it's decreasing will let me find the limit ? $\endgroup$ – GinKin Nov 6 '13 at 14:54
  • $\begingroup$ No, for instance $(\sin(n))_{n\in \Bbb N}$is bounded below and it doesn't converge. You need something more to be sure it converges. $\endgroup$ – Git Gud Nov 6 '13 at 14:56
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We prove by induction that:

  1. $1<x_n<3$
  2. $x_n$ is decreasing.

The base case is obvious. Now assume that $1<x_{n-1}<3$ for some $n$. Then $$ \frac{3}{4-1}< \frac{3}{4-x_{n-1}}<\frac{3}{4-3} $$ or, after simplifying, $1<x_n<3$, so $1.$ holds for $n$. Also, note that $1<x_{n-1}<3$ implies $$ (x_{n-1}-1)(x_{n-1}-3)<0\Rightarrow 3<4x_{n-1}-x_{n-1}^2 $$ so $$ x_n=\frac{3}{4-x_{n-1}}<x_{n-1} $$ So $2.$ holds as well. Now by the monotone convergence theorem, $x_n$ converges. With a little more work, we can show that this limit is actually $1$.

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  • $\begingroup$ Isn't it enough to show that it's bounded below by one and decreasing in order to prove the limit is 1 ? $\endgroup$ – GinKin Nov 8 '13 at 14:32
  • $\begingroup$ @GinKin No. It's also bounded below by $0$ or $1/2$; you have to prove that $1$ is the infimum. $\endgroup$ – egreg Nov 9 '13 at 16:00
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    $\begingroup$ Maybe not completely rigorous, but since the sequence converges, for sufficiently large $n$ we get $a_{n+1}=a_{n}$. Thus if the desired limit is $L$, solving $L=\frac{3}{4-L}$ yields $L=1$. $\endgroup$ – tc1729 Nov 10 '13 at 5:00
  • $\begingroup$ Thanks you two. I went throught all of my notes and searched online and still can't really figure out how to prove 1 is the infimum for this question in a rigorus way. $\endgroup$ – GinKin Nov 10 '13 at 15:35
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Claim: $(x_{n})$ is monotonically decreasing.

The base case clearly holds.

Now assume true for $n=k+1$ for some $k \in \mathbb{Z_{+}}$.

So $x_{k+1}<x_{k}$

We can rearrange the terms in terms of its predecessors. Then we get

$$x_{k+1}=4-\frac{3}{x_{k+2}} \text{ , } x_{k}=4-\frac{3}{x_{k+1}} $$

Rearranging gives the desired result.

Claim: $x_{n}$ is bounded below.

Suppose not. Then for all $n^*$ larger than some $N \in \mathbb{Z_{+}}$,

$x_{n^*}<-4$ $\space$ (since $x_{n}$ is decreasing)

Now, this implies $4-x_{n^*}>8$ but then $x_{n^*+1}>1$ which is impossible.

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