# Geometric series The geometric series 1/4 + 1/16 + 1/64 + 1/256 + ... shown as areas of purple squares. Each of the purple squares has 1/4 of the area of the next larger square (1/2×1/2 = 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 (coefficient 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 the sum of an infinite number of terms that have 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. In general, a geometric series is written as a + ar + ar2 + ar3 + ... , where a is the coefficient of each term and r is the common ratio between adjacent terms. Geometric series are among the simplest examples of infinite series and can serve as a basic introduction to Taylor series and Fourier series. Geometric series had an important role in the early development of calculus, are used throughout mathematics, and have important applications in physics, engineering, biology, economics, computer science, queueing theory, and finance.

The distinction between a progression and a series is that a progression is a sequence, whereas a series is a sum.

## Coefficient a The first nine terms of geometric series 1 + r + r2 + r3 ... drawn as functions (colored in the order red, green, blue, red, green, blue, ...) within the range -1 < r < 1. The closed form geometric series 1 / (1 - r) is the black dashed line.

The geometric series a + ar + ar2 + ar3 + ... is written in expanded form. Every coefficient in the geometric series is the same. In contrast, the power series written as a0 + a1r + a2r2 + a3r3 + ... in expanded form has coefficients ai that can vary from term to term. In other words, the geometric series is a special case of the power series. The first term of a geometric series in expanded form is the coefficient a of that geometric series.

In addition to the expanded form of the geometric series, there is a generator form of the geometric series written as

$\sum _{k=0}^{\infty }$ ark

and a closed form of the geometric series written as

a / (1 - r) within the range |r| < 1.

The derivation of the closed form from the expanded form is shown in this article's Sum section. The derivation requires that all the coefficients of the series be the same (coefficient a) in order to take advantage of self-similarity and to reduce the infinite number of additions and power operations in the expanded form to the single subtraction and single division in the closed form. However even without that derivation, the result can be confirmed with long division: a divided by (1 - r) results in a + ar + ar2 + ar3 + ... , which is the expanded form of the geometric series.

Typically a geometric series is thought of as a sum of numbers a + ar + ar2 + ar3 + ... but can also be thought of as a sum of functions a + ar + ar2 + ar3 + ... that converges to the function a / (1 - r) within the range |r| < 1. The adjacent image shows the contribution each of the first nine terms (i.e., functions) make to the function a / (1 - r) within the range |r| < 1 when a = 1. Changing even one of the coefficients to something other than coefficient a would (in addition to changing the geometric series to a power series) change the resulting sum of functions to some function other than a / (1 - r) within the range |r| < 1. As an aside, a particularly useful change to the coefficients is defined by the Taylor series, which describes how to change the coefficients so that the sum of functions converges to any user selected, sufficiently smooth function within a range.

## Common ratio r Close-up view of the cumulative sum of functions within the range -1 < r < -0.5 as the first 11 terms of the geometric series 1 + r + r2 + r3 + ... are added. The geometric series 1 / (1 - r) is the red dashed line. Play media
Complex geometric series (coefficient a = 1 and common ratio r = 0.5 e0t) converging to a circle. In the animation, each term of the geometric series is drawn as a vector twice: once at the origin and again within the head-to-tail vector summation that converges to the circle. The circle intersects the real axis at 2 (= 1/(1-1/2) when θ = 0) and at 2/3 (= 1/(1-(-1/2)) when θ = 180 degrees).

The geometric series a + ar + ar2 + ar3 + ... is an infinite series defined by just two parameters: coefficient a and common ratio r. Common ratio r is the ratio of any term with the previous term in the series. Or equivalently, common ratio r is the term multiplier used to calculate the next term in the series. The following table shows several geometric series:

a r Example series
4 10 4 + 40 + 400 + 4000 + 40,000 + ···
3 1 3 + 3 + 3 + 3 + 3 + ···
1 2/3 1 + 2/3 + 4/9 + 8/27 + 16/81 + ···
1/2 1/2 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ···
9 1/3 9 + 3 + 1 + 1/3 + 1/9 + ···
7 1/10 7 + 0.7 + 0.07 + 0.007 + 0.0007 + ···
1 −1/2 1 − 1/2 + 1/4 − 1/8 + 1/16 − 1/32 + ···
3 −1 3 − 3 + 3 − 3 + 3 − ···

The convergence of the geometric series depends on the value of the common ratio r:

• If |r| < 1, the terms of the series approach zero in the limit (becoming smaller and smaller in magnitude), and the series converges to the sum a / (1 - r).
• If |r| = 1, the series does not converge. When r = 1, all of the terms of the series are the same and the series is infinite. When r = −1, 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. See for example Grandi's series: 1 − 1 + 1 − 1 + ···.
• If |r| > 1, the terms of the series become larger and larger in magnitude. The sum of the terms also gets larger and larger, and the series does not converge to a sum. (The series diverges.)

The rate of convergence also depends on the value of the common ratio r. Specifically, the rate of convergence gets slower as r approaches 1 or −1. For example, the geometric series with a = 1 is 1 + r + r2 + r3 + ... and converges to 1 / (1 - r) when |r| < 1. However, the number of terms needed to converge approaches infinity as r approaches 1 because a / (1 - r) approaches infinity and each term of the series is less than or equal to one. In contrast, as r approaches −1 the sum of the first several terms of the geometric series starts to converge to 1/2 but slightly flips up or down depending on whether the most recently added term has a power of r that is even or odd. That flipping behavior near r = −1 is illustrated in the adjacent image showing the first 11 terms of the geometric series with a = 1 and |r| < 1.

The common ratio r and the coefficient a also define the geometric progression, which is a list of the terms of the geometric series but without the additions. Therefore the geometric series a + ar + ar2 + ar3 + ... has the geometric progression (also called the geometric sequence) a, ar, ar2, ar3, ... The geometric progression - as simple as it is - models a surprising number of natural phenomena,

As an aside, the common ratio r can be a complex number such as |r|eiθ where |r| is the vector's magnitude (or length), θ is the vector's angle (or orientation) in the complex plane and i2 = -1. With a common ratio |r|eiθ, the expanded form of the geometric series is a + a|r|eiθ + a|r|2ei2θ + a|r|3ei3θ + ... Modeling the angle θ as linearly increasing over time at the rate of some angular frequency ω0 (in other words, making the substitution θ = ω0t), the expanded form of the geometric series becomes a + a|r|eiω0t + a|r|2ei2ω0t + a|r|3ei3ω0t + ... , where the first term is a vector of length a not rotating at all, and all the other terms are vectors of different lengths rotating at harmonics of the fundamental angular frequency ω0. The constraint |r|<1 is enough to coordinate this infinite number of vectors of different lengths all rotating at different speeds into tracing a circle, as shown in the adjacent video. Similar to how the Taylor series describes how to change the coefficients so the series converges to a user selected sufficiently smooth function within a range, the Fourier series describes how to change the coefficients (which can also be complex numbers in order to specify the initial angles of vectors) so the series converges to a user selected periodic function.

## Sum

### Closed-form formula Offering a different perspective from the well-known algebraic derivation given in the adjacent section, the following is a geometric derivation of the closed form of the geometric series. As an overview, this geometric derivation represents terms of the geometric series as areas of overlapped squares and the goal is to transform those overlapped squares into an easily calculated non-overlapped area. The starting point is the partial sum 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 m + 1 to L-shaped area n + 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 scaling all the terms by coefficient a results in the general, closed-form, geometric series 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 = ar / (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) + ar / (1 - r2) = a (1 + r) / ((1 + r)(1 - r)) = a / (1 - r).

For $r\neq 1$ , the sum of the first n+1 terms of a geometric series, up to and including the r n term, is

$a+ar+ar^{2}+ar^{3}+\cdots +ar^{n}=\sum _{k=0}^{n}ar^{k}=a\left({\frac {1-r^{n+1}}{1-r}}\right),$ where r is the common ratio. One can derive that closed-form formula for the partial sum, s, by subtracting out the many self-similar terms as follows:

{\begin{aligned}s=a\ +\ &ar\ +\ ar^{2}\ +\ ar^{3}\ +\ \cdots \ +\ ar^{n},\\rs=\ &ar\ +\ ar^{2}\ +\ ar^{3}\ +\ \cdots \ +\ ar^{n}\ +\ ar^{n+1},\\s-rs=\ &a\ -\ ar^{n+1},\\s(1-r)=\ &a(1-r^{n+1}),\\s=\ &a\left({\frac {1-r^{n+1}}{1-r}}\right)\quad {\text{(if }}r\neq 1{\text{)}}.\end{aligned}} As n approaches 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 formula also holds for complex r, with the corresponding restriction, the modulus of r is strictly less than one.

As an aside, the question of whether an infinite series converges is fundamentally a question about the distance between two values: given enough terms, does the value of the partial sum get arbitrarily close to the value it is approaching? In the above derivation of the closed form of the geometric series, the interpretation of the distance between two values is the distance between their locations on the number line. That is the most common interpretation of distance between two values. However the p-adic metric, which has become a critical notion in modern number theory, offers a definition of distance such that the geometric series 1 + 2 + 4 + 8 + ... with a = 1 and r = 2 actually does converge to a / (1 - r) = 1 / (1 - 2) = -1 even though r is outside the typical convergence range |r| < 1.

### 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}} The second equality is true because if $|r|<1,$ then $r^{n+1}\to 0$ as $n\to \infty$ and

{\begin{aligned}(1+r+r^{2}+\cdots +r^{n})(1-r)&=((1-r)+(r-r^{2})+(r^{2}-r^{3})+...+(r^{n}-r^{n+1}))\\&=(1+(-r+r)+(-r^{2}+r^{2})+...+(-r^{n}+r^{n})-r^{n+1})\\&=1-r^{n+1}.\end{aligned}} 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

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

$g(0)={\frac {1}{1-r}}.$ ### Rate of convergence The cumulative sum of functions within the range -1 < r < -0.5 as the first 51 terms of the geometric series 1 + r + r2 + r3 + ... are added. The magnitude of slope at r = -1 increases very gradually with each added term. The geometric series 1 / (1 - r) is the red dashed line.

As shown in the above proofs, the closed form of the geometric series partial sum up to and including the n-th power of r is a(1 - rn+1) / (1 - r) for any value of r, and the closed form of the geometric series is the full sum a / (1 - r) within the range |r| < 1.

If the common ratio is within the range 0 < r < 1, then the partial sum a(1 - rn+1) / (1 - r) increases with each added term and eventually gets within some small error, E, ratio of the full sum a / (1 - r). Solving for n at that error threshold,

{\begin{aligned}a\left({\frac {1-r^{n+1}}{1-r}}\right)&\geq a\left({\frac {1-E}{1-r}}\right),\\1-r^{n+1}&\geq 1-E,\\r^{n+1}&\leq E,\\\ln(r^{n+1})&\leq \ln(E),\\(n+1)\ln(r)&\leq \ln(E),\\n+1&\geq \left\lceil {\frac {\ln(E)}{\ln(r)}}\right\rceil ,\end{aligned}} where 0 < r < 1, the ceiling operation $\lceil \rceil$ constrains n to integers, and dividing both sides by the natural log of r flips the inequality because it is negative. The result n+1 is the number of partial sum terms needed to get within aE / (1 - r) of the full sum a / (1 - r). For example to get within 1% of the full sum a / (1 - r) at r=0.1, only 2 (= ln(E) / ln(r) = ln(0.01) / ln(0.1)) terms of the partial sum are needed. However at r=0.9, 44 (= ln(0.01) / ln(0.9)) terms of the partial sum are needed to get within 1% of the full sum a / (1 - r).

If the common ratio is within the range -1 < r < 0, then the geometric series is an alternating series but can be converted into the form of a non-alternating geometric series by combining pairs of terms and then analyzing the rate of convergence using the same approach as shown for the common ratio range 0 < r < 1. Specifically, the partial sum

s = a + ar + ar2 + ar3 + ar4 + ar5 + ... + arn-1 + arn within the range -1 < r < 0 is equivalent to
s = a - ap + ap2 - ap3 + ap4 - ap5 + ... + apn-1 - apn with an n that is odd, with the substitution of p = -r, and within the range 0 < p < 1,
s = (a - ap) + (ap2 - ap3) + (ap4 - ap5) + ... + (apn-1 - apn) with adjacent and differently signed terms paired together,
s = a(1 - p) + a(1 - p)p2 + a(1 - p)p4 + ... + a(1 - p)p2(n-1)/2 with a(1 - p) factored out of each term,
s = a(1 - p) + a(1 - p)p2 + a(1 - p)p4 + ... + a(1 - p)p2m with the substitution m = (n - 1) / 2 which is an integer given the constraint that n is odd,

which is now in the form of the first m terms of a geometric series with coefficient a(1 - p) and with common ratio p2. Therefore the closed form of the partial sum is a(1 - p)(1 - p2(m+1)) / (1 - p2) which increases with each added term and eventually gets within some small error, E, ratio of the full sum a(1 - p) / (1 - p2). As before, solving for m at that error threshold,

{\begin{aligned}a(1-p)\left({\frac {1-p^{2m+2}}{1-p^{2}}}\right)&\geq a(1-p)\left({\frac {1-E}{1-p^{2}}}\right),\\1-p^{2m+2}&\geq 1-E,\\p^{2m+2}&\leq E,\\\ln(p^{2m+2})&\leq \ln(E),\\(2m+2)\ln(p)&\leq \ln(E),\\m+1&\geq \left\lceil {\frac {\ln(E)}{2\ln(p)}}\right\rceil ,\\m+1&\geq \left\lceil {\frac {\ln(E)}{2\ln(-r)}}\right\rceil ,\end{aligned}} where 0 < p < 1 or equivalently -1 < r < 0, and the m+1 result is the number of partial sum pairs of terms needed to get within a(1 - p)E / (1 - p2) of the full sum a(1 - p) / (1 - p2). For example to get within 1% of the full sum a(1 - p) / (1 - p2) at p=0.1 or equivalently r=-0.1, only 1 (= ln(E) / (2 ln(p)) = ln(0.01) / (2 ln(0.1)) pair of terms of the partial sum are needed. However at p=0.9 or equivalently r=-0.9, 22 (= ln(0.01) / (2 ln(0.9))) pairs of terms of the partial sum are needed to get within 1% of the full sum a(1 - p) / (1 - p2). Comparing the rate of convergence for positive and negative values of r, n + 1 (the number of terms required to reach the error threshold for some positive r) is always twice as large as m + 1 (the number of term pairs required to reach the error threshold for the negative of that r) but the m + 1 refers to term pairs instead of single terms. Therefore, the rate of convergence is symmetric about r = 0, which can be a surprise given the asymmetry of a / (1 - r). One perspective that helps explain this rate of convergence symmetry is that on the r > 0 side each added term of the partial sum makes a finite contribution to the infinite sum at r = 1 while on the r < 0 side each added term makes a finite contribution to the infinite slope at r = -1.

As an aside, this type of rate of convergence analysis is particularly useful when calculating the number of Taylor series terms needed to adequately approximate some user-selected sufficiently-smooth function or when calculating the number of Fourier series terms needed to adequately approximate some user-selected periodic function.

## Historic insights

### Zeno of Elea (c.495 – c.430 BC)

2,500 years ago, Greek mathematicians had a problem with walking from one place to another. Physically, they were able to walk as well as we do today, perhaps better. Logically, however, they thought that an infinitely long list of numbers greater than zero summed to infinity. Therefore, it was a paradox when Zeno of Elea pointed out that in order to walk from one place to another, you first have to walk half the distance, and then you have to walk half the remaining distance, and then you have to walk half of that remaining distance, and you continue halving the remaining distances an infinite number of times because no matter how small the remaining distance is you still have to walk the first half of it. Thus, Zeno of Elea transformed a short distance into an infinitely long list of halved remaining distances, all of which are greater than zero. And that was the problem: how can a distance be short when measured directly and also infinite when summed over its infinite list of halved remainders? The paradox revealed something was wrong with the assumption that an infinitely long list of numbers greater than zero summed to infinity.

### Euclid of Alexandria (c.300 BC) Elements of Geometry, Book IX, Proposition 35. "If there is any multitude whatsoever of continually proportional numbers, and equal to the first is subtracted from the second and the last, then as the excess of the second to the first, so the excess of the last will be to all those before it." For the same case of common ratio r>1, there are other geometric proofs of the geometric series partial sum closed form. One that represents terms as areas rather than lengths of line segments is summarized, as shown in the above figure, in three steps: (TOP) Represent the terms of a geometric series as the areas of overlapped similar triangles. (MIDDLE) From the largest to the smallest triangle, remove the overlapped left area portion (1/r) from the non-overlapped right area portion (1-1/r = (r-1)/r) and scale that non-overlapped trapezoid by r/(r-1) so its area is the same as the area of the original overlapped triangle. (BOTTOM) Notice that the area of the aggregate trapezoid is the area of a large triangle less the area of an empty small triangle at the large triangle's left tip. The large triangle is exactly the largest overlapped triangle scaled by r/(r-1). The empty small triangle started as a but that area was transformed into a non-overlapped scaled trapezoid leaving an empty left area portion (1/r). However, that empty triangle of area a/r must also be scaled by r/(r-1) so its slope matches the slope of all the non-overlapped scaled trapezoids. Therefore, Sn = area of large triangle - area of empty small triangle = arn+1/(r-1) - a/(r-1) = a(rn+1-1)/(r-1).

Euclid's Elements of Geometry Book IX, Proposition 35, proof (of the proposition in adjacent diagram's caption):

Let AA', BC, DD', EF be any multitude whatsoever of continuously proportional numbers, beginning from the least AA'. And let BG and FH, each equal to AA', have been subtracted from BC and EF. I say that as GC is to AA', so EH is to AA', BC, DD'.

For let FK be made equal to BC, and FL to DD'. And since FK is equal to BC, of which FH is equal to BG, the remainder HK is thus equal to the remainder GC. And since as EF is to DD', so DD' to BC, and BC to AA' [Prop. 7.13], and DD' equal to FL, and BC to FK, and AA' to FH, thus as EF is to FL, so LF to FK, and FK to FH. By separation, as EL to LF, so LK to FK, and KH to FH [Props. 7.11, 7.13]. And thus as one of the leading is to one of the following, so (the sum of) all of the leading to (the sum of) all of the following [Prop. 7.12]. Thus, as KH is to FH, so EL, LK, KH to LF, FK, HF. And KH equal to CG, and FH to AA', and LF, FK, HF to DD', BC, AA'. Thus, as CG is to AA', so EH to DD', BC, AA'. Thus, as the excess of the second is to the first, so is the excess of the last is to all those before it. The very thing it was required to show.

The terseness of Euclid's propositions and proofs may have been a necessity. As is, the Elements of Geometry is over 500 pages of propositions and proofs. Making copies of this popular textbook was labor intensive given that the printing press was not invented until 1440. And the book's popularity lasted a long time: as stated in the cited introduction to an English translation, Elements of Geometry "has the distinction of being the world's oldest continuously used mathematical textbook." So being very terse was being very practical. The proof of Proposition 35 in Book IX could have been even more compact if Euclid could have somehow avoided explicitly equating lengths of specific line segments from different terms in the series. For example, the contemporary notation for geometric series (i.e., a + ar + ar2 + ar3 + ... + arn) does not label specific portions of terms that are equal to each other.

Also in the cited introduction the editor comments,

Most of the theorems appearing in the Elements were not discovered by Euclid himself, but were the work of earlier Greek mathematicians such as Pythagoras (and his school), Hippocrates of Chios, Theaetetus of Athens, and Eudoxus of Cnidos. However, Euclid is generally credited with arranging these theorems in a logical manner, so as to demonstrate (admittedly, not always with the rigour demanded by modern mathematics) that they necessarily follow from five simple axioms. Euclid is also credited with devising a number of particularly ingenious proofs of previously discovered theorems (e.g., Theorem 48 in Book 1).

To help translate the proposition and proof into a form that uses current notation, a couple modifications are in the diagram. First, the four horizontal line lengths representing the values of the first four terms of a geometric series are now labeled a, ar, ar2, ar3 in the diagram's left margin. Second, new labels A' and D' are now on the first and third lines so that all the diagram's line segment names consistently specify the segment's starting point and ending point.

Here is a phrase by phrase interpretation of the proposition:

Proposition in contemporary notation
"If there is any multitude whatsoever of continually proportional numbers" Taking the first n+1 terms of a geometric series Sn = a + ar + ar2 + ar3 + ... + arn
"and equal to the first is subtracted from the second and the last" and subtracting a from ar and arn
"then as the excess of the second to the first, so the excess of the last will be to all those before it." then (ar-a) / a = (arn-a) / (a + ar + ar2 + ar3 + ... + arn-1) = (arn-a) / Sn-1, which can be rearranged to the more familiar form Sn-1 = a(rn-1) / (r-1).

Similarly, here is a sentence by sentence interpretation of the proof:

Proof in contemporary notation
"Let AA', BC, DD', EF be any multitude whatsoever of continuously proportional numbers, beginning from the least AA'." Consider the first n+1 terms of a geometric series Sn = a + ar + ar2 + ar3 + ... + arn for the case r>1 and n=3.
"And let BG and FH, each equal to AA', have been subtracted from BC and EF." Subtract a from ar and ar3.
"I say that as GC is to AA', so EH is to AA', BC, DD'." I say that (ar-a) / a = (ar3-a) / (a + ar + ar2).
"For let FK be made equal to BC, and FL to DD'."
"And since FK is equal to BC, of which FH is equal to BG, the remainder HK is thus equal to the remainder GC."
"And since as EF is to DD', so DD' to BC, and BC to AA' [Prop. 7.13], and DD' equal to FL, and BC to FK, and AA' to FH, thus as EF is to FL, so LF to FK, and FK to FH."
"By separation, as EL to LF, so LK to FK, and KH to FH [Props. 7.11, 7.13]." By separation, (ar3-ar2) / ar2 = (ar2-ar) / ar = (ar-a) / a = r-1.
"And thus as one of the leading is to one of the following, so (the sum of) all of the leading to (the sum of) all of the following [Prop. 7.12]." The sum of those numerators and the sum of those denominators form the same proportion: ((ar3-ar2) + (ar2-ar) + (ar-a)) / (ar2 + ar + a) = r-1.
"And thus as one of the leading is to one of the following, so (the sum of) all of the leading to (the sum of) all of the following [Prop. 7.12]." And this sum of equal proportions can be extended beyond (ar3-ar2) / ar2 to include all the proportions up to (arn-arn-1) / arn-1.
"Thus, as KH is to FH, so EL, LK, KH to LF, FK, HF."
"And KH equal to CG, and FH to AA', and LF, FK, HF to DD', BC, AA'."
"Thus, as CG is to AA', so EH to DD', BC, AA'."
"Thus, as the excess of the second is to the first, so is the excess of the last is to all those before it." Thus, (ar-a) / a = (ar3-a) / S2. Or more generally, (ar-a) / a = (arn-a) / Sn-1, which can be rearranged in the more common form Sn-1 = a(rn-1) / (r-1).
"The very thing it was required to show." Q.E.D.

### Archimedes of Syracuse (c.287 – c.212 BC)

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 1/4 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.

### Nicole Oresme (c.1323 – 1382) A two dimensional geometric series diagram Nicole Oresme used to determine that the infinite series 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + 6/64 + 7/128 + ... converges to 2.

Among his insights into infinite series, in addition to his elegantly simple proof of the divergence of the harmonic series, Nicole Oresme proved that the series 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + 6/64 + 7/128 + ... converges to 2. His diagram for his geometric proof, similar to the adjacent diagram, shows a two dimensional geometric series. The first dimension is horizontal, in the bottom row showing the geometric series S = 1/2 + 1/4 + 1/8 + 1/16 + ... , which is the geometric series with coefficient a = 1/2 and common ratio r = 1/2 that converges to S = a / (1-r) = (1/2) / (1-1/2) = 1. The second dimension is vertical, where the bottom row is a new coefficient aT equal to S and each subsequent row above it is scaled by the same common ratio r = 1/2, making another geometric series T = 1 + 1/2 + 1/4 + 1/8 + ... , which is the geometric series with coefficient aT = S = 1 and common ratio r = 1/2 that converges to T = aT / (1-r) = S / (1-r) = a / (1-r) / (1-r) = (1/2) / (1-1/2) / (1-1/2) = 2.

Although difficult to visualize beyond three dimensions, Oresme's insight generalizes to any dimension d. Using the sum of the d−1 dimension of the geometric series as the coefficient a in the d dimension of the geometric series results in a d-dimensional geometric series converging to Sd / a = 1 / (1-r)d within the range |r|<1. Pascal's triangle and long division reveals the coefficients of these multi-dimensional geometric series, where the closed form is valid only within the range |r|<1.

${\begin{matrix}{\text{ 1}}\\{\text{ 1}}\quad {\text{ 1}}\\{\text{ 1}}\quad {\text{ 2}}\quad {\text{ 1}}\\{\text{ 1}}\quad {\text{ 3}}\quad {\text{ 3}}\quad {\text{ 1}}\\{\text{ 1}}\quad {\text{ 4}}\quad {\text{ 6}}\quad {\text{ 4}}\quad {\text{ 1}}\end{matrix}}$ {\begin{aligned}&d\quad S^{d}/a\ {\text{(closed form)}}\quad &&S^{d}/a\ {\text{(expanded form)}}\\&1\quad 1/(1-r)\quad &&1+r+r^{2}+r^{3}+r^{4}+\cdots \\&2\quad 1/(1-r)^{2}\quad &&1+2r+3r^{2}+4r^{3}+5r^{4}+\cdots \\&3\quad 1/(1-r)^{3}\quad &&1+3r+6r^{2}+10r^{3}+15r^{4}+\cdots \\&4\quad 1/(1-r)^{4}\quad &&1+4r+10r^{2}+20r^{3}+35r^{4}+\cdots \\\end{aligned}} Note that as an alternative to long division, it is also possible to calculate the coefficients of the d-dimensional geometric series by integrating the coefficients of dimension d−1. This mapping from division by 1-r in the power series sum domain to integration in the power series coefficient domain is a discrete form of the mapping performed by the Laplace transform. MIT Professor Arthur Mattuck shows how to derive the Laplace transform from the power series in this lecture video, where the power series is a mapping between discrete coefficients and a sum and the Laplace transform is a mapping between continuous weights and an integral.

The closed forms of Sd/a are related to but not equal to the derivatives of S = f(r) = 1 / (1-r). As shown in the following table, the relationship is Sk+1 = f(k)(r) / k!, where f(k)(r) denotes the kth derivative of f(r) = 1 / (1-r) and the closed form is valid only within the range |r| < 1.

{\begin{aligned}&k\quad f^{(k)}(r)/a\ {\text{(closed form)}}\quad &&f^{(k)}(r)/a\ {\text{(generator form)}}\\&0\quad 1/(1-r)\quad &&\sum _{j=0}^{\infty }r^{j}\\&1\quad 1!/(1-r)^{2}\quad &&\sum _{j=1}^{\infty }jr^{j-1}\\&2\quad 2!/(1-r)^{3}\quad &&\sum _{j=2}^{\infty }j(j-1)r^{j-2}\\&3\quad 3!/(1-r)^{4}\quad &&\sum _{j=3}^{\infty }j(j-1)(j-2)r^{j-3}\\&4\quad 4!/(1-r)^{5}\quad &&\sum _{j=4}^{\infty }j(j-1)(j-2)(j-3)r^{j-4}\\\end{aligned}} ## 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.

### 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.

### 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.

### 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}} 