Convergence of a recurrence

We have $$ a_{n+1}-a_n = \frac{1}{2}\left(a_n-a_{n-1}\right) $$ hence $\{a_n\}_{n\geq 0}$ is a Cauchy sequence and $a_n\to L$. Such $L$ has to fulfill $$ L = \frac{L}{2}+4 $$ hence $L=\color{red}{8}$.

For the ((asub1/2+4)/2+4)... representation, set that equal to asubn=k as n approaches infinity. Then, consider that one less iteration (asubn-1) is also approaches k when n approaches infinity. Thus, k/2 +4=k, so k=8.

HINT: a more direct way to show the convergence of $a_n$ is

$a_{n+1} = \frac{a_n}{2} + 4 = \frac{a_n+8}{2}$

If $8<a_0$, then $8<\frac{a_n+8}{2}<a_n$

If $a_0<8$, then $a_n<\frac{a_n+8}{2}<8$

Do you see how to turn this into the function for how quickly $a_n$ converges?