The Sum of A Geometric Series of Lengths.

We consider a sequence of lengths where the ratio of each length to the previous length is a constant.
Examples:

1. If the ratio is 1/2 and the first length has a measurement of 4, then the sequence is 4,2,1,1/2,1/4,1/8,1/16,...
2. If the ratio is 1/5 and the first length has a measurement of 25, then the sequence is 25,5,1,1/5,1/25,1/125,1/625,...
3. If the ratio is 2/3 and the first length has a measurement of 9, then the sequence is 9,6,4,8/3,16/9,32/27,64/81,128/243,...
4. If the ratio is 4 and the first length has a measurement of 3, then the sequence is 3,12,48,152,...
5. If the ratio is r and the first length has a measurement of 1, then the sequence is 1, r, r2, r3, r4, r5,...
6. If the ratio is 4/3 and the first length has a measurement of 3, then the sequence is 3, 4, 16/3, 64/9, 256/27, 1024/81,...

A sequence of this type is called a geometric sequence.
Suppose we add the terms of a geometric sequence:

Example: 4 + 2+1 + 1/2 +1/4 + 1/8 +... = 7 + "1" = 8

BUT ... what about the other sequences?

Consider  the general  sequence 1, r, r2, r3, r4, r5,... and the sum of the first n terms, . Then

for any r where 0<r<1 and a large  value of n, S n is close to 1/(1-r).

Here is a visual argument to justify the previous statement:
In the figure below, consider the left  most square to have side of length 1. There are several similar squares with the ratios of the sides being r/1 in the figure where 0<r<1.
Using the large right triangle with base AB and the similar right triangle cut at the top of the left most square notice that AB/1 = 1/(1-r). But the sides of n of these  squares accumulate to give  a length of S n , and when n is large, this sum  is very close  AB  = 1/(1-r).