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.
Best Answer
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.