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 ® ¥ such that Xn+1 = Xn = L then, since
Xn+1 = 1
1+Xn
,
we can say that: L=1/(1+L) and therefore rearranging gives the quadratic equation L2+L-1=0. Solving this using the quadratic formula gives
L= -1+ Ö5
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 X2-X-1=0 I found the golden ratio
f = 1+ Ö5
2
and so
f-1 = -1+ Ö5
2
 = L
and
1
f
 = 2
(1+ Ö5)
 .
By multiplying the numerator and the denominator by 1-Ö5:
1
f
= 2(1-Ö5)
(-4)
= -1+ Ö5
2
 = L.
Therefore the limit L of the continued fraction Xn (as n® ¥) is equal to
1
f

where f 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): Xn = Fn / Fn+1
where Fn is a Fibonacci number from the sequence defined by the relation Fn+2=Fn+1+Fn with F1=1 and F2=1.

X1 = 1 and F1 /F2 = 1/1 = 1, therefore P(1) is true.

Assume that P(k) is true, then Xk = Fk / Fk+1. By the recurrence relation for the continued fractions Xk+1 = 1/ (1 + Xk). Hence
Xk+1
= 1
1+Xk
= 1
1+Fk/Fk+1
= Fk+1
Fk+1 + Fk
 .
But by the recurrence relation for the Fibonacci sequence Fk+2 = Fk+1 + Fk which gives
Xk+1 = Fk+1
Fk+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
Xn = Fn
Fn+1
 .
Since we have shown that the limit of Xn (as n® ¥) is 1/f we have now proved that the limit of Fn/Fn+1 = 1/f (as n® ¥) and so the limit (as n ® ¥) of Fn+1 / Fn is f, the golden ratio.