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.

  • $\begingroup$ Is $(x_n)_{n\in \Bbb N}$ decreasing? Is $(x_n)_{n\in \Bbb N}$ bounded below? $\endgroup$
    – Git Gud
    Nov 6, 2013 at 14:37
  • $\begingroup$ What I wrote is all I am given. $\endgroup$
    – GinKin
    Nov 6, 2013 at 14:45
  • 4
    $\begingroup$ 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, 2013 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, 2013 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, 2013 at 14:56

3 Answers 3


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 $1$.

  • $\begingroup$ 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, 2013 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, 2013 at 16:00
  • 1
    $\begingroup$ 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, 2013 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, 2013 at 15:35

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.


There is a theorem that says for any recursion $x_{n+1}=g(x_n)$, convergence is guaranteed whenever $| g'(x) |< 1$.

In our case, $g(x) = \frac{3}{4-x}$. This is clear since $x_{n+1} = g(x_n) = \frac{3}{4-x_n}$.

So lets evaluate the expression...

$g'(x) = \frac{3}{(4-x)^2}$, and we require $|g'(x)|<1$. Obviously $g'(x)$ is always positive.

We have $\frac{3}{(4-x)^2}<1$, which simplifies to $3 < (4-x)^2 = 16 - 8x + x^2$

If you solve the quadratic inequality you will see that $x>4+\sqrt{3}$ or $x<4-\sqrt{3}$.

The latter case, $x<4-\sqrt{3}\approx 2.26795\ldots$, is the case we are in. Since $x_1 = \frac32 = 1.5$, we are in the appropriate range for convergence.

If your initial term $x_1$ is not in the appropriate range, convergence can still happen. It's just not guaranteed and deeper analysis is required.

Also, just because $x_1$ isnt in the proper intervals doesnt necessarily mean that some other $x_i$ down the line wont be, and when/if that is the case, convergence is once again guaranteed. Once youre in the interval of convergence you stay there.


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