# Proving Convergence using Monotone Convergence Theorem

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$x_n$.

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.

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

• 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 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.

(a)
Let $$f(x):=x^2-4x+1.$$
Then, $$f(x_n)\leq 0$$ if and only if $$2-\sqrt{3}\leq x_n\leq 2+\sqrt{3}.$$
$$x_{n+1}-x_n=\frac{1}{4-x_n}-x_n=\frac{x_n^2-4x_n+1}{4-x_n}=\frac{f(x_n)}{4-x_n}.$$
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.

(b)
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 $$a$$.

(c)
$$\{x_{n+1}\}$$ 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, $$a=2-\sqrt{3}.$$