Proving $a_{0} = 1$, $a_{n + 1} = \sqrt{1 + a_{n}}$ converges to the golden ratio

Question

Define $a_{0} = 1$ and $a_{n + 1} = \sqrt{1 + a_{0}}$.

Then we have

$$\lim_{n\to\infty} a_{n} = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \ldots }}}} $$

So I let $x = \sqrt{1 + \sqrt{ 1 + \ldots }}$. Then we have

$$x = \sqrt{1 + x} \Longleftrightarrow x = \frac{1 \pm \sqrt{5}}{2}.$$

So I think it converges to the golden ratio. But how can I prove this? Clearly, it's monotone increasing. Now I can just show the golden ratio is a supremum for the sequence. But how can I do that? (Monotone Convergence Theorem)

Answer 1

You already know the sequence is increasing. In order to use the monotone convergence theorem, you need to show it's bounded as well. If $a_n < 2$ for some $n$, then

$$a_{n + 1}^2 = 1 + a_n < 3 < 4 \implies a_{n + 1} < 2$$

since the terms are positive. Hence, the sequence is bounded and convergent.

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Now since $a_{n + 1} = \sqrt{1 + a_n}$, it is valid to take a limit on both sides and find that the limit $L$ satisfies $L = \sqrt{1 + L}$, and you're pretty close to being done.

Answer 2

Define $$a_n=b_n+\phi$$where $\phi $ is the famous Golden Ratio. Therefore $$a_{n+1}=b_{n+1}+\phi=\sqrt{1+a_n}=\sqrt{b_n+\phi +1}$$where the third equality comes from the definition of $\phi$. This yields to $$b_{n+1}=\sqrt{b_n+\phi +1}-\phi ={b_n+\phi +1-\phi ^2\over \phi +\sqrt{b_n+\phi +1}}={b_n\over \phi +\sqrt{b_n+\phi +1}}$$therefore $$|b_{n+1}|\le {|b_n|\over \phi}$$which means that $b_n\to 0$ and $a_n\to \phi$

Answer 3

In this answer, Question on sequences and induction, I showed that if $a_{n+1} = \sqrt{a_n+k}$ where $k > 0$ then $\lim_{n \to \infty} a_n = d+1$ where $d^2+d = k$, so $d = \dfrac{\sqrt{4k+1}-1}{2}$ and $d+1 = \dfrac{\sqrt{4k+1}+1}{2}$.

When $k = 1$, $d+1 = \dfrac{\sqrt{5}+1}{2}$.