Prove that the sequence defined by $x_1 = 3$ and $x_{n+1} = \displaystyle\frac{1}{4-x_n}$ converges.

I know that the sequence looks like this: $\{3, 1, \displaystyle\frac{1}{3}, \frac{3}{11}...\}$ and the sequence is decreasing.

To use the Monotone Convergence Theorem I know that we have to show it is decreasing, thus show that $x_n \geq x_{n+1} (\forall n \in \mathbb{N}$) and bounded (above?).

So far I have come up with: We need to show that $x_n \geq x_{n+1}$ $\forall n \in \mathbb{N}$ and the sequence is bounded below. We have that $x_1 = 3$. Notice $x_2 = \displaystyle\frac{1}{4-x_1}=\frac{1}{4-3}=1$ and $3\geq 1$. So the statement $x_n \geq x_{n+1}$ holds when $n=1$.

I am having trouble with the proof from this point. Can someone walk me through step by step?

There are two other parts to this problem after proving this fact:

(b) Now that we know lim$x_n$ exists, explain why lim$x_{n+1}$ exists and equal to the same value.

(c) Take the limit of each side of the recursive sequence in part (a) to explicitly compute lim$x_n$.


3 Answers 3


We see that $x_{n+1}<x_{n}$ whenever $0<-1+4x_{n}-x_{n}^{2}=-(x_{n}-2-\sqrt{3})(x_{n}-2+\sqrt{3}).$ Since $x_{1}=3,$ which is between $2-\sqrt{3}$ and $2+\sqrt{3},$ this condition is satisfied, and thus the sequence is decreasing (note also that if $x_{n}>2-\sqrt{3},$ then $x_{n+1}=\frac{1}{4-x_{n}}>\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$).

(b) is always true, and for (c), if we take limits on both sides, we obtain $L=\frac{1}{4-L}.$ Solving, we see that we have $2\pm \sqrt{3}$ as possible solutions, however, we recall that $x_{1}=3<2+\sqrt{3},$ and the sequence is decreasing, so $L=2-\sqrt{3}$ is the only possibility.

  • $\begingroup$ can you go into further explanation for part (a)? how did you get $0 < -1+4x_n-x_n^2=\dots$?? $\endgroup$
    – rover2
    Oct 18, 2017 at 0:30
  • $\begingroup$ i know induction has to be used...i wrote above what i have for the base case. $\endgroup$
    – rover2
    Oct 18, 2017 at 0:31
  • $\begingroup$ I just took $\frac{1}{4-x_{n}}<x_{n}$ and simplified this a little. $\endgroup$ Oct 18, 2017 at 0:35

For a problem like this, where we are given a "recursive" definition of the sequence, my first thought is "proof by induction"!

Yes, we want to prove this sequence is decreasing that, for all n, $a_n> a_{n+1}$. Clearly, $a_1= 3$ and $a_2= 1$ so $a_1> a_2$.

Now we want to prove that, for any positive integer, k, if the statement is true for k, that is, if $a_k> a_{k+1}$, then the statement is true for k+ 1, that is, that $a_{k+1}< a_{k+ 2}$.

As for a lower bound it looks like these number will all be positive: that is $a_k> 0$ for all k. It is certainly true for k= 0, $a_0= 3$. Again, you now need to prove that if $a_k> 0$ then $a_{k+1}> 0$. It helps that, having proved this sequence is decreasing, it is clear that $a_k< 4$ for all k.

  • $\begingroup$ why is it $a_{k+1} < a_{k+2}$? shouldnt $a_{k+1}$ be greater than $a_{k+2}$ since it is a decreasing sequence the next term is less than the previous $\endgroup$
    – rover2
    Oct 18, 2017 at 0:38

I am reading "Understanding Analysis 2nd Edition" by Stephen Abbott.

This exercise is Exercise 2.4.1 on p.59 in this book.

Let $f(x):=x^2-4x+1.$
Then, $f(x_n)\leq 0$ if and only if $2-\sqrt{3}\leq x_n\leq 2+\sqrt{3}.$
If $2-\sqrt{3}\leq x_n\leq 2+\sqrt{3},$ then $2-\sqrt{3}\leq 4-x_n\leq 2+\sqrt{3}.$
If $2-\sqrt{3}\leq 4-x_n\leq 2+\sqrt{3},$ then $2-\sqrt{3}\leq \frac{1}{4-x_n}\leq 2+\sqrt{3}.$
Therefore, if $2-\sqrt{3}\leq x_n\leq 2+\sqrt{3},$ then $2-\sqrt{3}\leq x_{n+1}\leq 2+\sqrt{3}.$
And $2-\sqrt{3}\leq x_1=3\leq 2+\sqrt{3}.$
So, $2-\sqrt{3}\leq x_n\leq 2+\sqrt{3}$ for all $n\in\{1,2,\dots\}.$
So, $f(x_n)\leq 0$ for all $n\in\{1,2,\dots\}.$
So, $x_{n+1}-x_n\leq 0$ for all $n\in\{1,2,\dots\}.$
So, $\{x_n\}$ is a decreasing sequence.
And $2-\sqrt{3}$ is a lower bound of $\{x_n\}$.
So, $\{x_n\}$ converges.

Let $a:=\lim x_n$.
This means for any positive $\epsilon$, there exists a natural number $N$ such that if $n\geq N$, then $|x_n-a|<\epsilon.$
So, if $n\geq \max\{N-1, 1\}$, then $|x_{n+1}-a|<\epsilon.$
So, $\{x_{n+1}\}$ also converges to $a$.

$\{x_{n+1}\}$ converges to $a$.
By Algebraic Limit Theorem on p.50, $\{\frac{1}{4-x_n}\}$ converges to $\frac{1}{4-a}.$
Since $\{x_{n+1}\}=\{\frac{1}{4-x_n}\}$ and the limit of a sequence is unique by Theorem 2.2.7 on p.46, $a=\frac{1}{4-a}$.
So, $f(a)=0$.
So, $a=2+\sqrt{3}$ or $a=2-\sqrt{3}.$
Since $x_1=3<2+\sqrt{3}$ and $\{x_n\}$ is a decreasing sequence, $a=2-\sqrt{3}.$


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