# Geometric progression

Diagram illustrating three basic geometric sequences of the pattern 1(rn-1) up to 6 iterations deep. The first block is a unit block and the dashed line represents the infinite sum of the sequence, a number that it will forever approach but never touch: $2$, $3/2$, and $4/3$, respectively.

In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For example, the sequence 2, 6, 18, 54, ... is a geometric progression with common ratio 3. Similarly 10, 5, 2.5, 1.25, ... is a geometric sequence with common ratio 1/2. The sum of the terms of a geometric progression, or of an initial segment of a geometric progression, is known as a geometric series.

Thus, the general form of a geometric sequence is

$a,\ ar,\ ar^2,\ ar^3,\ ar^4,\ \ldots$

and that of a geometric series is

$a + ar + ar^2 + ar^3 + ar^4 + \cdots$

where r ≠ 0 is the common ratio and a is a scale factor, equal to the sequence's start value.

## Elementary properties

The n-th term of a geometric sequence with initial value a and common ratio r is given by

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

Such a geometric sequence also follows the recursive relation

$a_n = r\,a_{n-1}$ for every integer $n\geq 1.$

Generally, to check whether a given sequence is geometric, one simply checks whether successive entries in the sequence all have the same ratio.

The common ratio of a geometric series may be negative, resulting in an alternating sequence, with numbers switching from positive to negative and back. For instance

1, −3, 9, −27, 81, −243, ...

is a geometric sequence with common ratio −3.

The behaviour of a geometric sequence depends on the value of the common ratio.
If the common ratio is:

• Positive, the terms will all be the same sign as the initial term.
• Negative, the terms will alternate between positive and negative.
• Greater than 1, there will be exponential growth towards positive or negative infinity (depending on the sign of the initial term).
• 1, the progression is a constant sequence.
• Between −1 and 1 but not zero, there will be exponential decay towards zero.
• −1, the progression is an alternating sequence
• Less than −1, for the absolute values there is exponential growth towards positive and negative infinity (due to the alternating sign).

Geometric sequences (with common ratio not equal to −1,1 or 0) show exponential growth or exponential decay, as opposed to the Linear growth (or decline) of an arithmetic progression such as 4, 15, 26, 37, 48, … (with common difference 11). This result was taken by T.R. Malthus as the mathematical foundation of his Principle of Population. Note that the two kinds of progression are related: exponentiating each term of an arithmetic progression yields a geometric progression, while taking the logarithm of each term in a geometric progression with a positive common ratio yields an arithmetic progression.

An interesting result of the definition of a geometric progression is that for any value of the common ratio, any three consecutive terms a, b and c will satisfy the following equation:

$b^2=ac$

where b is considered to be the geometric mean between a and c.

## Geometric series

 2 + 10 + 50 + 250 = 312 − ( 10 + 50 + 250 + 1250 = 5 × 312 ) 2 − 1250 = (1 − 5) × 312

Computation of the sum 2 + 10 + 50 + 250. The sequence is multiplied term by term by 5, and then subtracted from the original sequence. Two terms remain: the first term, a, and the term one beyond the last, or arm. The desired result, 312, is found by subtracting these two terms and dividing by 1 − 5.

A geometric series is the sum of the numbers in a geometric progression. For example:

$2 + 10 + 50 + 250 = 2 + 2 \times 5 + 2 \times 5^2 + 2 \times 5^3. \,$

Letting a be the first term (here 2), m be the number of terms (here 4), and r be the constant that each term is multiplied by to get the next term (here 5), the sum is given by:

$\frac{a(1-r^m)}{1-r}$

In the example above, this gives:

$2 + 10 + 50 + 250 = \frac{2(1-5^4)}{1-5} = \frac{-1248}{-4} = 312.$

The formula works for any real numbers a and r (except r = 1, which results in a division by zero). For example:

$-2\pi + 4\pi^2 - 8\pi^3 = -2\pi + (-2\pi)^2 + (-2\pi)^3 = \frac{-2\pi(1 - (-2\pi)^3)}{1-(-2\pi)} = \frac{-2\pi(1 + 8\pi^3)}{1+2\pi} \approx -214.855.$

### Derivation

To derive this formula, first write a general geometric series as:

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

We can find a simpler formula for this sum by multiplying both sides of the above equation by 1 − r, and we'll see that

\begin{align} (1-r) \sum_{k=0}^{n} ar^k & = (1-r)(ar^0 + ar^1+ar^2+ar^3+\cdots+ar^n) \\ & = ar^0 + ar^1+ar^2+ar^3+\cdots+ar^n \\ & {\color{White}{} = ar^0} - ar^1-ar^2-ar^3-\cdots-ar^n - ar^{n+1} \\ & = a - ar^{n+1} \end{align}

since all the other terms cancel. If r ≠ 1, we can rearrange the above to get the convenient formula for a geometric series that computes the sum of n+1 terms:

$\sum_{k=0}^{n} ar^k = \frac{a(1-r^{n+1})}{1-r}.$

### Related formulas

If one were to begin the sum not from k=0, but from a different value, say m, then

$\sum_{k=m}^n ar^k=\frac{a(r^m-r^{n+1})}{1-r}.$

Differentiating this formula with respect to r allows us to arrive at formulae for sums of the form

$\sum_{k=0}^n k^s r^k.$

For example:

$\frac{d}{dr}\sum_{k=0}^nr^k = \sum_{k=1}^n kr^{k-1}= \frac{1-r^{n+1}}{(1-r)^2}-\frac{(n+1)r^n}{1-r}.$

For a geometric series containing only even powers of r multiply by  1 − r2  :

$(1-r^2) \sum_{k=0}^{n} ar^{2k} = a-ar^{2n+2}.$

Then

$\sum_{k=0}^{n} ar^{2k} = \frac{a(1-r^{2n+2})}{1-r^2}.$

Equivalently, take  r2  as the common ratio and use the standard formulation.

For a series with only odd powers of r

$(1-r^2) \sum_{k=0}^{n} ar^{2k+1} = ar-ar^{2n+3}$

and

$\sum_{k=0}^{n} ar^{2k+1} = \frac{ar(1-r^{2n+2})}{1-r^2}.$

### 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 ( | r | < 1 ). Its value can then be computed from the finite sum formulae

$\sum_{k=0}^\infty ar^k = \lim_{n\to\infty}{\sum_{k=0}^{n} ar^k} = \lim_{n\to\infty}\frac{a(1-r^{n+1})}{1-r}= \lim_{n\to\infty}\frac{a}{1-r} - \lim_{n\to\infty}{\frac{ar^{n+1}}{1-r}}$
Diagram showing the geometric series 1 + 1/2 + 1/4 + 1/8 + ... which converges to 2.

Since:

$r^{n+1} \to 0 \mbox{ as } n \to \infty \mbox{ when } |r| < 1.$

Then:

$\sum_{k=0}^\infty ar^k = \frac{a}{1-r} - 0 = \frac{a}{1-r}$

For a series containing only even powers of $r$,

$\sum_{k=0}^\infty ar^{2k} = \frac{a}{1-r^2}$

and for odd powers only,

$\sum_{k=0}^\infty ar^{2k+1} = \frac{ar}{1-r^2}$

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

$\sum_{k=m}^\infty ar^k=\frac{ar^m}{1-r}$

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

$\frac{d}{dr}\sum_{k=0}^\infty r^k = \sum_{k=0}^\infty kr^{k-1}= \frac{1}{(1-r)^2}$

This formula only works for | r | < 1 as well. From this, it follows that, for | r | < 1,

$\sum_{k=0}^{\infty} k r^k = \frac{r}{\left(1-r\right)^2} \,;\, \sum_{k=0}^{\infty} k^2 r^k = \frac{r \left( 1+r \right)}{\left(1-r\right)^3} \, ; \, \sum_{k=0}^{\infty} k^3 r^k = \frac{r \left( 1+4 r + r^2\right)}{\left( 1-r\right)^4}$

Also, the infinite series 1/2 + 1/4 + 1/8 + 1/16 + · · · is an elementary example of a series that converges absolutely.

It is a geometric series whose first term is 1/2 and whose common ratio is 1/2, so its sum is

$\frac12+\frac14+\frac18+\frac{1}{16}+\cdots=\frac{1/2}{1-(+1/2)} = 1.$

The inverse of the above series is 1/2 − 1/4 + 1/8 − 1/16 + · · · is a simple example of an alternating series that converges absolutely.

It is a geometric series whose first term is 1/2 and whose common ratio is −1/2, so its sum is

$\frac12-\frac14+\frac18-\frac{1}{16}+\cdots=\frac{1/2}{1-(-1/2)} = \frac13.$

### Complex numbers

The summation formula for geometric series remains valid even when the common ratio is a complex number. In this case the condition that the absolute value of r be less than 1 becomes that the modulus of r be less than 1. It is possible to calculate the sums of some non-obvious geometric series. For example, consider the proposition

$\sum_{k=0}^{\infty} \frac{\sin(kx)}{r^k} = \frac{r \sin(x)}{1 + r^2 - 2 r \cos(x)}$

The proof of this comes from the fact that

$\sin(kx) = \frac{e^{ikx} - e^{-ikx}}{2i} ,$

which is a consequence of Euler's formula. Substituting this into the original series gives

$\sum_{k=0}^{\infty} \frac{\sin(kx)}{r^k} = \frac{1}{2 i} \left[ \sum_{k=0}^{\infty} \left( \frac{e^{ix}}{r} \right)^k - \sum_{k=0}^{\infty} \left(\frac{e^{-ix}}{r}\right)^k\right]$.

This is the difference of two geometric series, and so it is a straightforward application of the formula for infinite geometric series that completes the proof.

## Product

The product of a geometric progression is the product of all terms. If all terms are positive, then it can be quickly computed by taking the geometric mean of the progression's first and last term, and raising that mean to the power given by the number of terms. (This is very similar to the formula for the sum of terms of an arithmetic sequence: take the arithmetic mean of the first and last term and multiply with the number of terms.)

$\prod_{i=0}^{n} ar^i = \left( \sqrt{a_1 \cdot a_{n+1}}\right)^{n+1}$ (if $a,r > 0$).

Proof:

Let the product be represented by P:

$P=a \cdot ar \cdot ar^2 \cdots ar^{n-1} \cdot ar^{n}$.

Now, carrying out the multiplications, we conclude that

$P=a^{n+1} r^{1+2+3+ \cdots +(n-1)+(n)}$.

Applying the sum of arithmetic series, the expression will yield

$P=a^{n+1} r^{\frac{n(n+1)}{2}}$.
$P=(ar^{\frac{n}{2}})^{n+1}$.

We raise both sides to the second power:

$P^2=(a^2 r^{n})^{n+1}=(a\cdot ar^n)^{n+1}$.

Consequently

$P^2=(a_1 \cdot a_{n+1})^{n+1}$ and
$P=(a_1 \cdot a_{n+1})^{\frac{n+1}{2}}$,

which concludes the proof.

## Relationship to geometry and Euclid's work

Books VIII and IX of Euclid's Elements analyze geometric progressions and give several of their properties.

A geometric progression is given this name because each term is the geometric mean of the two adjacent terms.

### Elements, Book IX

The geometric progression 1, 2, 4, 8, 16, 32, … (or, in the binary numeral system, 1, 10, 100, 1000, 10000, 100000, … ) is important in number theory. Book IX, Proposition 36 of Elements proves that if the sum of the first n terms of this progression is a prime number, then this sum times the nth term is a perfect number. For example, the sum of the first 5 terms of the series 1 + 2 + 4 + 8 + 16 = 31, which is a prime number. The sum 31 multiplied by 16 (the 5th term in the series) equals 496, which is a perfect number.

Book IX, Proposition 35, proves that in a geometric series if the first term is subtracted from the second and last term in the sequence then as the excess of the second is to the first, so will the excess of the last be to all of those before it. (This is a restatement of our formula for geometric series from above.) Applying this to the geometric progression 31, 62, 124, 248, 496 (which results from 1, 2, 4, 8, 16 by multiplying all terms by 31), we see that 62 minus 31 is to 31 as 496 minus 31 is to the sum of 31, 62, 124, 248. Therefore the numbers 1, 2, 4, 8, 16, 31, 62, 124 and 248 add up to 496 and further these are all the numbers which divide 496. For suppose that p divides 496 and it is not amongst these numbers. Assume pq equals 16 × 31, or 31 is to q as p is to 16. Now p cannot divide 16 or it would be amongst the numbers 1, 2, 4, 8 or 16. Therefore 31 cannot divide q. And since 31 does not divide q and q measures 496, the fundamental theorem of arithmetic implies that q must divide 16 and be amongst the numbers 1, 2, 4, 8 or 16. Let q be 4, then p must be 124, which is impossible since by hypothesis p is not amongst the numbers 1, 2, 4, 8, 16, 31, 62, 124 or 248.