Prove that $a_n
So I got this question and I was able to do up to (III) (a) however part b has been pretty difficult for me to complete and I can't seem to get it right . Any help ? 
I tried considering $a_n - a_{n+1}$ and I got the required proof but I don't feel as though this method is logical enough 
$a_n - a_{n+1} < 0$
$a_n < a_{n+1}$
 A: Guide:


*

*Prove that $a_n>0$ if $a_1>0$.

*Compute $a_{n+1}-a_n$ in terms of $a_n$, your goal is to check that the expression is positive.

*check that the denominator is positive.

*Show that the numerator is positive using the property that $0<a_n<2$.

A: Alt. hint: $\;a_{n+1}=\dfrac{4(a_n+1\color{red}{+3-3})}{a_n+4}=4 - \dfrac{12}{a_n+4}\,$, and therefore:
$$\require{cancel}
a_{n+1}-a_{n}=\left(\cancel{4} - \dfrac{12}{a_n+4}\right) - \left(\cancel{4} - \dfrac{12}{a_{n-1}+4}\right) = \dfrac{12(a_n-a_{n-1})}{(a_{n}+4)(a_{n-1}+4)}
$$
It follows that $\,a_{n+1}-a_n\,$ and $\,a_n-a_{n-1}\,$ have the same sign, so the entire sequence is monotonic, and the direction of monotonicity is given by the sign of $\,a_2-a_1\,$, in this case positive so the sequence is increasing.
A: Item III.a. we use II. result $a_{n+1} - 2 = 2 - \frac{12}{4 + a_n}$, and $a_n < 2 \iff a_n + 4 < 6$, then $\frac{12}{4 + a_n} > 2$ Therefore $a_{n+1} - 2 = 2 - \frac{12}{4 + a_n} < 0 \iff a_{n+1} < 2$.
A: So you can check your solutions.
We get $a_2=\frac{20}{11}$ and $a_3=\frac{31}{16}$.
For II):
It is $a_{n+1}-2=\frac{4(1+a_n)}{4+a_n}-2=\frac{4+4a_n-8-2a_n}{4+a_n}=\frac{2a_n-4}{4+a_n}=\frac{2(a_n-2)}{4+a_n}$
For III):
It is $a_{n+1}=\frac{4(1+a_n)}{4+a_n}\stackrel{a_n<2}{<}\frac{4(1+2)}{4+a_n}=\frac{12}{4+a_n}\stackrel{a_n<2}{<}\frac{12}{4+2}=2$. Hence $a_{n+1}<2$.
Finally:
We show, that $a_{n+1}-a_n>0$
We have $\frac{4(1+a_n)}{1+a_n}-a_n=\frac{4(1+a_n)-a_n(1+a_n)}{1+a_n}=\frac{4-a_n^2}{4+a_n}=\frac{(2+a_n)(2-a_n)}{1+a_n}>0$
Because $2+a_n>0$ and $1+a_n>0$ since $a_n>0$ (sequence of positiv numbers)
Also $2-a_n>0\Leftrightarrow 2>a_n$ which is true.
Every factor greater than zero means the product is greater than zero.
