# Prove recursively defined sequence converges

I would like some advice on how to solve problems like the following: Let $(x_n)$ be a sequence defined by $x_1= 3$ and $x_{n+1} = \frac{1}{4-x_n}$. Prove that the sequence converges.

My strategy is to use the Monotone Convergence Theorem, but I am having trouble showing that the sequence is decreasing and bounded below. Here's my work so far:

Decreasing: The first 4 values are $3,1,1/3,3/11$, so let's assume $x_n \leq x_{n-1} \leq \ldots \leq x_1$. Want to show $x_{n+1} \leq x_n$. We have $x_{n+1} = \frac{1}{4-x_n}$. I want to have an upper bound for the RHS, but can't find one and don't really know where to go from here.

Bounded below: I wanted to show that all values are positive, but if we assume $x_n > 0$, that doesn't rule out $x_{n+1}$ from being negative.

• To prove the values are positive, it is easier to prove that all values are in the interval $(2-\sqrt{3}, 2+\sqrt{3})$. – vadim123 Apr 8 '16 at 23:48
• It's not apparent to me why anyone would think to use that approach without already knowing something about this sequence. – rorty Apr 8 '16 at 23:54
• You are lead to that interval by looking at fixed points. That is, if $f(x)=\frac 1{4-x}$ then $f(x)=x\implies x=2\pm \sqrt 3$. So...if the series converges, the limit is one of those two values. – lulu Apr 8 '16 at 23:58

assume that it convergences to x. In which case, $x = \frac{1}{4-x}$ solve for x. There are two possible solutions... one of those will prove to be unstable.

Now that you know what x should equal it should be simple to show that $x_1>x_2>...>x$ your sequence is bounded below and monotonically decreasing and therefore convergent.

Suppose {$x_n$} converges to $x$.

$x = \frac{1}{4-x}\\ 4x - x^2 = 1\\ - (x^2 -4x + 1) =0\\ - (x - 2+\sqrt3)(x - 2-\sqrt3)=0\\$

I claim that $\forall x_n\in(2-\sqrt3,2+\sqrt3), 2-\sqrt3<x_{n+1}<x_n$

to show that $x_{n+1}<x_n$, we can show that $x_n - x_{n+1}>0$

$x_n - x_{n+1} = x_n - \frac{1}{4-x_n} = \frac{-x_n^2 + 4x_n - 1}{4-x_n}$ and $\frac{-x_n^2 + 4x_n - 1}{4-x_n} > 0$ when x is in the interval.

And to show that $x_{n+1} > 2-\sqrt3$, when $x_n>2-\sqrt3, 4-x_n < 2+\sqrt3$ and $\frac{1}{4-x_n} > \frac{1}{2+\sqrt3}.$

$x_2 \in(2-\sqrt3,2+\sqrt3)$ then for all $n>2, 2-\sqrt3<x_{n+1}<x_n$

$x_n$ is monotonically decreasing and is bounded below.

• Allowing $x = \frac{1}{4-x}$ is a heuristic just to lead to possible solutions right? I can't see how you can rigorously allow that equation from the definition of the limit. Does this usually work with recursively defined sequences? – rorty Apr 9 '16 at 0:12
• If the sequence has a limit $L$, then $L$ must satisfy $L=\frac{1}{4-L}$. – vadim123 Apr 9 '16 at 0:16
• Indeed. We make a leap of faith that this is the lower bound, and then we can rigorously show that it in fact is. – Doug M Apr 9 '16 at 0:16
• @vadim123 I now see your point. What you say is true because $\lim x_n = \lim x_{n+1}$ combined with the algebraic limit theorem. – rorty Apr 9 '16 at 0:58
• @DougM Can you elaborate on how knowing the limit helps make the argument that the sequence is decreasing? I found that the sequence is bounded below. – rorty Apr 9 '16 at 1:14