Problem: Prove that, $x_{n+1}=a-\frac{1}{x_n}$ is convergent for $a\ge 4$ where $x_0=a$
My progress: The only technique I know to show a sequence convergent, is through Monotone convergent theorem. So I tried to show that the sequence is bounded, but messed up. (I tried it by induction). Not I can show that it is Monotone (and the sequence doesn't seem to be monotone).
Please help me. If it is posted before I am unaware of that (I searched it but couldn't find). Thank you in advance.
Best Answer
Bounded:
Base case: $x_0=a\geq 3$.
Inductive proof: if $x_n\geq 3$ holds true for $n=k$, then for $n=k+1$:
$x_k\geq 3\Longrightarrow -\frac{1}{x_k}\geq -\frac{1}3\Longrightarrow x_{k+1}=a-\frac{1}{x_k}\geq a-\frac{1}3\geq 4-\frac{1}3\geq 3$.
By induction we proved that $x_n\geq3$ holds true for all $n\in\mathbb{N}$. In other words, the sequence is bounded.
Monotony:
Base case: $x_1=a-\frac{1}a<x_0=a$.
Inductive proof: if $x_{n+1}<x_n$ holds true for $n=k$, then for $n=k+1$:
$x_{k+1}<x_k\Longrightarrow -\frac{1}{x_{k+1}}<-\frac{1}{x_k}\Longrightarrow a-\frac{1}{x_{k+1}}<a-\frac{1}{x_k}\Longrightarrow x_{k+2}<x_{k+1}$.
By induction we proved that the sequence $x_n$ is strictly decreasing.