# 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

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## 3 Answers

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