How to show that $a_{n+1} = \sqrt{12+4a_n}$ is upper bounded?

Question

I am trying to prove that the sequence given by the recursive relation $a_{n+1} = \sqrt{12+4a_n}$ is convergent, $a_1 = 1$.

I managed to prove using induction that the sequence is stricly increasing.

I'm using the theorem that states that

If a sequence is monotonous and bounded then it is convergent.

Therefore, I must now prove there exists an upper bound

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My attempt

I will use induction again.

• For $n = 1: a_1 = 5 < M\in\mathbb{R}$
• For $n = k: a_k < M\in\mathbb{R}$
• For $n = k+1: a_{k+1} = \sqrt{12+4a_k} = 2\sqrt{3+a_k} < M \iff a_k < \frac{M^2}{2} -3 < \frac{M^2}{2} < M^2 $

I've shown that $a_{k+1}$ is less that $M^2$ whereas in the induction step I stated that $a_{k+1}$ is less than $M$. The square kinda confuses me, and I am not sure if I indeed proved boundness here, thus I ask this question

Answer 1

Note that $a_{k+1}=2 \sqrt{3+a_{k}}<M \iff a_{k}<\frac{M^2}{4}-3$. Then you could do $M=\frac{M^2}{4}-3$ which indeed gives $M=6$ as a solution.

Answer 2

Hint : prove by induction that $a_n \leq 6$ for all $n$.

Answer 3

The way to approach this sort of problems is usually as follows.

Imagine you have already proven that the sequence converges... so $\lim_{n\to\infty}a_n=a\in\mathbb R$. Wouldn't you be interested to find what's $a$? The way to do it is: in the equation $a_{n+1}=\sqrt{12+4a_n}$ you calculate limits of the left- and right-hand side when $n\to\infty$. You get:

$$a=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\sqrt{12+4a_n}=\sqrt{12+\lim_{n\to\infty}a_n}=\sqrt{12+4a}$$

so $a=\sqrt{12+4a}$ which implies $a=6$.

So what you have proven is that, if $a_n$ converges, it has to converge to $6$ and no other number. Also you know it converges (as you wouldn't be asked to prove it if it didn't!) so knowing that it is monotonically increasing, you immediately see that $a_n\lt 6$, approaching $6$ "from below", and actually $6=\sup\{a_n:n\in\mathbb N\}$.

Thus, maybe it pays off to now try to forget all that we said until this point, and prove that $a_n\lt 6$, which will immediately mean that your sequence is monotonically increasing and bounded - hence convergent.

And, indeed (proof by induction), $a_1=5\lt 6$ and if $a_n\lt 6$, then $a_{n+1}=\sqrt{12+4a_n}\lt\sqrt{12+4\cdot 6}=6$.