Abbott Exercise: Prove that $ x_1 = 3 $, $ x_{n+1} = \frac {1} {4-x_n} $ converges.

Question

In Understanding Analysis by Stephen Abbott there is an exercise:
$$\text{Prove that $ x_1 = 3 $, $ x_{n+1} = \frac {1} {4-x_n} $ converges}$$

In the solution Abbott says that the series is bounded by the following logic:

• Need to prove that if $ x_n > x_{n+1} $, then $ x_{n+1} > x_{n+2} $.
• $x_n > x_{n+1} $ implies that $ −x_n <−x_{n+1} $. Then we have $ 4−x_n <4−x_{n+1} $.

The conclusion of the above is that $x_n$ is decreasing.

I can follow. But how can one just assume $x_n > x_{n+1}$ by only the first and second term? I feel somewhat uncomfortable with this, as in it's not a complete proof, why is this ok?

I mean he used $ x_n > x_{n+1} $ as a fact and base, though it was something that is needed to be proven to show that the sequence is bounded no?

Thank you, I'm probably asking silly obvious things again.

Answer 1

As already said in a comment, this is just an induction step to show that the sequence is decreasing.

But if this is all what is written, it is definitely not a complete proof of the convergence of the sequence.

A possible way to see the boundedness of the sequence you may argue in a similar way as in the book:

• For $0\leq x\leq 3 \Rightarrow 0\geq -x\geq -3 \Rightarrow 4 \geq 4-x \geq 1 \Rightarrow \boxed{\frac 14 \leq \frac 1{4-x}\leq 1}$

So, boundedness is established.

For monotonicity you can just invoke the derivative of $f(x)=\frac 1{4-x}$:

• $\boxed{f'(x)= -\frac{1}{(4-x)^2} < 0}$ on $[0,3]$. So, $f(x)$ is strictly decreasing.

Now, since $x_2 <x_1=3$, you can conclude using strict monotonicity and induction that

$$x_{n+1}=f^{n-1}(x_2) < f^{n-1}(x_1)=x_n \mbox{ for all } n \in \mathbb{N},$$

where $f^{n-1}$ denotes the $(n-1)$-fold composition of $f$.