Continued fractions and the Fibonacci sequence


The convergents of the continued fraction

1+11+11+11+...1 + \Large \frac {1}{1+ \Large\frac{1}{1 + \Large\frac {1}{1+...}}} = 1+11+11+11+11+11+11+11+...1+ \large\frac{1}{1+}\normalsize \large\frac{1}{1+}\normalsize \large\frac{1}{1+}\normalsize \large\frac{1}{1+}\normalsize \large\frac{1}{1+}\normalsize \large\frac{1}{1+}\normalsize \large\frac{1}{1+}\normalsize ...

are

21,32,53,85,138,2113,3421,5534,8955,14489,233144,337233,...\large\frac{2}{1}\normalsize , \large\frac{3}{2}\normalsize , \large\frac{5}{3}\normalsize , \large\frac{8}{5}\normalsize , \large\frac{13}{8}\normalsize , \large\frac{21}{13}\normalsize , \large\frac{34}{21}\normalsize , \large\frac{55}{34}\normalsize , \large\frac{89}{55}\normalsize , \large\frac{144}{89}\normalsize , \large\frac{233}{144}\normalsize , \large\frac{337}{233}\normalsize , ...

where each fraction is composed of successive terms of the Fibonacci sequence.

The sequence of convergents tend to the (so-called) golden ratio ϕ=12(1+5)=1.618033...\phi = \large\frac{1}{2}\normalsize (1+√5) = 1.618033 ...