<s>[INST] 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!
 [/INST] 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.
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