Proof Using the Monotone Convergence Theorem for the sequence $a_{n+1} = \sqrt{4 + a_n}$ Consider the recursively defined sequence $a_0 = 1$
$a_{n+1} = \sqrt{4 + a_n}$
How to prove that the sequence converges using the Monotone Convergence Theorem, and find the limit?

Step 1. $\{a_n\}$ is strictly increasing, i.e., $a_{n+1}> a_n$.
This is shown inductively. For $n=1$, We have that $$a_2=\sqrt{4+a_1}=\sqrt{4+1}=\sqrt{5}>1=a_1.$$
Assume now that it is true for $n=k$, i.e., $a_{k+1}>a_{k}$. Then $a_{k+1}+4>a_{k}+4$, and hence
$$
a_{k+2}=\sqrt{4+a_{k+1}}>\sqrt{4+a_{k}}=a_{k+1},
$$
and thus it is true for $n=k+1$. Hence indeed, $\{a_n\}$ is increasing.
Step 2. $\{a_n\}$ is upper bounded. 
In particular, we shall show that $a_n<3$ inductively. So, for $k=1$, we have that $a_1=1<3$.
Assume that that it is true for $n=k$ and $a_k<3$. Then $a_{k+1}=\sqrt{4+a_k}<\sqrt{4+3}=\sqrt{7}<3$. Thus $\{a_n\}$ is upper bounded by 3.
Step 1.+Step 2. imply that $\{a_n\}$ converges, as an increasing and upper bounded sequence does converge.
Step 3. Finding of the limit of $\{a_n\}$.
Assume that $a_n\to a$, then $a_n+4\to a+4$ and $a_{n+1}=\sqrt{a_n+4}\to\sqrt{a+4}$.
But $\{a_n\}$ and $\{a_{n+1}\}$ have the same limit, and hence
$$
a=\sqrt{a+4},
$$
which implies that
$$
a^2-a-4=0,
$$
and hence
$$
a=\frac{1\pm\sqrt{17}}{2}.
$$
The negative root is rejected as the terms of $\{a_n\}$ are positive.
Thus
$$
a_n\to\frac{1+\sqrt{17}}{2}.
$$
