Copyright © University of Cambridge. All rights reserved.
'Golden Fractions' printed from https://nrich.maths.org/
This solution came from Joseph of Colyton Grammar School. Also
Yosef who has just graduated from High School and Andrei from Tudor
Vianu National College, Bucharest, Romania sent in good solutions.
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 $$X_{n+1} = {1\over 1+X_n },$$ 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 $$L={-1+ \sqrt5
\over 2}$$ 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 $$\phi = {1+ \sqrt 5\over 2}$$ and so $$\phi -1 = {-1+
\sqrt5 \over 2} = L$$ and $${1\over \phi} = {2\over (1+ \sqrt
5)}.$$ By multiplying the numerator and the denominator by $1-\sqrt
5$: $$\eqalign { {1\over \phi} &= {2(1-\sqrt 5)\over(-4)} \cr
&= {-1+ \sqrt 5\over 2} = L.}$$ Therefore the limit $L$ of the
continued fraction $X_n$ (as $n\to \infty$) is equal to ${1\over
\phi}$ where $\phi$ 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 $$P(n): X_n = F_n / F_{n+1}$$ 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
$$\eqalign{ X_{k+1}&= {1\over 1+X_k} \cr &= {1\over
1+F_k/F_{k+1}} \cr &= {F_{k+1}\over F_{k+1} + F_k}.}$$
But by the recurrence relation for the Fibonacci sequence $F_{k+2}
= F_{k+1} + F_k$ which gives $$X_{k+1}= {F_{k+1}\over F_{k+2}}$$
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 $$X_n = {F_n\over
F_{n+1}}.$$ Since we have shown that the limit of $X_n$ (as $n\to
\infty$) is $1/\phi$ we have now proved that the limit of
$F_n/F_{n+1}= 1/\phi$ (as $n\to \infty$) and so the limit (as $n
\to \infty$) of $F_{n+1} / F_n$ is $\phi$, the golden ratio.