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Geometric Series/Sum

August 13th, 2010

A geometric series is is a series where each of the terms is a multiple of the previous one. Some examples are

1 + 2 + 4 + 8 + 16 + …       each term is twice the previous one
4 + 12 + 36 + 108 + …       each term is three times the previous one

In general the nth term of a geometric series is written as ar^{n-1} where a is the first term and r is the factor we multiply each term by. If a is 1 then the terms are just the powers of r.

We are often interested at find the value of a geometric sum, that is the sum of all the values of a geometric series to a certain point. We can also find the value of the sum of infinitely many terms if |r|<1. To find both of these first let the sum of the first n terms be denoted Sn.

So S_n = a + ar + ar^2 + ar^3 + = \sum^n_{i=1}{ar^{n-1}}

The formula for a finite sum that we will try to prove is

\Large S_n = \frac{a(r^n - 1)}{r-1}


Proof of finite sum

Consider
rS_n - S_n = r\sum^n_{i=1}{ar^{i-1}} - \sum^n_{i=1}{ar^{i-1}} from the definition of Sn


We can now factorise the left-hand side and tidy up the right multiply the r through the first sum and change the index down on the second,


S_n(r-1) = \sum^n_{i=1}{ar^i} - \sum^{n-1}_{i=0}{ar^i}


We can now break up the sums into parts in order that we cancel bits of them in the next step


S_n(r-1) = ar^n + \sum^{n-1}{i=1}{ar^i} - \sum^{n-1}{i=1}{ar^i} + ar^0


So cancelling out the two sums and factorising the a noting that anything the the power of 0 is 1 gives


S_n(r-1) = a(r^n - 1)


and then dividing through by (r-1) gives us our required result


S_n = \frac{a(r^n - 1)}{r-1} .

Infinite Sums

As noted earlier is |r|<1 we can also find the value of an infinite sum. It may seem like a strange idea to be able to find the value of the sum of an infinite number of things but if |r|<1 the size of each term keeps getting smaller so the amount the sum increases by each time gets smaller and smaller.

A simple way of visualising how this can work is to imagine a piece of rope two meters long. We are going to cut this rope up into an infinite number of pieces with the length of each piece representing a term in the sum. Firstly cut the rope in two. Put one piece to the side as this is our first term so a=1 as this piece is 1 meter long. r=1/2 because each piece of rope will be half the length of the previous one. Now cut the remaining piece of the rope in half and again put one piece to the side as the second term and cut the remainder in half. Theoretically we will always be able to repeat this process because you can always halve something. This will mean that we will now have an infinite number of pieces of rope with lengths 1m, 1/2m, 1/4m, 1/8m, … so the lengths of the pieces of rope form a geometric progression and we know that the sum of their lengths must be less than two as we only started out with two meters of rope. In fact it is exactly two which we can prove.

In general,


\Large S_\infty = \sum^\infty_{i=0}{ar^i} = \frac{a}{1-r} if |r|<1


To show this simply note that rn tends to 0 as n tends to infinity. Then using the algebra of limits and multiplying the top and bottom y -1 we find the above expression to be true.

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