<s>[INST] Verification of infinite square root equation proof 
b) Show also that the infinite square root
  $$\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}}=\frac{1+\sqrt5}{2}$$

Now, I've devised two proofs for this statement (both of which include using substitution), and it entails assuming that the case is true, and then restating the problem.
Proof 1:
let $\displaystyle u=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}}$
hence $$u=\frac{1+\sqrt5}{2}$$ from here, I went on to express the equation as a quadratic as follows: $$2u-1=\sqrt5$$ $$4u^2-4u+1=5$$ $$4u^2-4u-4=0$$ and then, naturally, just substituted the new values into the quadratic formula: $$u=\frac{4\pm\sqrt{4^2-4(4)(-4)}}{8}$$ $$u=\frac{4\pm4\sqrt{5}}{8}$$ $$u=\frac{1\pm\sqrt5}{2}$$ the only problem I found here is that, evidently, the quadratic formula leaves the $\pm$ in its answer, as opposed to the desired $+$, so the second proof is as follows:
Proof 2:
let $\displaystyle u=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}}$
hence, $$\sqrt{1+u}=\frac{1+\sqrt5}{2}$$ and from here, I simply proceeded to solve for $u$, so $$1+u=\frac{(1+\sqrt5)^2}{4}$$ $$u=\frac{6+2\sqrt5}{4}-1$$ $$u=\frac{3+\sqrt5}{2}-\frac{2}{2}$$ $$u=\frac{1+\sqrt5}{2}$$
which (I think) completes the proof. The question is, are either of these approaches valid? Are there any better approaches that someone could list? Any responses are appreciated.
 [/INST] Let's show that the sequence $(a_n)_{n=1}^\infty$ given by $a_1 = 1$, $a_{n+1} = \sqrt{a_n + 1}$ is monotonically increasing and bounded from above.
We shall prove by induction that $a_n \leq \frac{1+\sqrt{5}}{2}$.
$$a_1 = 1 < \frac{1+\sqrt{5}}{2}$$
Assume $a_n \leq \frac{1+\sqrt{5}}{2}$, for some $n\in\mathbb{N}$.
$$a_{n+1} = \sqrt{1 + a_n} \leq \sqrt{1 + \frac{1+\sqrt{5}}{2}} = \sqrt{\left(\frac{1+\sqrt{5}}{2}\right)^2} = \frac{1+\sqrt{5}}{2}$$
Now we shall prove that $(a_n)_{n=1}^\infty$ is increasing:
$$a_{n+1} \geq a_n \iff \sqrt{a_n + 1} \geq a_n \iff 1+a_n - a_n^2 \geq 0 \iff a_n \in \left[\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right]$$
The last statement holds, so we have $a_{n+1}\geq a_n$.
Thus, $(a_n)_{n=1}^\infty$ is increasing and bounded from above, hence it is convergent.
Set $u = \lim_{n\to\infty} a_n$.
Letting $n\to\infty$ in the recurrence relation $a_{n+1} = \sqrt{a_n + 1}$ gives $u = \sqrt{1 + u}$.
Thus, $u = \frac{1\pm\sqrt{5}}{2}$. Since $a_n \geq 0, \forall n \in \mathbb{N}$, we have $u \geq 0 \implies u = \frac{1+\sqrt{5}}{2}$.
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