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.

Examples of a geometric sequence are powers r^k of a fixed number r, such as 2^k and 3^k. The general form of a geometric sequence is

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

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 sequence 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 (unsigned) 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 ar^m. 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), n 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:

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

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:

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

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

${\begin{aligned}(1-r)\sum _{k=1}^{n}ar^{k-1}&=(1-r)(ar^{0}+ar^{1}+ar^{2}+ar^{3}+\cdots +ar^{n-1})\\&=ar^{0}+ar^{1}+ar^{2}+ar^{3}+\cdots +ar^{n-1}-ar^{1}-ar^{2}-ar^{3}-\cdots -ar^{n-1}-ar^{n}\\&=a-ar^{n}\end{aligned}}$

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 terms:

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

Related formulas

If one were to begin the sum not from k=1, 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:

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

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

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

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,

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

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

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

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

${\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+\cdots ={\frac {1/2}{1-(+1/2)}}=1.$

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

${\frac {1}{2}}-{\frac {1}{4}}+{\frac {1}{8}}-{\frac {1}{16}}+\cdots ={\frac {1/2}{1-(-1/2)}}={\frac {1}{3}}.$

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

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

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

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

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.)

$\prod _{i=0}^{n}ar^{i}=\left({\sqrt {a_{0}\cdot a_{n}}}\right)^{n+1}$ (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,

$P^{2}=(a_{0}\cdot a_{n})^{n+1}$ and

$P=(a_{0}\cdot a_{n})^{\frac {n+1}{2}}$,

which concludes the proof.

Relationship to geometry and Euclid's work

Books VIII and IX of Euclid's Elements analyzes geometric progressions (such as the powers of two, see the article for details) and give several of their properties.^[1]

See also

• Arithmetic progression
• Arithmetico-geometric sequence
• Linear difference equation
• Exponential function
• Harmonic progression
• Harmonic series
• Infinite series
• Preferred number
• Thomas Robert Malthus
• Geometric distribution

References

1. • Heath, Thomas L. (1956). The Thirteen Books of Euclid's Elements (2nd ed. [Facsimile. Original publication: Cambridge University Press, 1925] ed.). New York: Dover Publications.

• Hall & Knight, Higher Algebra, p. 39, ISBN 81-8116-000-2

External links

• Hazewinkel, Michiel, ed. (2001), "Geometric progression", Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4
• Derivation of formulas for sum of finite and infinite geometric progression at Mathalino.com
• Geometric Progression Calculator
• Nice Proof of a Geometric Progression Sum at sputsoft.com
• Weisstein, Eric W. "Geometric Series". MathWorld.