Prove that the sequence $(a_n)$ defined by $a_0 = 1$, $a_{n+1} = 1 + \frac 1{a_n}$ is convergent in $\mathbb{R}$ I will post the exercise below: 

Prove that the sequence $(a_n)$ defined by $a_0 = 1$, $a_{n+1} = 1 + \frac 1{a_n}$ for $n \in \mathbb N$ is convergent in $\mathbb R$ with the Euclidean metric, and determine afterwards is limit. Can you intepret the limit geometrically (hint: Golden ratio)?

So I need to prove that the sequence is convergent in $\mathbb{R}$  with the Euclidean metric, and how do I prove that? The limit must be $1$, but how to interpret it geometrically?

Let $\phi$ be the golden ratio, which is the only positive real such that $\phi =  1 + 1/\phi$.  Now consider the sequence $\epsilon_n = a_n - \phi$.  We seek to prove the invariant:
$$|\epsilon_n/\epsilon_0| \le \phi^{-n}.$$
For $n=0$, this is clear by inspection.  With some algebra we see that:
$$
   \epsilon_{n+1} = 1 + {1\over a_n}  - \phi = -{\epsilon_n\over a_n\phi}.
$$
Now our invariant already implies that $a_n = \phi + \epsilon_n \ge \phi - \phi^{-n}\epsilon_0 \ge 1.$  Therefore 
$$|\epsilon_{n+1}| \le  \left|{\epsilon_n\over \phi}\right| \le \phi^{-(n+1)}.$$
So our invariant remains after $n \rightarrow n+1$.  And therefore for all nonnegative integers $n$ by induction.  But since $\phi > 1$, that invariant means that  $\epsilon_n$ approaches zero or $a_n$ approaches $\phi$ as $n$ goes to infinity.
