Verification of infinite square root equation proof

Question

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.

Answer 1

As written, neither of your approaches are valid. In each, you are assuming the conclusion you want to prove, and using it to show your conclusion. Or at least you're assuming something very close to your conclusion, like $\sqrt{1 + u} = \frac{1 + \sqrt 5}{2}$; that's not at all part of the given assumption.

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As for an approach that does work: What you have to start with is the identity

$$\sqrt{1 + u} = u$$

This can be rearranged as a quadratic equation with two roots, similar to your first approach. The negative root cannot be correct (make sure you understand why!), so the positive root is it.

Answer 2

The hint:

Our sequence it's $a^2_{n+1}=1+a_n$, $a_1=1$.

Prove that $a$ converges and calculate the limit.

Answer 3

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 $u \geq 0 \implies u = \frac{1+\sqrt{5}}{2}$.