# Solution to $x=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ldots}}}$ [duplicate]

Compute $x=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\ldots}}}$

Thank you

• The answers from your other question can be adapted to this problem. Jun 9, 2012 at 16:56
• You might have heard of golden ratio.en.wikipedia.org/wiki/Golden_ratio This might be interesting for you . Jun 9, 2012 at 16:57

Look at the equation $$x^2-x-1=0$$

It is clear is solution is not zero. Thus, write

$$x^2=x+1$$

This is equivalent to

$$x=1+\frac 1 x$$

Using this recursively

$$x=1+\frac 1 {1+\frac 1 x}$$

Thus, define $\{ x_n\}$ recursively as $x_0=1$ and $x_n=1+\dfrac{1}{x_{n-1}}$

Show the sequence is positive, increasing for $n>3$ and bounded, thus it converges to $\ell$. You can then show that $\lim x_n=\ell = x$, where $x$ is the positive solution of the first equation discussed, namely

$$\phi = \frac{\sqrt 5 +1}{2}$$

Denote by $x=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ldots}}}$. Because the fraction goes on forever, the denominator of the second term in the RHS is identical to $x$. Thus, we have $x=1+1/x$ or the same as $x^2=1+x$. Thus, $x=\frac{\sqrt 5-1}{2}$.

• Strictly speaking, you need to show that the continued fraction converges. (Fortunately this is not hard. The Banach fixed point theorem, for example, will take care of it.) Jun 9, 2012 at 16:57
• Your stated value of $x = (\sqrt{5}-1)/2$ is smaller than 1. Jun 9, 2012 at 17:45
• The value of $x$ you give does not solve the equation, rather it satifies $x^2=1-x$. Did you mean $x=(1-\sqrt5)/2$, which does solve the equation? Indeed, even though the latter value is negative, the supposition that $1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ldots}}}$ denotes this value is consistent. Conventionally however such continued fractions are taken to denote the positive solution of the quadratic equation, can you see why? Jun 10, 2012 at 19:50
• This is definitely not a good explanation/solution of the problem. (-1). "goes on forever" is not a satisfactory term here.
– Pedro
Jun 11, 2012 at 2:46

Here is another, more pedestrian, and sketchy, way of doing this:

Let the sequence $\frac{p_n}{q_n}$ be defined as follows:

$$p_1=q_1=1$$

$$\frac{p_n}{q_n}=1+\frac 1{\frac{p_{n-1}}{q_{n-1}}} = \frac{p_{n-1}+q_{n-1}}{p_{n-1}}$$

It is clear that this computes terminated versions of the continued fraction and we have $$q_n=p_{n-1} \text{ and } p_n=p_{n-1}+q_{n-1}=p_{n-1}+p_{n-2}$$

It follows that numerator and denominator are successive Fibonacci numbers, and the limit ratio is the golden ratio.

There is an efficient discussion of continued fractions in Hardy & Wright "An Introduction to the Theory of Numbers" which shows that successive estimates from terminated fractions lie either side of the limit and converge to it.

• "terminated versions of the continued fraction" - convergents is the usual term of art. Basically, one is saying here that the $n$-th convergent of the continued fraction is $\dfrac{F_{n+1}}{F_n}$. For proving that the limit of that as $n\to\infty$ is $\phi$, see this. Jun 9, 2012 at 18:15