# Proof Using the Monotone Convergence Theorem for the sequence $a_{n+1} = \sqrt{4 + a_n}$ [closed]

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?

## closed as off-topic by Jendrik Stelzner, metamorphy, Xander Henderson, A. Pongrácz, José Carlos SantosJun 2 at 20:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jendrik Stelzner, metamorphy, Xander Henderson, A. Pongrácz, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.

It's easy to see that $a_n>0$ for all $n\geq 1$. Use mathematical induction to prove that $a_n\leq 4$ for all $n\geq1$ and $a_n$ is increasing. Hence $a_n$ is convergent, say its limit is $A$, then $A=\sqrt{A+4}$ and $A>0$, which give us that $A=\frac{1+\sqrt{17}}{2}$.

Hint:

• show (by induction) that for all $n \geq 0$, $1\leq a_n < \frac{1+\sqrt{17}}{2}$;
• use the right inequality to show that $a_n$ is an increasing sequence;
• use these two points (monotone convergence) to argue that $a_n\nearrow a\in\left(1,\frac{1+\sqrt{17}}{2}\right]$;
• using the fact that the recurrence relation has only two fixed points, and that the limit has to be one of them, find $a$.

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 $$a_n\to\frac{1+\sqrt{17}}{2}.$$