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}$ .

Answer:

  1. I was able to prove this part by mathematical induction.
  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.