# Geometric progression

(Redirected from Geometric series)

In mathematics, a geometric progression (also known as a geometric sequence, and, inaccurately, as a geometric series; see below) is a sequence of numbers such that the quotient of any two successive members of the sequence is a constant called the common ratio of the sequence.

Thus without loss of generality a geometric sequence can be written as

$a_1=ar^0=a,ar^1=ar,ar^2,ar^3,...\,$

where r ≠ 0 is the common ratio and a is a scale factor. Thus the common ratio gives a family of geometric sequences whose starting value is determined by the scale factor. Pedantically speaking, the case r = 0 ought to be excluded, since the common ratio is not even defined; but the sequence that is always 0 is included, by convention.

## Formulae

Progressions allow the use of a few simple formulae to find each term. The nth term can be defined as

$a_n = a\,r^{n-1} \quad \mbox{where n is an integer such that }n \ge 1$

The common ratio is then

$r=\left(\frac{a_n}{a}\right)^{1/(n-1)} \quad \mbox{where n is an integer such that }n \ge 2$

and the scale factor is

$a=\frac{a_n}{r^{n-1}}.$

## Examples

A sequence with a common ratio of 2 and a scale factor of 1 is

1, 2, 4, 8, 16, 32, ....

A sequence with a common ratio of 2/3 and a scale factor of 729 is

729 (1, 2/3, 4/9, 8/27, 16/81, 32/243, 64/729, ....) = 729, 486, 324, 216, 144, 96, 64, ....

A sequence with a common ratio of −1 and a scale factor of 3 is

3 (1, −1, 1, −1, 1, −1, 1, −1, 1, −1, ....) = 3, −3, 3, −3, 3, −3, 3, −3, 3, −3, ....

A non-zero geometric progression shows exponential growth or exponential decay.

If the common ratio is:

Compare this with an arithmetic progression showing linear growth (or decline) such as 4, 15, 26, 37, 48, .... Note that the two kinds of progression are related: taking the logarithm of each term in a geometric progression yields an arithmetic one.

## Geometric series

A geometric series is, strictly speaking, the sum of the numbers in a geometric progression:

$\sum_{k=0}^{n} x^k = x^0+x^1+x^2+x^3+...+x^n \,$

We can find a simpler formula for this sum by multiplying both sides of the above equation by $(1-x)\,$ and we'll see that

$(1-x) \sum_{k=0}^{n} x^k = 1-x^{n+1}\,$

since all the other terms cancel. Rearranging gives the convenient formula for a geometric series:

$\sum_{k=0}^{n} x^k = \frac{1-x^{n+1}}{1-x}$

Note: If one were to begin the sum not from 0, but from a higher term, say m, then

$\sum_{k=m}^n x^k=\frac{x^m-x^{n+1}}{1-x}$

Differentiating the sum with respect to x allows us to arrive at formulae for sums of the form

$\sum_{k=0}^n k^s x^k$

For example:

$\frac{d}{dx}\sum_{k=0}^nx^k = \sum_{k=0}^nkx^{k-1}= \frac{1-x^{n+1}}{(1-x)^2}-\frac{(n+1)x^n}{1-x}$

### Infinite geometric series

An infinite geometric series is an infinite series whose successive terms have a common ratio. Such a series converges if and only if the absolute value of the common ratio is less than one. Its value can then be computed from the finite sum formulae by setting terms containing xn to zero:

$\sum_{k=0}^\infty x^k=\frac{1}{1-x}$

or, in cases where the sum does not start at k = 0,

$\sum_{k=a}^\infty x^k=\frac{x^a}{1-x}$

Both are valid only for |x| < 1. This last formula is actually valid in every Banach algebra, as long as the norm of x is less than one, and also in the field of p-adic numbers if |x|p < 1. As in the case for a finite sum, we can differentiate to calculate formulae for related sums. For example:

$\frac{d}{dx}\sum_{k=0}^\infty x^k = \sum_{k=0}^\infty kx^{k-1}= \frac{1}{(1-x)^2}$

This formula only works for |x| < 1, as well.