Geometric progression

In mathematics, a geometric progression (also known as a geometric sequence or a geometric series) is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed number called the common ratio.

example: 1,3,9,27... common ratio is 3.

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

$a,ar,ar^{2},ar^{3},ar^{4},...$

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.

Euclid's books VIII and IX analyze a geometric progression. A geometric progression gains its geometric character from the very important mathematical theorem that two similar plane figures or numbers are in duplicate ratio to their corresponding sides. Further two similar solid figures and numbers are in triplicate ratio of their corresponding sides.

For example the square numbers 4 and 9 are in the ratio 4 to 6 to 9, which is the duplicate ratio of their sides, 2 and 3. The cube numbers 8 and 125 are in the ratio of 8 to 20 to 50 to 125, which is the triplicate ratio of their sides. In the geometric progression 4,6,9 the number 6 is the mean proportional between 4 and 9, and in the geometric progression 8,20,50,125 the numbers 20 and 50 are the mean proportionals between 8 and 125. All squares have one mean proportional between them all cubes have two mean proportionals between and so on to higher dimensions.

Formulae

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

$a_{n}=a_{1}\,r^{n-1}$	where n is an integer such that $n\geq 1$
The common ratio is then
$r=\left({\frac {a_{n}}{a}}\right)^{\frac {1}{n-1}}$ or $r={\sqrt[{n-1}]{\frac {a_{n}}{a}}}$	where n is an integer such that $n\geq 2$
and the scale factor is
$a_{1}={\frac {a_{n}}{r^{n-1}}}$	where n is an integer such that $n\geq 1.$

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Geometric series

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

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

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

$(1-r)\sum _{k=0}^{n}ar^{k}=a-ar^{n+1}\,$

since all the other terms cancel. Rearranging (for $r\neq 1$) gives the convenient formula for a geometric series:

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

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

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

Differentiating the sum 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:

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

For a geometric series containing only even powers of $r$ multiply by $(1-r^{2})$:

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

Then

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

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

$\sum _{k=0}^{\infty }ar^{k}=\lim _{n\to \infty }{\sum _{k=0}^{n}ar^{k}}=\lim _{n\to \infty }{\frac {a(1-r^{n+1})}{1-r}}=\lim _{n\to \infty }{\frac {a}{1-r}}-\lim _{n\to \infty }{\frac {ar^{n+1}}{1-r}}={\frac {a}{1-r}}$

For example,

$\sum _{k=0}^{\infty }(191)\left({\frac {6}{7}}\right)^{k}={\frac {191}{1-{\frac {6}{7}}}}=1337$

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,

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

Above formulae are valid only for | r | < 1. The latter formula is actually 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,

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

Complex numbers

The summation formula for geometric series remains valid even when the common ratio is a complex number. This fact can be used, along with Euler's formula, to calculate some sums of non-obvious geometric series, such as:

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

It is clear that this is just the difference of two geometric series. From here, it is straightforward formula application to calculate that

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

Product

The product of a geometric progression is the product of all terms, and can be mathematically defined as

$\prod _{i=0}^{n}ar^{i}=(a_{1}\cdot a_{n+1})^{\frac {n+1}{2}}$.

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.

Book IX, Proposition 36

The geometric progression 1,2,4,8,16... or in binary 1,10,100,1000,10000... is very interesting and important in number theory. In Euclid's book IX, proposition 36 it is proved that if the sum of the first n terms of this progression is prime, then this sum times the nth term is a perfect number.

In order to facilitate an understanding of this theorem, let's prove that 496 is perfect number using Euclid's method. The series 1,2,4,8,16 sums to 31 (prime) and this number multiplied by 16 equals 496. The geometric progression 31,62,124,248,496 is a progression with same ratio as 1,2,4,8,16. In Book IX, proposition 35 Euclid 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. Therefore 496 minus 31 is to the sum of 31,62,124,248 as 62 minus 31 is to 31. therefore the numbers 1,2,4,8,16,31,62,124,248 add up to 496 and further these are all the numbers which measure 496. For suppose that P measures 496 and it is not amongst these numbers. Let P measure it according to Q and therefore P×Q equals 16×31, or 31 is to Q as P is to 16. Now P cannot measure 16 or it would be amongst the numbers 1,2,4,8,16. Therefore 31 cannot measure Q. And since 31 does not measure Q and Q measures 496, the fundamental theory of arithmetic implies that Q must divide 16 and be amongst the numbers 1,2,4,8,16. Let Q be 4, then P must be 124, which is impossible since by hypothesis P is not amongst the numbers 1,2,4,8,16,31,62,124,248.

See also

• Arithmetic progression
• Exponential function
• Harmonic series
• Infinite series
• Thomas Robert Malthus