Limit of sequence of continued fractions

Question

Using fixed point theorem evaluate limit of sequence of continued fractions: $$2,\quad 2+\dfrac{1}{2},\quad 2+\dfrac{1}{2 + \dfrac{1}{2}}, \cdots$$

How to prove it?

Answer 1

In addition to the other answer you need to prove that $f(x)=2+\frac 1 x$ is a contraction in order to use the fixed point theorem.

$$f'(x) = - \frac 1 {x^2}$$

The mean value theorem says

$$\left| \frac{f(b)-f(a)}{b-a} \right| = |f'(x_0)| \text{ for some } x_0 \in [a,b]$$

We know that $|f'(x_0)| = \frac 1 {x^2} < 1 \iff |x_0|>1$ This means $f$ is a contraction in $(-\infty,1)$ as well as in $(1,\infty)$, because we can use $|f'(x_0)| =: K$ as our constant $K$ that we need for the contration property (if $x_0$ is in one of these intervals), with this choice we have the contraction property:

$$|f(b)-f(a)| \leq K|b-a|$$ which is what we need.

Answer 2

Suppose that the sequence has a limit. Let it be $S$. Then

$$S=2+\frac{1}{S}$$

$$S^2=2S+1$$

$$S^2-2S-1=0$$

We use the quadratic formula:

$$S = \frac{2\pm \sqrt{8}}{2}$$

We discard the negative solution, so

$$S = 1 + \sqrt{2}$$