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?

$$\begin{align}
a_{n+1}&=3+\frac {a_n}2\\
a_{n+1}-6&=\frac 12 \left(a_n-6\right)\\
b_n&=\frac 12 b_{n-1}=\cdots =\frac 1{2^{n-1}}b_1&&(b_n=a_n-6)\\
a_n-6&=\frac 1{2^{n-1}}(\underbrace{\;a_1\;}_{=2}-6)\\
\color{red}{a_n}&\color{red}{=6-\frac 8{2^n}}
\end{align}$$
which is increasing and bounded (at $6$) as $n\to \infty$. 
