Differents ways to evaluate the sum $\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$ 
Evaluate $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$$
My approach:
Let $$x=\sqrt{12+ \sqrt{12+\sqrt{12+\cdots}}}$$
so, we have that $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}\iff \sqrt{12+x}=x \implies 12+x=x^{2} \iff (x+3)(x-4)=0$$
So, the answer is $\boxed{4}$.


Is correct my solution?
Can you show other ways for to solve this problem?
Can you suggest me any  textbooks with similar problems?
Thank you so much!
 A: Let $a_1=\sqrt{12}$ and $a_n=\sqrt{12+a_{n-1}}$. We must show that $a_n\to 4$. In order to do this, it is sufficient to show that
$1)\ \lim_{n\to\infty}a_n\text{ exists}$
$2)\ \text{This limit is }4$
You have basically skipped step $1)$ and gone straight to step $2)$. To fill in the first step, first note that $0<a_n<4$ for all $n$. We proceed by induction:
$$0<a_1=\sqrt{12}<\sqrt{16}=4$$
$$0<\sqrt{12}\leq a_n=\sqrt{12+a_{n-1}}<\sqrt{12+4}=\sqrt{16}=4$$
Second, we may now show that $a_n$ is increasing. This is obvious as
$$\frac{a_{n+1}}{a_n}=\frac{\sqrt{12+a_n}}{a_n}=\sqrt{\frac{12}{a_n^2}+\frac{1}{a_n}}$$
Since $a_n<4$, this becomes
$$\sqrt{\frac{12}{a_n}^2+\frac{1}{a_n}}>\sqrt{\frac{12}{16}+\frac{1}{4}}=\sqrt{\frac{3}{4}+\frac{1}{4}}=\sqrt{1}=1$$
Thus, $a_{n+1}>a_n$. Since we have already established that $a_n<4$ we may conclude that
$$\lim_{n\to\infty}a_n=L$$
exists. This concludes step $1)$. As you have already proved step $2)$, we may conclude that
$$\lim_{n\to\infty}a_n=4$$
The reason we may discard the $-3$ answer is that $0<a_n<4$ so $L\in[0,4]$.

EDIT: Edited as OP requested more resources for these types of infinite radicals. The best place to start (at least in my mind) is Herschfeld's Convergence Theorem. The theorem states that
$$\lim_{k\to\infty} \left[x_0+(x_1+(x_2+(\cdots +(x_k)^p)^p)^p)^p\right]$$
exists if and only if $(x_n)^{p^n}$ is bounded. We see that at each step, $x_n=12$. Since $12^{2^{-n}}$ is bounded, we may conclude that the series converges.
A: Method 1 - contraction mapping theorem.
Let $g : [0,\infty) \to [0,\infty)$ be the map $x \mapsto \sqrt{x+12}$. For any $x, y \in [0,\infty)$, we have
$$\begin{align} & g(x) - g(y) = \sqrt{x+12} - \sqrt{y+12} = \frac{x - y}{\sqrt{x+12} + \sqrt{y+12}}\\
\implies & |g(x)-g(y)| \le \frac{|x-y|}{2\sqrt{12}}
\end{align}
$$
This means $g$ is a contraction mapping over $[0,\infty)$.
By Contraction mapping theorem, $g(x)$ have a unique fixed point over $[0,\infty)$. Furthermore, if one pick any $z \in [0,\infty)$ and construct a sequence $z_n$ by $$z_n = \begin{cases}z, &n = 0\\g(z) = \sqrt{z+12}, & n > 0\end{cases}$$
$z_n$ will converges to that fixed point.
Since $g(4) = 4$, that unique fixed point is $4$. By setting $z$ to $0$, we find
$$\begin{array}{rcl}
z_1 &=& \sqrt{12},\\
z_2 &=& \sqrt{12+\sqrt{12}},\\ 
z_3 &=& \sqrt{12+\sqrt{12+\sqrt{12}}}\\
&\vdots&
\end{array}$$
converges to $4$.
Method 2 - explicit bound.
Define $z_n$ as in method $1$ and let $y_n = 4 - z_n$ for $n \ge 1$.
Notice
$$y_{n+1} = 4 - z_{n+1} = 4 - \sqrt{12 + z_n}
= 4 - \sqrt{16-y_n}
= \frac{y_n}{4 + \sqrt{16-y_n}}\tag{*1}
$$
$y_{n+1}$ has same sign as $y_n$. Since $y_1 = 4 - \sqrt{12}> 0$, all $y_n$ are positive.
Notice $\sqrt{16-y_n} = \sqrt{12+z_n} \ge 12$, $(*1)$ implies
$$0 < y_{n+1} < \frac{y_n}{4+\sqrt{12}}$$
Replace $n$ by $1, 2, \ldots, m-1$ and combine the inequalities, we find for $m \ge 1$,
$$0 < y_m \le \frac{y_1}{(4+\sqrt{12})^{m-1}}\quad
\implies\quad 4 - \frac{4}{(4+\sqrt{12})^m}
\le z_m < 4$$
By squeezing,
$$\lim_{m\to\infty} z_m = \lim_{m\to\infty}
\underbrace{\sqrt{12 + \sqrt{12 + \sqrt{12 + \cdots}}}}_{m\text{ times}} = 4$$
A: Edit
First see the comment of NinadMunshi immediately following this answer.  I have edited the answer accordingly.

Continuing Infinity_hunter's answer, and letting
$$x_n~ \text{denote}~ 
\sqrt{12 + \sqrt{12 +  \cdots  \text{n times} }}
$$
To show convergence, all that is necessary is to show that

*

*the sequence is bounded

*the sequence is strictly increasing.

Clearly, $0 < x_1 < 4.$  Assume that $0 < x_n < 4.$  Then
$$x_{(n+1)} = \sqrt{12 + x_n} < \sqrt{12 + 4} < 4$$
and
$$x_{(n+1)} = \sqrt{12 + x_n} > \sqrt{12 + 0} > 0.$$
Therefore,
$$0 < x_n < 4 \implies 0 < x_{(n+1)} < 4.$$
To show that the sequence is (therefore) strictly increasing:
Since
$$ 0 < x_n < 4$$
and
$$[(x_n)^2 - x_n - 12] = (x_n - 4)(x_n + 3)$$
I conclude that
$$[(x_n)^2 - x_n - 12] < 0 \implies (x_n)^2 < x_n + 12.$$
However, by the definition of the sequence
$$\left[x_{(n+1)}\right]^2 = x_n + 12.$$
Therefore
$$\left[x_{(n+1)}\right]^2 > (x_n)^2.$$
Therefore, since each element in the sequence is positive,
$$x_{(n+1)} > (x_n).$$
Therefore, the sequence is strictly increasing.
A: Your solution is correct. The other answers have addressed formality and rigour as a concern, but from the sound of your comments, you're more interested in the kind of problem more than the right answer, so I hope the following helps.
This kind of problem is an infinitely nested radical. As far as I know, there's no other way as straightforward as recognizing the infinitely nested square root is just a quadratic in disguise. In fact, an infinitely nested cube root would be a cubic in disguise.
Solution of Polynomial Equations with Nested Radicals on Arxiv actually derives a generic nested radical as a solution to a generalized quadratic equation, and dives deeper into different kinds of polynomials.
On Infinitely Nested Radicals is a less crazier article from a magazine that discusses their convergence and what type of numbers could be represented by them.
A Chronology of Continued Square Roots and Other Continued Compositions [...] on Arxiv is exactly what it sounds like. At a glance it's written pretty well.
A: You need to show the sequence
$$
x_n = \sqrt{12 + \sqrt{12 +  \cdots  \text{n times} }}
$$ converges. Only then you can take $x = \sqrt{12 + \sqrt{12 +\cdots}} $ and continue to other steps.
It is easy to see that $x_n = \sqrt{12 + x_{n-1}}$ with $x_1 = \sqrt {12}$.
As answered by @user2661923 and @QC_QAOA the sequence $x_n$ is bounded. Here is an easy way of showing $x_n$ is strictly increasing. Observe that $x_1 < x_2$ and we assume that $x_{n-1} < x_n$. Then
$$x_{n+1} = \sqrt{12 +x_n} > \sqrt{12 + x_{n-1}} = x_n$$
which completes our inductive argument and we conclude that $x_n$ converges.
Observing that $x_n > 0$ for any $n$ we can conclude that $x \ge0$ by limit theorems. So we must have $x = 4$
