Proving a sequence converges to the golden ratio I have a sequence defined as such

$$a_1 = 1 \text{ and } a_{n+1} = \sqrt{1+a_n}$$

And I'm asked to prove that the sequence converges to the golden ratio. My first guess was to try $|a_n - L|<\epsilon$, as per usual when trying to determine convergence, but I don't actually have a definition for any $a_n$ (or do I?). If I rearrange for $a_n$, I get
$$a_n = a_{n+1}^2-1$$
But does it make sense to define $a_n$ in terms of its later terms? It doesn't to me.
 A: Here are the steps you need to prove.


*

*First prove that $1 \leq a_n < \dfrac{\sqrt5 + 1}2$ using induction.

*Nest prove that $a_n$ is a monotone increasing sequence using induction.

*Now recall that due to completeness of the real numbers a monotone increasing sequence bounded above converges. Call the limit as $L$.

*Make use of limit laws/rules to conclude that $L = \sqrt{1+L}$

*Prove that $L = \dfrac{\sqrt{5}+1}2$

A: Notice the golden ratio $\varphi$ is a fix point of the map $x \mapsto \sqrt{x+1}$.
If $a_m = \varphi$ for some $m$, then 
$$a_n = \varphi\;\;\text{ for all } n \ge m\quad\implies\quad\lim_{n\to\infty} a_n = \varphi$$
We will consider the case what if $a_n \ne \varphi$ for all $n$ below.
One way to check the sequence converges to $\varphi$ is show near $\varphi$, the
map $x \mapsto \sqrt{x+1}$ is a contraction mapping and $a_n$ does get close to $\varphi$ for some $n$. 
We can do this using mean value theorem
$$|a_{n+1} - \varphi | = | \sqrt{a_n+1} - \sqrt{\varphi+1} | = \frac{1}{2\sqrt{\xi+1}} | a_n - \varphi | \le \frac12 |a_n - \varphi |$$
where $\xi$ is some number between $a_n$ and $\varphi$. We can also do this pure algebraically
$$\frac{a_{n+1}-\varphi}{a_n - \varphi} = \frac{\sqrt{1+a_n} - \sqrt{1+\varphi}}{a_n - \varphi} = \frac{1}{\sqrt{1+a_n}+\sqrt{1+\varphi}} < \frac12$$
Both approaches lead to $$|a_n - \varphi| \le 2^{-(n-1)} |a_1 - \varphi|\;\;\text{ for all }n \quad\implies\quad \lim_{n\to\infty} a_n = \varphi\;\;\text{ again! }$$
