As has been pointed out elsewhere there is a fascinating link between the reciprocal of 89 and the sum of the decimal point sequence made up Fibonacci terms so that

1/89 = .01 + .001 + .0002 + .00003 + .000005 + .0000008 + .00000013 + .000000021 + .0000000034 + …

What is involved here is a well-ordered numerical pattern whereby in base 10, 89 = 100 - 10 - 1. (89 is a term in the Fibonacci Sequence)

This can be then generalized for any base

Thus in base 8 for example, 100 - 10 - 1 = 67. (67 = 55 base 10 is a term in the Fibonacci Sequence).

Therefore the reciprocal of 67 is equal to the same decimal point sequence of Fibonacci terms (expressed in base 8).

Thus

1/67 = .01 + .001 + .0002 +.00003 + .000005 + +.0000010 + .00000015 + .000000025 + .0000000042 + …

This relationship of course also holds in base 2 where 100 - 10 - 1 = 1.

So we have the remarkable result that sum of this decimal point sequence of Fibonacci terms = 1.

Converting denary to binary terms, 1 = 1, 2 = 10, 3 = 11, 5 = 101, 8 = 1000, 13 = 1101, 21 = 10101, 34 = 100010 etc.

Therefore

1 = .01 + .001 + .0010 + .00011 + .000101 + .0001000 + .00001101 + .000010101 + .0000100010 + …

It is fitting that this decimal sum of Fibonacci terms (which starts with the binary digits  0, 1)  = 1.

Similar interesting connections can also be made to the Lucas sequence

1, 3, 4, 7, 11, 18, 29, 47, 76, ….

When we add up a similar decimal point sequence of these terms the sum is equal to the reciprocal of 89 *5/6.  This multiple of 89 is directly related to the base number in the form n/2 divided by (n/2) + 1.

Therefore 6/(89*5) = .01 + .003 + .0004 + .00007 + .000011 + .0000018 + .00000029 + .000000047 + ….

In base 8 the decimal sum of Lucas terms = 67*(4/5) = 54.

Therefore (in base 8)

1/54 = .01 + .003 + .0004 + .00007 + .000013 + .0000022 + .00000035 + .000000057 + …
 

Thus for base 6, the multiple is 3/4, and for base 4 2/3.

When we continue into base 2, the multiple is 1/2
 

Therefore (in binary), this decimal sum of Lucas terms = the reciprocal of .1 = 10 (2 in denary).

Therefore 10 = .01 + .11 + .100 + .0111 + .01011 + .010010 + .0011101 + .00101111+…

So again remarkably, the sum of the point series of the Lucas series (in binary) is double that of the corresponding series of Fibonacci terms.