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Geometric Series
A geometric series is a sum of numbers, such as
1 + 1/2 + 1/4 + 1/8 + 1/16 + ...
where the first term is not zero and each consecutive pair of terms in the sum has the same ratio, called the common ratio. In the example above we have 1/(1/2) = 2, (1/2)/(1/4) = 2, etc... While seemingly fairly simple this type of sum is found extensively in many areas including geometry of real numbers, fractals, probability, economics, converting rational numbers from decimal form to fraction form (see below) and more.
We will show how to sum a geometric series when possible, give the formulas and then give some examples of this below.
It is common to see the common ratio between terms denoted by r, so if a geometric series starts with first term a, the series is
a + a*r + a*r^2 + a*r^3 + a*r^4 + ...
When summing an infinite number of terms (such as 1 + 1 + 1 + 1 + ...) it is not clear the sum should be a number, but a remarkable property about geometric series is that the sum is always a number when -1 < r < 1 and otherwise the sum is not a number (e.g., our 1 + 1 + 1 + ... series tends to infinite, so is not a number). To see why, suppose we are only interested in the first n + 1 terms of the series,
a + a*r + a*r^2 + ... + a*r^n
This is a sum of a n + 1 numbers, so is certainly a real number itself. If we call this number S_n, so
S_0 = a,
S_1 = a + a*r,
S_2 = a + a*r + a*r^2,
etc...
we have
S_n = a + a*r + a*r^2 + ... + a*r^n
and multiplying both sides by r gives
S_n*r = a*r + a*r^2 + a*r^3 + ... + a*r^n + a*r^(n+1)
which looks almost like our original sum S_n on the right. We have an extra a*r^(n+1) and we are missing the first term a, so writing
S_n*r = 0 + a*r + a*r^2 + a*r^n + a*r^(n+1)
= (a - a) + a*r + a*r^2 + ... + a*r^n + a*r^(n+1)
= a + a*r + a*r^2 + ... + a*r^n + a*r^(n+1) - a
we can now see S_n on the right hand side again. Indeed, we see that
S_n*r = S_n + a*r^(n+1) - a
and
a - a*r(n+1) = S_n - S_n*r = S_n*(1 - r)
so that
S_n = [a - a*r^(n+1)]/(1 - r) = a*[1 - r^(n+1)]/(1 - r).
when r is not equal to 1 (we cannot divide by 0!). Now if -1 < r < 1 then we know that r^(n+1) tends to 0 as n tends to infinity, so our sum S_n tends to
S = a*[1 - 0]/(1 - r) = a/(1 - r).
On the other hand, when r > 1 or r < 1 we know that r^(n_1) does not tend to a finite number, so the S_n do not tend to a finite number either. The only case we have not mentioned yet is when r = 1, but then the series is
S_n = a + a*1 + a*1^2 + ... + a*1^n = a + a + ... + a = (n + 1)*a
which certainly does not equal a number when n tends to infinity.
Formula for geometric series
If we have some real number a and some r such that -1 < r < 1, then the finite geometric series
a + a*r + a*r^2 + ... + a*r^(n+1)
has the sum
a*[1 - r^(n+1)]/(1 - r)
while the infinite geometric series
a + a*r + a*r^2 + ...
has the sum
a/(1 - r).
Examples
For example, the series with a = 1 and r = 1/2 above has sum
1/(1 - 1/2) = 1/(1/2) = 2.
(This particular sum is well known for one or two jokes.)
The geometric series with first term a = 2.5 and common ratio r = -1/3 has a sum of
2.5/[1 - (-1/3)] = 2.5/[1 + 1/3] = (5/2)/(4/3) = (5/2)*(3/4) = 15/8
whereas if we started with a = -2.5 instead, the series has a sum of
-2.5/[1 - (-1/3)] = -15/8.
Application for converting decimals to fractions
To see how this can be useful converting decimal numbers to fractions, consider the number 0.1111111... with 1s repeating. As
0.1 = 1/10,
0.01 = 1/100,
0.001 = 1/1000,
etc...
we can rewrite this as
0.11111... = 1/10 + 1/100 + 1/1000 + ...
which is a geometric sum with first term a = 1/10 and common ratio r = 1/10, so using our formula for the sum above,
0.111... = (1/10)/(1 - 1/10) = (1/10)/(9/10) = (1/10)*(10/9) = 1/9.
For a more complicated decimal, consider 0.124312431243... where the repeating part is 1243. The decimal is equivalent to
1243/10000 + 1243/10000^2 + 1243/10000^3 + ...
and so our simple sum formula gives
0.124312431243... = (1243/10000)/(1 - 1/10000)
= (1243/10000)*(10000/9999)
= 1243/9999 = 113/909
For the most complicated type of repeating decimal, consider one that starts with something different from the repeating pattern, e.g., 0.42567676767... where after the initial 425 we repeat with 67 forever. This is simply
0.425 + 0.000676767...
and so we have
0.425676767... = 425/1000 + 67/100000 + 67/(100000*100) + 67/(100000*100^2) + ...
= 425/1000 + (67/100000)/(1 - 1/100)
= 425/1000 + 67/99000)
= (425*99 + 67)/99000 = 21071/49500.
This may be less pleasant without a calculator, but just as simple as the examples above with our wonderful formula!
This also gives a nice validation that 0.999... = 1. Indeed, we have
0.999 = 9/10 + 9/100 + 9/1000 + ...
= (9/10)/(1 - 1/10)
= (9/10)/(9/10) = 1.