Jtdylktuy

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

Then we have

$$lim_{ntoinfty} a_{n} = sqrt{1 + sqrt{1 + sqrt{1 + sqrt{1 + ldots }}}} $$

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)

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.

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.

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 $$b_{n+1}=sqrt{b_n+phi +1}-phi ={b_n+phi +1-phi ^2over phi +sqrt{b_n+phi +1}}={b_nover phi +sqrt{b_n+phi +1}}$$therefore $$|b_{n+1}|le {|b_n|over phi}$$which means that $b_nto 0$ and $a_nto phi$

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