# Prove that this sequence converges to the golden ratio…

Question: Consider the sequence $$\{x_n\}$$ defined by $$x_{n + 1} = 1 + 1/x_n$$ for all $$n \in \mathbb{N}$$ and with $$x_1 = 1$$.

1. Prove that $$0 < x_{2n + 2} \leq x_{2n}$$ and $$x_{2n + 1} \geq x_{2n – 1} > 0$$.
2. Prove that the subsequences $$\{x_{2n}\}$$ and $$\{x_{2n – 1}\}$$ both converge to the same limit.
3. Prove that the sequence converges to $$\frac{1 + \sqrt{5}}{2}$$.

2. This is where I am getting stuck. My initial idea was to show that the sequence $$\{x_{2n} – x_{2n – 1}\} \to 0$$ as $$n \to \infty$$, but this didn’t seem to get me anywhere. Here is my attempt.
Proof: Fix $$\epsilon > 0$$. Consider the following
$$|x_{2n} – x_{2n – 1}| = \bigg|\frac{1}{x_{2n – 1}} – \frac {1}{x_{2n – 2}}\bigg| \leq \bigg|\frac{1}{x_{2n – 1}}\bigg| + \bigg|\frac{1}{x_{2n – 2}}\bigg|.$$
Now consider the sequence $$\{1/x_{2n – 1}\}$$. Since $$\{x_{2n – 1}\}$$ is a positive increasing sequence, we have that $$1/x_{2n – 1} < 1$$ for all $$n \in \mathbb{N}$$. Thus, the sequence $$\{1/x_{2n – 1}\}$$ converges by the monotone convergence theorem to some limit $$\alpha$$. This is where I was getting stuck. Any help would be nice.
1. I saw that this question was linked here: Prove that the sequence $(a_n)$ defined by $a_0 = 1$ , $a_{n+1} = 1 + \frac 1{a_n}$ is convergent in $\mathbb{R}$ , but I was wondering if someone could elaborate more.