Contractive condition implies the sequence is Cauchy

Question

$a_n$ is a sequence that satisfies the following contractive condition;

$|a_{n+2}-a_{n+1}| \le c|a_{n+1}-a_n|, n\ge 1 $ for some $ 0 \lt c \lt 1.$ Show the sequence $a_n$ is a Cauchy sequence and converges.

Consider the sequence $x_n$ given by $x_1=2$ and $x_{n+1} = 2+ \frac{1}{x_n}$

1. Show that $x_n$ is bounded below by 2
2. Show that $x_n$ satisfies the contractive condition in the first part and converges
3. Find lim $x_n$

Not looking for the answers. Just strong hints to guide me to the answer. The first part of the problem itself, makes sense. I'm unsure on the steps needed to prove this though. The second part for (1), would I want to use induction to show that it keeps getting closer to 2 but never reaches it. For part (2), I may be way off, but as simple as plugging into the contractive condition above? I am confused on part 3.

A lot here, I appreciate any sort of help. Thanks

Answer 1

The hypothesis ensures by induction that $|a_{n+p} - a_n|\leq \sum_{k=0}^p c^k |a_1 - a_0| = |a_1 - a_0| \frac{a^n-c^{p+1}}{1-c}$, and the latter can be made $\leq \varepsilon$ when $p\geq 0$ and $n\geq N$ for a well chosen $N$. This is the exact Cauchy criterion.

1. You show it easily by induction.
2. I leave it out to you
3. For the part 3, if $l$ is the limit, by continuity of the map $x\mapsto 2 + \frac{1}{x}$, the number $l$ must satisfy $l = 2 + \frac{1}{l}$. This is equivalent to a polynomial equation of degree $2$ in $l$, that has two roots, but only one root which is $\geq 2$. This unique root is the limit you are looking for. (I leave you solving the polynomial.)

Hope this helps.