First we have to schow that the numbers F2m/F2m+1 are numbers in a Farey sequence.
This is the case, because fractions of the form Fn/Fn+1 cannot be reduced, which is clear from A special property
or from the formula (Fn)2 + Fn+1Fn - (Fn+1)2 = (-1)n+1 (or inductively from Fn+1 = Fn + Fn-1).
Next we want to be sure that the numbers 1/2 + 1/8 + ... + 1/22m-1 are numbers of a dyadic sequence. This is easy because you can write this number
as a dyadic number, i.e. as a number with as denominator a power of 2.
Next, we prove (by induction) that for each n Fn/Fn+1 is adjacent to Fn+1/Fn+2
in the Fn+2-th Farey sequence.
The statement is correct for n=1, for F1/F2 (=1) and F2/F3 (=1/2) are adjacent in the F3-th (=2nd) sequence.
Suppose that we have already proven that for certain r>0 Fr/Fr+1 and Fr+1/Fr+2 are adjacent in the Fr+2-th sequence,
then we will show that Fr+1/Fr+2 and Fr+2/Fr+3 are adjacent in the Fr+3-th sequence.
That proves the statement inductively.
Well, in the Fr+3-th sequence Fr+2/Fr+3 = (Fr+Fr+1)/(Fr+1+Fr+2) will be
inserted between Fr/Fr+1 and Fr+1/Fr+2. Apparently Fr+2/Fr+3 and Fn+1/Fn+2
are adjacent in the Fr+3-th sequence.
(Inductively). If F2m/F2m+1 corresponds to am = 1/2 + 1/8 + ... + 1/22m-1 and F2m+1/F2m+2 to am + 1/22m (which is true for m=1), then F2m+2/F2m+3 (=(F2m+F2m+1)/(F2m+1+F2m+2)) corresponds to (am + (am + 1/22m))/2
= am+1 and F2m+3/F2m+4 to (am + 1/22m) + am+1 =
am+1 + 1/22.(m+1). This completes the proof.