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Fibonacci sequence, golden ratio and Egyptian fractions

It turns out that there's a way to use the Fibonacci sequence to write the golden ratio in terms of Egyptian fractions. Here's how.

If we write the Fibonacci sequence as

f(n + 2) = f(n + 1) + f(n)

then we can write

f(n) = (anbn) / (ab)

where

a = (1 + √5) / 2

and

b = (1 – √5) / 2

Using this form of f(n) it's easy to show that

f(n)2fn+1 fn-1 = (-1)n-1

which we can rewrite as

fn+1 / fn = fn / fn-1 + (-1)n-1 / (fn fn-1)

But we also have that

fn / fn-1 = fn-1 / fn-2 + (-1)n-2 / (fn-1 fn-2)

so that

fn+1 / fn = fn-1 / fn-2 + (-1)n-2 / (fn-1 fn-2) + (-1)n-1 / (fn fn-1)

If we keep making similar substitutions, we find that we get that

fn+1 / fn = f2 / f1 + 1 / f1 f2 – 1 / f2 f3 + … + (-1)n-1 / fn-1 fn-2

The golden ratio a is the limit of fn+1 / fn as n gets large, so we have that

a = f2 / f1 + 1 / f1 f2 - 1 / f2 f3 + 1 / f3 f4 + …

= 1 + 1 – 12 + 16115 + …

Each of the terms in this sum happen to be fractions of the form 1n for some integer n, which are exactly the so-called Egyptian fractions. Apparently, the classical Egyptians used only fractions of that form, plus the single additional fraction 23, to represent all rational numbers.

If you try to actually do a few computations with Egyptian fractions, you'll probably come to the same conclusion that I did: that representing rational numbers this way is probably one of the greatest technical blunders of all time.

In any event, the above representation for the golden ratio that we got from the Fibonacci sequence is actually a way to represent it as a sum of Egyptian fractions. Yet another aspect of the Fibonacci sequence that I didn't expect.