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This is part of a larger problem:

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

I want to show that it is bounded below (by $0$ or something) and that it is decreasing, so by the Monotone Convergence Theorem it converges.

Does this seem like a valid strategy? My main concern is that I am not sure how to show the sequence is decreasing. I was thinking induction, but I don't know how to set that up for this case.

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Claim. $x_n > x_{n+1}$ $\forall n \in \mathbb{N} = \{1, 2, \dots\}$.

Proof. For $n = 1$, $x_1 = 3$, and $x_2 = \dfrac{1}{4-3} = 1$, so clearly, $x_1 > x_2$.

Suppose for an arbitrary $k \in \mathbb{N}$ that $x_k > x_{k+1}$.

Then observe that (as long as $x_{j} \neq 4$ for all $j \in \mathbb{N}$)

$$\begin{align} x_{k} > x_{k+1}\Leftrightarrow -x_{k} < -x_{k+1} \Leftrightarrow 4-x_{k}<4-x_{k+1} \Leftrightarrow \dfrac{1}{4-x_{k}}>\dfrac{1}{4-x_{k+1}}\Leftrightarrow x_{k+1}>x_{k+2}\text{.} \end{align}$$ Hence by induction, the claim is proven.

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    $\begingroup$ You need to make sure that $4-x_k$ and $4-x_{k+1}$ have the same sign. (If they have the same sign, and $a<b$, then $\frac1a>\frac1b$... but if they have different signs, and $a<b$, then $\frac1a<\frac1b$.) It's easy to show, but you should include it for the sake of completeness. $\endgroup$ – Akiva Weinberger Oct 13 '14 at 3:44
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Your sequence is the sequence of $x$-coordinates on a zigzag path between the graphs of $y=\dfrac{1}{4-x}$ and $y=x$.

enter image description here

The path begins in a closed region between the graphs on which $x>\dfrac{1}{4-x}$, $(3,1)$. The geometry makes it clear the iterated values decrease.

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Assuming $x_n$ positive we have

$$x_{n+1} \leq x_n \Leftrightarrow {1\over{4-x_n}} \leq x_n \Leftrightarrow x_n^2-4x_n+1 \leq 0 \Leftrightarrow x_n \in [2-\sqrt{3}, 2+\sqrt{3}] $$

Try to prove by induction that $x_{n+1} \leq x_n$ and that $x_n \geq 2-\sqrt{3}$.

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Yes that's a valid strategy.
Here is an alternative way to do it only using definition of converge sequence.
First solve $x=\frac{1}{4-x}$ you get $x=2+\sqrt{3}, 2-\sqrt{3}$, write down the first few terms of the sequence it's not hard to guess it should converge to $2-\sqrt{3}$. Now we prove it indeeds converge to that. To simplify the notation, let $a=2-\sqrt3$ , clearly $a^2-4a+1=0$.
Observe $$|x_{n+1}-a|=\\|\frac{1}{4-x_n}-a|=\\|\frac{1-4a+ax_n}{4-x_n}|=\\|\frac{-a^2+ax_n}{4-x_n}|=\\ \frac{|a||x_n-a|}{|4-x_n|}\leq \\|a||x_n-a|$$ For last inequality, all you need to do is proving every term is positive and less or equal to 3, by induction. Because |a|<1, the sequence is geometrically or faster getting close to $a$, therefore converges to $a$(i.e. for every $\epsilon$ find large enough N s.t. $|a|^N$ is small enough which makes $|x_n-a|$ small for $n\geq N$,...you know the rest :p).

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