Proof verification: $a_1=2$, $a_{n+1}=3+\frac{a_n}{2}$ is increasing and bounded For base case, $a_2=3+\frac{a_1}{2}=4>a_1$
For $n=k$, let $a_{k+1}>a_k$
Adding 3 on both sides after dividing by 2,
$$3+\frac{a_{k+1}}{2}>3+\frac{a_k}{2}$$
$$a_{k+2}>a_{k+1}$$
Hence the sequence is increasing.
How can I show that the sequence is bounded?

Let $a_n=b_n+6$. Then
$$ b_{n+1}+6=3+\frac{b_n}{2}+3 $$
is equivalent to $b_{n+1}=\frac{b_n}{2}$ or to $b_n=\frac{b_1}{2^{n-1}}$. Since $a_1=2$ we have $b_1=-4$, hence
$ b_n = -\frac{8}{2^n} $  and
$$\boxed{ a_n = \color{red}{6-\frac{8}{2^n}} }$$
Now the claim is trivial.
