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?

Writing out the first few terms will yield a very interesting pattern:
$a_0 = \frac{1}{1}, a_1 =\frac{2}{1}, a_2=\frac{3}{2}, a_3=\frac{5}{3} ...$
Do u see it now. Every term is a ratio of fibonacci nos. i.e. if $T_n$ is the nth fibonacci number starting from 1, then $a_n = \frac{T_{n+2}}{T_{n+1}}$. The limit of this is indeed the golden ratio. Try solving the difference equation to see it in case you are unaware of this result.
As for the proof that the observation is correct and not a coincidence, use induction.
We verified the base cases above, so assume 
$$ a_n = \frac{T_{n+2}}{T_{n+1}}$$
Now $$a_{n+1} = 1+ \frac{1}{a_n} = 1 + \frac{T_{n+1}}{T_{n+2}}= \frac{T_{n+2} + T_{n+1}}{T_{n+2}} = \frac{T_{n+3}}{T_{n+2}}$$
Hence proved. $\blacksquare$
