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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.
Taylor McMillan
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3 Answers3

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An iteration has a limit only when $x_{t+1}=x_t$.

Set $x_t=x_{t+1}=T$ and you get: $$T=1+\frac 1T\implies T^2-T-1=0\implies T=\phi^{\pm}$$

$\phi^- <0$, and no term in this sequence will be, since $x_1 >0$ and $y>0\implies \frac1y>0$

So just prove that it does converge, then it must converge to $\phi^+=\frac{1+\sqrt 5}{2}$

Rhys Hughes
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Denote $y_n=x_{2n}$. Then you have $$y_{n+1}=x_{2n+2}=1+\frac{1} {x_{2n+1}} =1+\frac{1}{(1+1/x_{2n})} =\frac{y_n +1} {y_n +2} $$ From this you can show two things : the first is that $y_n$ is bounded (by induction on $n$), and then since it is increasing and bounded you can compute its limit $L$. Since $y_n$ converges, so do $y_{n+1}$, and to the same limit. On the other hand you can apply arithmetic of limits and get that $L=(L+1)/(L+2)$ so $L^2 + L - 1=0$ and by looking at the sequence you can deduce which root of the polynomial is the limit.
You can proceed analogously for $z_n=x_{2n+1}$ and you will see that the limit is the same.

Binyamin Riahi
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It’s easy to see that $x_n\geq 1$ from the original recurrence. You can show $x_n\leq2$ as follows. It’s true for $n=1,2$. And $x_{n+1}\leq 1+1/x_n\leq 1+1/1=2$.

Since you’ve shown the even/odd subsequences are monotonic, and bounded, they both converge to some limit in $[1,2]$. Now let $y_n:=x_{2n}$. Then:

$$y_{n+1}=1+1/(1+1/y_n)$$

Let $L$ be the limit of $y_n$. Taking limits of both sides gives:

$$L=1+1/(1+1/L)$$

Now solve for $L$. Repeat for $x_{2n+1}$.

Alex R.
  • 32,771