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 
$a_n - a_{n+1} < 0$
$a_n < a_{n+1}$

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.
