A geometric sequence is a sequence in which the ratio consecutive terms is constant.

Since this ratio is common to all consecutive pairs of terms, it is called the common ratio. It is denoted by r. If the ratio between consecutive terms is not constant, then the sequence is not geometric.

The formula for the common ratio of a geometric sequence is r = a_{n+1} /
a_{n}

A geometric sequence is an exponential function. Instead of y=a^{x}, we write
a_{n}=cr^{n} where r is the common ratio and c is a constant (not the first term of
the sequence, however).

A recursive definition, since each term is found by multiplying the previous term by the common
ratio, a_{k+1}=a_{k }* r.

The idea here is similar to that of the arithmetic sequence, except each term is multiplied by an additional factor of r. The exponent on the r will be one less than the term number. The first term hasn't been multiplied by r at all (exponent on r is 0). The second term has been multiplied by r once. The third term has been multiplied by r twice, and so on.

The formula for the general term of a geometric sequence is a_{n} = a_{1} r^{n-1}

A series is a sum of a sequence. We want to find the n^{th} partial sum or
the sum of the first n terms of the sequence. We will denote the n^{th}
partial sum as S_{n}.

Consider the geometric series S_{5} = 2 + 6 + 18 + 54 + 162. There
is a trick that can be used to find the sum of the series. In this series,
r=3.

S_{5} = 2 + 6 + 18 + 54 + 162

Multiply both sides of the equation by -r = -3

-3 S_{5} = - 6 - 18 - 54 - 162 - 486

Now add the two equations together. Notice that when you do that, all but the first and last terms cancel out.

S_{5} - 3S_{5} = 2 - 486

You can factor an S_{n} out of the left hand side and factor a 2 out
of the right hand side.

(1-3) S_{5} = 2 (1 - 243 )

Now divide both sides by 1-3

S_{5} = 2 ( 1 - 243 ) / ( 1 - 3 )

Yes, I realize I could have simplifed this. If you do, you get 2 ( -242 ) / ( -2) = 242. You're probably wondering why I didn't go ahead and simplify. Well, when you're trying to make a conjecture about the general case, it doesn't usually help to simplify and lose the original values.

Now, if we try to figure out where the different parts of that formula come
from, we can
conjecture about a formula for the n^{th} partial sum. The 2 in the
numerator is the first term, a_{1}.
The 243 in the numerator is the ratio times the n^{th} term - that makes
it the n+1 term, a_{1}*r^{n}. Since
both terms in the numerator have an a_{1} in them, that can be factored
out. The 1 in the denominator is always 1 and the 3 in denominator was the ratio,
r.

That makes the sum of the first n terms S_{n} = a_{1} (1-r^{n}) / (1-r).

There is an implied domain that r cannot equal 1, but since it is implied, it does not need to be stated.

The formula for the n^{th} partial sum of a geometric series is S_{n} =
a_{1} (1-r^{n}) / (1-r)

There is another type of geometric series, and infinite geometric series. An infinite geometric series is the sum of an infinite geometric sequence.

When the ratio has a magnitude greater than 1, the terms in the sequence will get larger and larger, and the if you add larger and larger numbers forever, you will get infinity for an answer. So, we don't deal with infinite geometric series when the magnitude of the ratio is greater than one.

The magnitude of the ratio can't equal one because that the series wouldn't be geometric and the sum formula would have division by zero.

The only case left, then, is when the magnitude of the ratio is less than one. Consider r=1/2. A sequence might be 1, 1/2, 1/4, 1/8, 1/16, 1/32, 1/64, 1/128, 1/256, 1/512, 1/1024, 1/2048, 1/4096, 1/8192, 1/16384, 1/32768, 1/65536, .... As the sequence goes on, the terms are getting smaller and smaller, approaching zero.

I said earlier that a geometric sequence was an exponential function. What happens to an
exponential function when the base is between 0 and 1? It was asymptotic to the x-axis (y=0) to
the right. That is, as x approached infinity, y approached 0. Well, the same thing happens here,
as n approaches infinity, r^{n} approaches 0. So, if you replace r^{n} with 0 in the summation formula,
the 1-r^{n} part just becomes 1, and the numerator just becomes a_{1}.

The formula for the sum of an infinite geometric series is S_{∞} = a_{1} / (1-r ).