Contractive Principle to Prove Convergence How do I use the contraction principle to show ${a_{n}}$ converges?
$a_{1}=1 \:$ and$\: a_{n+1}= 1+ \frac{1}{1+a_{n}}.$
I know I need to show that $\left | a_{n+2}-a_{n+1} \right |=k\left | a_{n+1}-a_{n} \right |$ with $k \in (0,1)$
So far I have:
$\left | a_{n+2}-a_{n+1} \right |= \left | \left ( 1+ \frac{1}{1+a_{n+1}} \right )-\left ( 1+ \frac{1}{1+a_{n}}  \right ) \right | =\left | \frac{1}{1+a_{n+1}} -\frac{1}{1+a_{n}}\right |=\left | \frac{a_{n}-a_{n+1}}{(1+a_{n+1})(1+a_{n})} \right |$
I am stuck from here...
I see the $\lim_{n\to\infty} a_{n}= \sqrt 2$ and I believe $\frac{1}{2}\leq {a_{n}} \leq \frac{3}{2}.$
Can I get some nudges to push me along?
 A: One possible approach, since you know the limit is $\sqrt{2} \approx 1.4142$. It is reasonable to propose the following claim after trying a few values.
Claim: $$\frac{7}{5} \leq a_n \leq \frac32$$ for $n \geq 2$.
This is true for $n=2$ since $a_2 = 1.5$.
Suppose that $$\frac75 \leq a_k \leq \frac32$$
then $$\frac{12}{5} \leq 1+a_k \leq \frac52$$
and hence then $$\frac25 \leq \frac{1}{1+a_k} \leq \frac5{12}$$
hence
$$\frac75 \leq a_{k+1} \leq \frac{17}{12}$$
Since $\frac{17}{12} \leq \frac32$, we have
$$\frac75 \leq a_{k+1} \leq \frac32$$
Hence we have the following claim: $$\frac{7}{5} \leq a_n \leq \frac32$$ for $n \geq 2.$
In fact, for simplicity, We can just say that $1 \leq a_n$ for all $n \geq 1$.
Now using your new tool , try to  obtain an upper bound for $$\frac{1}{(a_n+1)(a_{n+1}+1)}$$
A: Show that $f(x) = 1 + {1\over 1 +x}$ maps $[1,\infty) \to [1, \infty)$.
Show that $|f'(x)| \le {1 \over 4} < 1$ for $x \ge 1$.
Show that $|f(x)-f(y)| \le {1 \over 4} |x-y|$ for $x,y \ge 1$.
Note that $a_{n+1} = f(a_n)$ and the contraction mapping theorem shows that there is a unique $a$ such that $a_n \to a$. Since $a=f(a)$ we can solve for $a$ to get $a^2=2$ and so $a= \sqrt{2}$ (we are living in $[1,\infty)$.
