Why is the continued fraction [1,1,1,1,1...] equal to ϕ?

Question

I wanted to know what the continued fraction $1 + \frac 1{1 + \frac 1{1+ \frac 1{1+\frac 1{1+ \frac 1{1 + \ldots}}}}}$ would euqal to,

so I chose $x=1+\frac1{x}$ because you can put the euqation in for x infinitly times and you get the continued fraction I wanted to solve. When I solved for $x$ I got two solutions $\frac {1+\sqrt{5}}{2}$ and $\frac {1-\sqrt{5}}{2}$ which both would satisfy my equation.

But I looked it up and everyone says its only $\frac {1+\sqrt{5}}{2}$, the golden ratio $\phi$ and I wonder why do you only take the positive solution to the equation $x=1+\frac1{x}$?

Answer 1

When I solved for $x$ I got two solutions $\frac {1+\sqrt{5}}{2}$ and $\frac {1-\sqrt{5}}{2}$ which both would satisfy my equation.

They both satisfy the equation, but the first value is positive and the second value is negative.

The continued fraction is obviously positive, so only the first value remains.

Apart from that, i order to be strict, you'll have to show that the recursion $x\mapsto 1+1/x$ converges. Only if it converges one can conclude that the limit exists and equals the mentioned value.

Answer 2

If $x_n\ge1$, then $x_{n+1}=1+\frac1{x_n}\gt1$. Furthermore, $x_nx_{n-1}=x_{n-1}+1\ge2$. Therefore, $$ \begin{align} |x_{n+1}-x_n| &=\left|\,\left(1+\frac1{x_n}\right)-\left(1+\frac1{x_{n-1}}\right)\,\right|\tag{1a}\\ &=\left|\,\frac{x_{n-1}-x_n}{x_nx_{n-1}}\,\right|\tag{1b}\\ &\le\frac{|x_{n-1}-x_n|}2\tag{1c} \end{align} $$ Thus, the limit exists since the following sum converges by the ratio test: $$ \begin{align} x_\infty &=\lim_{n\to\infty}x_n\tag{2a}\\ &=x_1+\sum_{k=1}^\infty(x_{k+1}-x_k)\tag{2b} \end{align} $$ Now we can use the fact that the limit obeys $x_\infty\gt1$ and $$ x_\infty=1+\frac1{x_\infty}\tag3 $$ to get $$ \begin{align} x_\infty &=\frac{1+\sqrt5}2\tag{4a}\\ &=\phi\tag{4b} \end{align} $$