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.
Best Answer
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.