<s>[INST] 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?
 [/INST] 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 $[1,\infty)$.
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