# Using the contraction principle to prove that the sequence $a_n=f(a_{n-1})$ is convergent where $f(x)=1+1/x$

Define function $f(x)$ from $\mathbb R^1$ to $\mathbb R^1$ by $f(x) = 1+1/x$. Define $a_n$ inductively by $a_1=1$ and $a_n=f(a_{n-1})$ for $n \geq 2$.

1. Prove using the contraction principle that $(a_n)$ is convergent
2. Prove that if $\lim_{n\to \infty}$ of $a_n= L$ then $L = F(L)$; hint Q2C
3. Determine $\lim_{n\to \infty}$ of $a_n$

### My thoughts

I honestly have no idea what the contraction principle is, I have googled it and have found no results. And I also don't know what Q2C means.

• What are your thoughts on the problem? That is, please edit your question and include a brief summary of what you've tried so far, or any ideas you might have towards the solution. Also, out of courtesy to those helping you, you should format the mathematical expressions in your question properly. See this page for a tutorial on how to do so. – Omnomnomnom Nov 29 '14 at 20:52
• proof-theory is a technical branch of mathematical logic. Do not use this tag unless this is the intended sense. – Andrés E. Caicedo Nov 29 '14 at 22:00
• I googled contraction principle, and the first link that shows on top point to contraction principle on wikipedia, where it branches, and one of the two branches is Banach contraction principle, also known as Banach fixed-point theorem It is likely it has something to do with the problem you posted above, try to find out what, study it. (I couldn't believe you googled contraction principle and found nothing about it.) – Mirko Nov 30 '14 at 0:47

## 1 Answer

The Banach contraction principle (which I state here only for the reals) says that if $f$ is a contraction mapping, that is if there is $q$ with $0\le q<0$ such that $|f(x)-f(y)|\le q |x-y|$ for all $x,y$ then $f$ has a unique fixed point $L$, that is $f(L)=L$, and moreover taking any $a_1$ the sequence $a_1,f(a_1),f(f(a_1)),...$ converges to $L$.

So, $a_1=1$ and $a_2=1+\frac11=2$. We will prove by induction that $\frac32\le a_n \le 2$ whenever $n\ge 2$. Indeed this is true for $n=2$. Assuming it is valid for some $n\ge2$, we have $\frac12\le \frac1{a_n} \le \frac23$, so $1+\frac12=\frac32 \le 1+\frac1{a_n}=a_{n+1} \le 1+\frac23=\frac53<2$, which completes this proof by induction.

So consider the restriction of $f$ to the interval $[\frac32,2]$, we will show that it is a contraction mapping with $q=\frac49$. Indeed, take any $x,y\in[\frac32,2]$. Then $|f(x)-f(y)|=|1+\frac1x-(1+\frac1y)|= |\frac1x-\frac1y|=|\frac{y-x}{x y}|\le|y-x|\cdot(\frac23)^2=\frac49|x-y|$.

So, since $f$ is a contraction mapping on $[\frac32,2]$, the sequence $a_2,f(a_2),f(f(a_2)),...$ converges to some limit $L\in[\frac32,2]$.

To find the value of $L$ use $Q2C$. If you don't know what $Q2C$ means (I do not), then use common sense :) In the equality $a_n=f(a_{n-1})$ take limit as $n\to\infty$, we obtain $L=f(L)=1+\frac1L$ (note that this is exactly what your hint says), so we have the equation $L^2=L+1$, or $L^2-L-1=0$, with roots $L_1=\frac{1-\sqrt{5}}2$ and $L_2=\frac{1+\sqrt{5}}2$. Note that $0>L_1\not\in[\frac32,2]$ hence we must have that $L=L_2=\frac{1+\sqrt{5}}2$.