1. ## Terms in sequence

Define a sequence, $\displaystyle a_1=\frac{1}{2}$ and $\displaystyle a_{n+1}=\frac{1}{2+a_n}$
Prove that $\displaystyle |a_{n+1}-a_{n+2}|<\frac{|a_n-a_{n+1}|}{4}$

Proof by induction

When n=1, it is true
$\displaystyle \frac{1}{60}<0.04$

Assume true for n=k
$\displaystyle |a_{k+1}-a_{k+2}|<\frac{|a_k-a_{k+1}|}{4}$

Test n=k+1
$\displaystyle |a_{k+2}-a_{k+3}|<\frac{|a_{k+1}-a_{k+2}|}{4}$

$\displaystyle 4|\frac{1}{2+a_{k+1}}-\frac{1}{2+a_{k+2}}|<|\frac{1}{2+a_k}-\frac{1}{2+a_{k+1}}|$

$\displaystyle 4|\frac{a_{k+1}-a_{k+2}}{(2+a_{k+1})(2+a_{k+2})}|<|\frac{a_{k}-a_{k+1}}{(2+a_{k})(2+a_{k+1})}|$

By hypothesis, this is true for the numerators, now testing the denominators
if
$\displaystyle (2+a_{k+1})(2+a_{k+2})>(2+a_k)(2+a_{k+1})$ it is always true
Simplifying
$\displaystyle a_{k+2}>a_k$

However, this is not always true.

Now I need help. I am unsure how to proceed, or any of the preceding steps are wrong and I need to find a new method t solve the problem.
All help greatly appreciated

2. To compare them, right both in terms of $\displaystyle a_{n+1}$ only.

Since $\displaystyle a_{n+1}= \frac{1}{2+ \frac{1}{a_n}}$, for all n, $\displaystyle a_{n+2}= \frac{1}{2+ \frac{1}{a_{n+1}}}$ so $\displaystyle a_{n+1}- a_{n+2}= \frac{(a_{n+1})(2+ \frac{1}{a_{n+1}})}- 1}{2+ \frac{1}{a_{n+1}}$$\displaystyle = \frac{2a_{n+1}}{2+ \frac{1}{a_{n+1}}}$.

$\displaystyle a_{n+1}= \frac{1}{2+ \frac{1}{a_n}}= \frac{a_n}{2a_n+ 1}$
$\displaystyle 2a_na_{n+1}+ a_{n+1}= a_n$
$\displaystyle 2a_na_{n+1}- a_n= -a_{n+1}$
$\displaystyle a_n(1- 2a_{n+1})= a_{n+1}$
$\displaystyle a_n= \frac{a_{n+1}{1- 2a_{n+1}}}$
so that
$\displaystyle a_n- a_{n+1}= \frac{a_{n+1}}{1- 2a_{n+1}}- a_{n+1}= \frac{-2a_{n+1}^2}{1- 2a_{n+1}}$

Now compare those.

3. Thanks