Supposing that the sequence of continued fractions does tend to a limit $L$ as $n\to \infty$ such that $X_{n+1}=X_n=L$ then, since

${\displaystyle X_{n+1}=\frac{1}{1+X_n},}$

we can say that: $L=1/(1+L)$ and therefore rearranging gives the quadratic equation $L^2+L-1=0$. Solving this using the quadratic formula gives

${\displaystyle L=\frac{-1+\sqrt{5}}{2}}$

I discarded the negative root because the limit must be positive since we are summing positive fractions.

By taking the positive root of the equation $X^2-X-1=0$ I found the golden ratio

${\displaystyle \varphi =\frac{1+\sqrt{5}}{2}}$

and so

${\displaystyle \varphi -1=\frac{-1+\sqrt{5}}{2}=L}$

and

${\displaystyle \frac{1}{\varphi }=\frac{2}{(1+\sqrt{5})}.}$

By multiplying the numerator and the denominator by $1-\sqrt{5}$:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{\frac{1}{\varphi }}& =\frac{2(1-\sqrt{5})}{(-4)}\\ \multicolumn{1}{c}{}& =\frac{-1+\sqrt{5}}{2}=L.\end{array}}$

Therefore the limit $L$ of the continued fraction $X_n$ (as $n\to \infty$) is equal to $\frac{1}{\varphi }$ where $\varphi$ is the golden ratio.

I used induction to prove each continued fraction is equal to the ratio of two Fibonacci numbers, that is to prove the statement $P(n)$ given by

${\displaystyle P(n):X_n=F_n/F_{n+1}}$

where $F_n$ is a Fibonacci number from the sequence defined by the relation $F_{n+2}=F_{n+1}+F_n$ with $F_1=1$ and $F_2=1$.

$X_1=1$ and $F_1/F_2=1/1=1$, therefore $P(1)$ is true.

Assume that $P(k)$ is true, then $X_k=F_k/F_{k+1}$. By the recurrence relation for the continued fractions $X_{k+1}=1/(1+X_k)$. Hence

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{X_{k+1}}& =\frac{1}{1+X_k}\\ \multicolumn{1}{c}{}& =\frac{1}{1+F_k/F_{k+1}}\\ \multicolumn{1}{c}{}& =\frac{F_{k+1}}{F_{k+1}+F_k}.\end{array}}$

But by the recurrence relation for the Fibonacci sequence $F_{k+2}=F_{k+1}+F_k$ which gives

${\displaystyle X_{k+1}=\frac{F_{k+1}}{F_{k+2}}}$

and hence $P(k+1)$ is true.

Therefore if $P(k)$ is true then $P(k+1)$ is true. But $P(1)$ is true and so, by the axiom of mathematical induction, $P(n)$ is true for all positive integers. So we have proved

${\displaystyle X_n=\frac{F_n}{F_{n+1}}.}$

Since we have shown that the limit of $X_n$ (as $n\to \infty$) is $1/\varphi$ we have now proved that the limit of $F_n/F_{n+1}=1/\varphi$ (as $n\to \infty$) and so the limit (as $n\to \infty$) of $F_{n+1}/F_n$ is $\varphi$, the golden ratio.