 Geometric Series
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Geometric Series Each of the purple squares has 1/4 of the area of the next larger square (1/2× = 1/4, 1/4×1/4 = 1/16, etc.). The sum of the areas of the purple squares is one third of the area of the large square. Another geometric series (common scale a = 4/9 and common ratio r = 1/9) shown as areas of purple squares. The total purple area is S = a / (1 - r) = (4/9) / (1 - (1/9)) = 1/2, which can be confirmed by observing that the outer square is partitioned into an infinite number of L-shaped areas each with four purple squares and four yellow squares, which is half purple.

In mathematics, a geometric series is a series with a constant ratio between successive terms. For example, the series

${\frac {1}{2}}\,+\,{\frac {1}{4}}\,+\,{\frac {1}{8}}\,+\,{\frac {1}{16}}\,+\,\cdots$ is geometric, because each successive term can be obtained by multiplying the previous term by 1/2.

Geometric series are among the simplest examples of infinite series with finite sums, although not all of them have this property. Historically, geometric series played an important role in the early development of calculus, and they continue to be central in the study of convergence of series. Geometric series are used throughout mathematics, and they have important applications in physics, engineering, biology, economics, computer science, queueing theory, and finance.

## Common ratio

The terms of a geometric series form a geometric progression, meaning that the ratio of successive terms in the series is constant. This relationship allows for the representation of a geometric series using only two terms, r and a. The term r is the common ratio, and a is the first term of the series. As an example the geometric series given in the introduction,

${\frac {1}{2}}\,+\,{\frac {1}{4}}\,+\,{\frac {1}{8}}\,+\,{\frac {1}{16}}\,+\,\cdots$ may simply be written as

$a+ar+ar^{2}+ar^{3}+\cdots$ , with $a={\frac {1}{2}}$ and $r={\frac {1}{2}}$ .

The following table shows several geometric series with different start terms and common ratios:

Start term, a Common ratio, r Example series
4 10 4 + 40 + 400 + 4000 + 40,000 + ···
9 1/3 9 + 3 + 1 + 1/3 + 1/9 + ···
7 1/10 7 + 0.7 + 0.07 + 0.007 + 0.0007 + ···
3 1 3 + 3 + 3 + 3 + 3 + ···
1 −1/2 1 − 1/2 + 1/4 − 1/8 + 1/16 − 1/32 + ···
3 –1 3 − 3 + 3 − 3 + 3 − ···

The behavior of the terms depends on the common ratio r:

If r is between −1 and +1, the terms of the series approach zero in the limit (becoming smaller and smaller in magnitude), and the series converges to a sum. In the case above, where r is 1/2, the series converges to 1.
If r is greater than one or less than minus one the terms of the series become larger and larger in magnitude. The sum of the terms also gets larger and larger, and the series has no sum. (The series diverges.)
If r is equal to one, all of the terms of the series are the same. The series diverges.
If r is minus one the terms take two values alternately (for example, 2, −2, 2, −2, 2,... ). The sum of the terms oscillates between two values (for example, 2, 0, 2, 0, 2,... ). This is a different type of divergence and again the series has no sum. See for example Grandi's series: 1 − 1 + 1 − 1 + ···.

## Sum

The sum of a geometric series is finite as long as the absolute value of the ratio is less than 1; as the numbers near zero, they become insignificantly small, allowing a sum to be calculated despite the series containing infinitely many terms. The sum can be computed using the self-similarity of the series.

### Example

Consider the sum of the following geometric series:

$s\;=\;1\,+\,{\frac {2}{3}}\,+\,{\frac {4}{9}}\,+\,{\frac {8}{27}}\,+\,\cdots .$ This series has common ratio 2/3. If we multiply through by this common ratio, then the initial 1 becomes a 2/3, the 2/3 becomes a 4/9, and so on:

${\frac {2}{3}}s\;=\;{\frac {2}{3}}\,+\,{\frac {4}{9}}\,+\,{\frac {8}{27}}\,+\,{\frac {16}{81}}\,+\,\cdots .$ This new series is the same as the original, except that the first term is missing. Subtracting the new series (2/3)s from the original series s cancels every term in the original but the first,

$s\,-\,{\frac {2}{3}}s\;=\;1,\;\;\;{\mbox{so }}s=3.$ A similar technique can be used to evaluate any self-similar expression.

### Formula The following is a geometric derivation of the closed form formula for the partial geometric series, S = rm + rm+1 + ... + rn-1 + rn when m < n and common ratio r > 1. Each term of the series ri is represented by the area of an overlapped square of area Ai that can be transformed into a non-overlapped L-shaped area Li = Ai - Ai-1 or, equivalently, Li+1 = Ai+1 - Ai. Due to being a geometric series, Ai+1 = r Ai. Therefore, Li+1 = Ai+1 - Ai = (r - 1) Ai, or Ai = Li+1 / (r - 1).

In words, each square is overlapped but can be transformed to a non-overlapped L-shaped area at the next larger square (next power of r) and scaled by 1 / (r - 1) so that the transformation from overlapped square to non-overlapped L-shaped area maintains the same area. Therefore the sum S = Am + Am+1 + ... + An-1 + An = (Lm+1 + Lm+2 + ... + Ln + Ln+1) / (r - 1). Observe that the non-overlapped L-shaped areas from L-shaped area n + 1 to L-shaped area m + 1 are a partition of the non-overlapped square An+1 less the upper right square notch Am, because there are no overlapped smaller squares to be transformed into that notch of area Am. Therefore, substituting Ai = ri and applying a common scale a results in the closed form S = (rn+1 - rm) a / (r - 1) when m < n and r > 1.

Although the above geometric proof assumes r > 1, the same closed form formula can be shown to apply to any value of r with the possible exception of r = 0 (depending on how you choose to define zero to the power of zero). For example for the case of r = 1, S = (1n+1 - 1m) a / (1 - 1) = 0 / 0. However, applying L'Hôpital's rule results in S = (n + 1 - m) a when r = 1.

For the case of 0 < r < 1, start with S = (rn+1 - rm) a / (r - 1) when m < n, r > 1 and let m = -? and n = 0 so S = a r / (r - 1) when r > 1. Dividing the numerator and denominator by r gives S = a / (1 - (1/r)) when r > 1, which is equivalent to S = a / (1 - r) when 0 < r < 1 because inverting r reverses the order of the series (biggest to smallest instead of smallest to biggest) but does not change the sum.

The range 0 < r < 1 can be extended to the range -1 < r < 1 by applying the derived formula, S = a / (1 - r) when 0 < r < 1, separately to two partitions of the geometric series: one with even powers of r (which cannot be negative) and the other with odd powers of r (which can be negative). The sum over both partitions is S = a / (1 - r2) + a r / (1 - r2) = a (1 + r) / ((1 + r)(1 - r)) = a / (1 - r).

For $r\neq 1$ , the sum of the first n terms of a geometric series is

$a+ar+ar^{2}+ar^{3}+\cdots +ar^{n-1}=\sum _{k=0}^{n-1}ar^{k}=a\left({\frac {1-r^{n}}{1-r}}\right),$ where a is the first term of the series, and r is the common ratio. One can derive the formula for the sum, s, as follows:

{\begin{aligned}s&=a+ar+ar^{2}+ar^{3}+\cdots +ar^{n-1},\\rs&=ar+ar^{2}+ar^{3}+\cdots +ar^{n-1}+ar^{n},\\s-rs&=a-ar^{n},\\s(1-r)&=a(1-r^{n}),\\s&=a\left({\frac {1-r^{n}}{1-r}}\right)\quad {\text{(if }}r\neq 1{\text{)}}.\end{aligned}} As n goes to infinity, the absolute value of r must be less than one for the series to converge. The sum then becomes

$a+ar+ar^{2}+ar^{3}+ar^{4}+\cdots =\sum _{k=0}^{\infty }ar^{k}={\frac {a}{1-r}},{\text{ for }}|r|<1.$ When a = 1, this can be simplified to

$1\,+\,r\,+\,r^{2}\,+\,r^{3}\,+\,\cdots \;=\;{\frac {1}{1-r}},$ the left-hand side being a geometric series with common ratio r.

The formula also holds for complex r, with the corresponding restriction, the modulus of r is strictly less than one.

### Proof of convergence

We can prove that the geometric series converges using the sum formula for a geometric progression:

{\begin{aligned}1+r+r^{2}+r^{3}+\cdots \ &=\lim _{n\rightarrow \infty }\left(1+r+r^{2}+\cdots +r^{n}\right)\\&=\lim _{n\rightarrow \infty }{\frac {1-r^{n+1}}{1-r}}.\end{aligned}} Since (1 + r + r2 + ... + rn)(1−r)

= ((1-r) + (r - r2) + (r2 - r3) + ... + (rn - rn+1))

= ((1-r) + (r - r2) + (r2 - r3) + ... + (rn - rn+1))

= 1−rn+1 and for | r | < 1.

Convergence of geometric series can also be demonstrated by rewriting the series as an equivalent telescoping series. Consider the function,

$g(K)={\frac {r^{K}}{1-r}}.$ Note that

$1=g(0)-g(1)\ ,\ r=g(1)-g(2)\ ,\ r^{2}=g(2)-g(3)\ ,\ldots$ Thus,

$S=1+r+r^{2}+r^{3}+\cdots =(g(0)-g(1))+(g(1)-g(2))+(g(2)-g(3))+\cdots .$ If

$|r|<1$ then

$g(K)\longrightarrow 0{\text{ as }}K\to \infty .$ So S converges to

$g(0)={\frac {1}{1-r}}.$ ## Applications

### Repeating decimals

A repeating decimal can be thought of as a geometric series whose common ratio is a power of 1/10. For example:

$0.7777\ldots \;=\;{\frac {7}{10}}\,+\,{\frac {7}{100}}\,+\,{\frac {7}{1000}}\,+\,{\frac {7}{10000}}\,+\,\cdots .$ The formula for the sum of a geometric series can be used to convert the decimal to a fraction,

$0.7777\ldots \;=\;{\frac {a}{1-r}}\;=\;{\frac {7/10}{1-1/10}}\;=\;{\frac {7/10}{9/10}}\;=\;{\frac {7}{9}}.$ The formula works not only for a single repeating figure, but also for a repeating group of figures. For example:

$0.123412341234\ldots \;=\;{\frac {a}{1-r}}\;=\;{\frac {1234/10000}{1-1/10000}}\;=\;{\frac {1234/10000}{9999/10000}}\;=\;{\frac {1234}{9999}}.$ Note that every series of repeating consecutive decimals can be conveniently simplified with the following:

$0.09090909\ldots \;=\;{\frac {09}{99}}\;=\;{\frac {1}{11}}.$ $0.143814381438\ldots \;=\;{\frac {1438}{9999}}.$ $0.9999\ldots \;=\;{\frac {9}{9}}\;=\;1.$ That is, a repeating decimal with repeat length n is equal to the quotient of the repeating part (as an integer) and 10n - 1.

### Archimedes' quadrature of the parabola

Archimedes used the sum of a geometric series to compute the area enclosed by a parabola and a straight line. His method was to dissect the area into an infinite number of triangles.

Archimedes' Theorem states that the total area under the parabola is 4/3 of the area of the blue triangle.

Archimedes determined that each green triangle has 1/8 the area of the blue triangle, each yellow triangle has 1/8 the area of a green triangle, and so forth.

Assuming that the blue triangle has area 1, the total area is an infinite sum:

$1\,+\,2\left({\frac {1}{8}}\right)\,+\,4\left({\frac {1}{8}}\right)^{2}\,+\,8\left({\frac {1}{8}}\right)^{3}\,+\,\cdots .$ The first term represents the area of the blue triangle, the second term the areas of the two green triangles, the third term the areas of the four yellow triangles, and so on. Simplifying the fractions gives

$1\,+\,{\frac {1}{4}}\,+\,{\frac {1}{16}}\,+\,{\frac {1}{64}}\,+\,\cdots .$ This is a geometric series with common ratio and the fractional part is equal to

$\sum _{n=0}^{\infty }4^{-n}=1+4^{-1}+4^{-2}+4^{-3}+\cdots ={4 \over 3}.$ The sum is

${\frac {1}{1-r}}\;=\;{\frac {1}{1-{\frac {1}{4}}}}\;=\;{\frac {4}{3}}.$ This computation uses the method of exhaustion, an early version of integration. Using calculus, the same area could be found by a definite integral.

### Fractal geometry

In the study of fractals, geometric series often arise as the perimeter, area, or volume of a self-similar figure.

For example, the area inside the Koch snowflake can be described as the union of infinitely many equilateral triangles (see figure). Each side of the green triangle is exactly 1/3 the size of a side of the large blue triangle, and therefore has exactly 1/9 the area. Similarly, each yellow triangle has 1/9 the area of a green triangle, and so forth. Taking the blue triangle as a unit of area, the total area of the snowflake is

$1\,+\,3\left({\frac {1}{9}}\right)\,+\,12\left({\frac {1}{9}}\right)^{2}\,+\,48\left({\frac {1}{9}}\right)^{3}\,+\,\cdots .$ The first term of this series represents the area of the blue triangle, the second term the total area of the three green triangles, the third term the total area of the twelve yellow triangles, and so forth. Excluding the initial 1, this series is geometric with constant ratio r = 4/9. The first term of the geometric series is a = 3(1/9) = 1/3, so the sum is

$1\,+\,{\frac {a}{1-r}}\;=\;1\,+\,{\frac {\frac {1}{3}}{1-{\frac {4}{9}}}}\;=\;{\frac {8}{5}}.$ Thus the Koch snowflake has 8/5 of the area of the base triangle.

The convergence of a geometric series reveals that a sum involving an infinite number of summands can indeed be finite, and so allows one to resolve many of Zeno's paradoxes. For example, Zeno's dichotomy paradox maintains that movement is impossible, as one can divide any finite path into an infinite number of steps wherein each step is taken to be half the remaining distance. Zeno's mistake is in the assumption that the sum of an infinite number of finite steps cannot be finite. This is of course not true, as evidenced by the convergence of the geometric series with $r=1/2$ .

This, however, is not a complete resolution to Zeno's dichotomy paradox. Strictly speaking, unless we allow for time to move in reverse, where the step size begins with $r=1/2$ and approaches zero as a limit, this infinite series would otherwise have to begin with an infinitesimally small step. Treating infinitesimals in this way is typically not something which is rigorously defined mathematically, outside of Nonstandard Calculus. So, while it is true that the entire infinite summation yields a finite number, we can not create a simple ordering of the terms when starting from an infinitesimal, and therefore we can not adequately describe the first step of any given action.

### Euclid

Book IX, Proposition 35 of Euclid's Elements expresses the partial sum of a geometric series in terms of members of the series. It is equivalent to the modern formula.

### Economics

In economics, geometric series are used to represent the present value of an annuity (a sum of money to be paid in regular intervals).

For example, suppose that a payment of $100 will be made to the owner of the annuity once per year (at the end of the year) in perpetuity. Receiving$100 a year from now is worth less than an immediate $100, because one cannot invest the money until one receives it. In particular, the present value of$100 one year in the future is $100 / (1 + $I$ ), where $I$ is the yearly interest rate. Similarly, a payment of$100 two years in the future has a present value of $100 / (1 + $I$ )2 (squared because two years' worth of interest is lost by not receiving the money right now). Therefore, the present value of receiving$100 per year in perpetuity is

$\sum _{n=1}^{\infty }{\frac {\100}{(1+I)^{n}}},$ which is the infinite series:

${\frac {\100}{(1+I)}}\,+\,{\frac {\100}{(1+I)^{2}}}\,+\,{\frac {\100}{(1+I)^{3}}}\,+\,{\frac {\100}{(1+I)^{4}}}\,+\,\cdots .$ This is a geometric series with common ratio 1 / (1 + $I$ ). The sum is the first term divided by (one minus the common ratio):

${\frac {\100/(1+I)}{1-1/(1+I)}}\;=\;{\frac {\100}{I}}.$ For example, if the yearly interest rate is 10% ($I$ = 0.10), then the entire annuity has a present value of $100 / 0.10 =$1000.

This sort of calculation is used to compute the APR of a loan (such as a mortgage loan). It can also be used to estimate the present value of expected stock dividends, or the terminal value of a security.

### Geometric power series

The formula for a geometric series

${\frac {1}{1-x}}=1+x+x^{2}+x^{3}+x^{4}+\cdots$ can be interpreted as a power series in the Taylor's theorem sense, converging where $|x|<1$ . From this, one can extrapolate to obtain other power series. For example,

{\begin{aligned}\tan ^{-1}(x)&=\int {\frac {dx}{1+x^{2}}}\\&=\int {\frac {dx}{1-(-x^{2})}}\\&=\int \left(1+\left(-x^{2}\right)+\left(-x^{2}\right)^{2}+\left(-x^{2}\right)^{3}+\cdots \right)dx\\&=\int \left(1-x^{2}+x^{4}-x^{6}+\cdots \right)dx\\&=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+\cdots \\&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}x^{2n+1}.\end{aligned}} By differentiating the geometric series, one obtains the variant

$\sum _{n=1}^{\infty }nx^{n-1}={\frac {1}{(1-x)^{2}}}\quad {\text{ for }}|x|<1.$ Similarly obtained are:

$\sum _{n=2}^{\infty }n(n-1)x^{n-2}={\frac {2}{(1-x)^{3}}}\quad {\text{ for }}|x|<1,$ and
$\sum _{n=3}^{\infty }n(n-1)(n-2)x^{n-3}={\frac {6}{(1-x)^{4}}}\quad {\text{ for }}|x|<1.$ 