FURTHER POINT SERIES CONNECTIONS
 
 
 

We can derive similar results for sums of Fibonacci decimal type sequences where we combine multiples of terms.

The first case again is when we combine multiples of the last term (a) with the first term.

Again the simplest example is when a = 2 so that we repeatedly combine twice the last term with the previous term in deriving our sequence.

This results in the series

1, 2, 5, 12, 29, 70, 169, 408, ….

The sum of the corresponding decimal point series = the reciprocal of 79.

Thus 1/79 = .01 + .002, + .0005 + .00012 + .000029 + .0000070 + .00000169 + .000000408 + ….

Thus whereas the Fibonacci point series = 100 - 10 - 1 (where a = 1), the Pell is 100 - 20 - 1 (where a = 2).

Thus in general terms where we use a multiple of the second term (i.e. a>1 ) , the sum of the point series is given by 100 - 10a - 1.

Thus where a = 3, the sum of the point series = 100 - 30 - 1 = 69.

This series where we continually combine 3 times the last with the second last term is

1, 3, 10, 33, 109, 360, 1189, 3927….

Thus 1/69 = .01 + .003 + .0010 + .00033 + .000109 + .0000360 + .00001189 + .000003927 …

This procedure applies across bases n, where a < n.

Thus using the previous series (where a = 3), the sum of the point sequence in base 8 = 100 - 30 - 1 = 47.

Therefore (n base 8)

1/47 = .01 + .003 + .0012 + .00041 + .000156 + .0000550 + .00002245 + .000007527 + ….
 
 

We can also generate Fibonacci type extensions where we vary the value of the second term (while keeping the first term constant). So here the value of b varies while a = 1.

So again in the simplest case where we continually combine the first term with the second term we have the sequence

1, 1, 3, 5, 11, 21, 43, 85, 171, 341….

The sum of decimal point terms for this sequence = the reciprocal of 88.

Thus 1/88 = .01 + .001 + .0003 + .00005 + .000011 + .0000021 + .00000043 + .000000085 + …

So here (where b = 2), the sum of the series = the reciprocal of 100 - 10 - 2.

Therefore in general terms the sum of terms of this point series = 100 - 10 - b where b < n (i.e. number base).

Again this result applies across number bases.

So in base 8, the sum of the point series (where a = 1, b = 2) = 100 - 10 - 2 = reciprocal of 66.

Thus (in base 8)

1/66 = .01 + .001 + .0003 + .00005 + .000013 + .0000025 + .00000053 + .000000105 + ….
 
 

When we combine multiples of the two terms then the sum of terms of the point series is given in any base n, by 100 - 10a - b (where a and b < n).

Thus for example in the series were a = 3, b = 2 the sum of decimal point terms (in base 10) = the reciprocal of 100 - 30 - 2 = 68.

This series is 0, 1, 3, 11, 39, 139, 495, 1763,…

Therefore 1/68 = .01 + .003 + .0011 + .00039 + .000139 + .0000495 + .00001763 + …