Geometric series

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 = r^m + r^m+1 + ... + r^n-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 A_i that can be transformed into a non-overlapped L-shaped area L_i = A_i - A_i-1 or, equivalently, L_i+1 = A_i+1 - A_i. Due to being a geometric series, A_i+1 = r A_i. Therefore, L_i+1 = A_i+1 - A_i = (r - 1) A_i, or A_i = L_i+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 = A_m + A_m+1 + ... + A_n-1 + A_n = (L_m+1 + L_m+2 + ... + L_n + L_n+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 A_n+1 less the upper right square notch A_m (because there are no overlapped smaller squares to be transformed into that notch of area A_m). Therefore, substituting A_i = rⁱ and scaling all the terms by coefficient a results in the general, closed-form, geometric series S = (rⁿ⁺¹ - r^m) 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 - 1^m) 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 - r^m) 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 ${\displaystyle r\neq 1}$, the sum of the first n+1 terms of a geometric series, up to and including the r ⁿ term, is

${\displaystyle a+ar+ar^{2}+ar^{3}+\cdots +ar^{n}=\sum _{k=0}^{n}ar^{k}=a\left({\frac {1-r^{n+1}}{1-r}}\right),}$

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:[3][4][5]

${\displaystyle {\begin{aligned}s=a\ +\ &ar\ +\ ar^{2}\ +\ ar^{3}\ +\ \cdots \ +\ ar^{n},\\rs=\ &ar\ +\ ar^{2}\ +\ ar^{3}\ +\ \cdots \ +\ ar^{n}\ +\ ar^{n+1},\\s-rs=\ &a\ -\ ar^{n+1},\\s(1-r)=\ &a(1-r^{n+1}),\\s=\ &a\left({\frac {1-r^{n+1}}{1-r}}\right)\quad {\text{(if }}r\neq 1{\text{)}}.\end{aligned}}}$

As n approaches infinity, the absolute value of r must be less than one for the series to converge. The sum then becomes

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

When a = 1, this can be simplified to

${\displaystyle 1\,+\,r\,+\,r^{2}\,+\,r^{3}\,+\,\cdots \;=\;{\frac {1}{1-r}}.}$

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.

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

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

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 rⁿ⁺¹ → 0 for | r | < 1.

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

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

Note that

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

Thus,

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

If

${\displaystyle |r|<1}$

then

${\displaystyle g(K)\longrightarrow 0{\text{ as }}K\to \infty .}$

So S converges to

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

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,

${\displaystyle {\begin{aligned}a\left({\frac {1-r^{n+1}}{1-r}}\right)&\geq a\left({\frac {1-E}{1-r}}\right),\\1-r^{n+1}&\geq 1-E,\\r^{n+1}&\leq E,\\ln(r^{n+1})&\geq ln(E),\\(n+1)ln(r)&\geq ln(E),\\n+1&\geq \left\lceil {\frac {ln(E)}{ln(r)}}\right\rceil ,\end{aligned}}}$

where 0 < r < 1, the ceiling operation ${\displaystyle \lceil \rceil }$ constrains n to integers, and the natural log operation ln flips the inequality because it negates both sides of the inequality (because both sides are less than one). 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 + ... + ar^n-1 + arⁿ within the range -1 < r < 0 is equivalent to

s = a - ap + ap² - ap³ + ap⁴ - ap⁵ + ... + ap^n-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) + ... + (ap^n-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)p^2(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)p^2m 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 - p^2(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,

${\displaystyle {\begin{aligned}a(1-p)\left({\frac {1-p^{2m+2}}{1-p^{2}}}\right)&\geq a(1-p)\left({\frac {1-E}{1-p^{2}}}\right),\\1-p^{2m+2}&\geq 1-E,\\p^{2m+2}&\leq E,\\ln(p^{2m+2})&\geq ln(E),\\(2m+2)ln(p)&\geq ln(E),\\m+1&\geq \left\lceil {\frac {ln(E)}{2ln(p)}}\right\rceil ,\\m+1&\geq \left\lceil {\frac {ln(E)}{2ln(-r)}}\right\rceil ,\end{aligned}}}$

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.

Book IX, Proposition 35[6] 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.

Archimedes' dissection of a parabolic segment into infinitely many triangles

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:

${\displaystyle 1\,+\,2\left({\frac {1}{8}}\right)\,+\,4\left({\frac {1}{8}}\right)^{2}\,+\,8\left({\frac {1}{8}}\right)^{3}\,+\,\cdots .}$

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

${\displaystyle 1\,+\,{\frac {1}{4}}\,+\,{\frac {1}{16}}\,+\,{\frac {1}{64}}\,+\,\cdots .}$

This is a geometric series with common ratio 1/4 and the fractional part is equal to

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

The sum is

${\displaystyle {\frac {1}{1-r}}\;=\;{\frac {1}{1-{\frac {1}{4}}}}\;=\;{\frac {4}{3}}.}$

This computation uses the method of exhaustion, an early version of integration. Using calculus, the same area could be found by a definite integral.

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

${\displaystyle 0.7777\ldots \;=\;{\frac {7}{10}}\,+\,{\frac {7}{100}}\,+\,{\frac {7}{1000}}\,+\,{\frac {7}{10000}}\,+\,\cdots .}$

The formula for the sum of a geometric series can be used to convert the decimal to a fraction,

${\displaystyle 0.7777\ldots \;=\;{\frac {a}{1-r}}\;=\;{\frac {7/10}{1-1/10}}\;=\;{\frac {7/10}{9/10}}\;=\;{\frac {7}{9}}.}$

The formula works not only for a single repeating figure, but also for a repeating group of figures. For example:

${\displaystyle 0.123412341234\ldots \;=\;{\frac {a}{1-r}}\;=\;{\frac {1234/10000}{1-1/10000}}\;=\;{\frac {1234/10000}{9999/10000}}\;=\;{\frac {1234}{9999}}.}$

Note that every series of repeating consecutive decimals can be conveniently simplified with the following:

${\displaystyle 0.09090909\ldots \;=\;{\frac {09}{99}}\;=\;{\frac {1}{11}}.}$

${\displaystyle 0.143814381438\ldots \;=\;{\frac {1438}{9999}}.}$

${\displaystyle 0.9999\ldots \;=\;{\frac {9}{9}}\;=\;1.}$

That is, a repeating decimal with repeat length n is equal to the quotient of the repeating part (as an integer) and 10ⁿ - 1.

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 + ${\displaystyle I}$ ), where ${\displaystyle I}$ is the yearly interest rate.

Similarly, a payment of $100 two years in the future has a present value of $100 / (1 + ${\displaystyle I}$)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

${\displaystyle \sum _{n=1}^{\infty }{\frac {\$100}{(1+I)^{n}}},}$

which is the infinite series:

${\displaystyle {\frac {\$100}{(1+I)}}\,+\,{\frac {\$100}{(1+I)^{2}}}\,+\,{\frac {\$100}{(1+I)^{3}}}\,+\,{\frac {\$100}{(1+I)^{4}}}\,+\,\cdots .}$

This is a geometric series with common ratio 1 / (1 + ${\displaystyle I}$ ). The sum is the first term divided by (one minus the common ratio):

${\displaystyle {\frac {\$100/(1+I)}{1-1/(1+I)}}\;=\;{\frac {\$100}{I}}.}$

For example, if the yearly interest rate is 10% (${\displaystyle I}$ = 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.

The interior of the Koch snowflake is a union of infinitely many triangles.

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

${\displaystyle 1\,+\,3\left({\frac {1}{9}}\right)\,+\,12\left({\frac {1}{9}}\right)^{2}\,+\,48\left({\frac {1}{9}}\right)^{3}\,+\,\cdots .}$

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

${\displaystyle 1\,+\,{\frac {a}{1-r}}\;=\;1\,+\,{\frac {\frac {1}{3}}{1-{\frac {4}{9}}}}\;=\;{\frac {8}{5}}.}$

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 ${\displaystyle r=1/2}$.

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 ${\displaystyle r=1/2}$ 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.

The formula for a geometric series

${\displaystyle {\frac {1}{1-x}}=1+x+x^{2}+x^{3}+x^{4}+\cdots }$

can be interpreted as a power series in the Taylor's theorem sense, converging where ${\displaystyle |x|<1}$. From this, one can extrapolate to obtain other power series. For example,

${\displaystyle {\begin{aligned}\tan ^{-1}(x)&=\int {\frac {dx}{1+x^{2}}}\\&=\int {\frac {dx}{1-(-x^{2})}}\\&=\int \left(1+\left(-x^{2}\right)+\left(-x^{2}\right)^{2}+\left(-x^{2}\right)^{3}+\cdots \right)dx\\&=\int \left(1-x^{2}+x^{4}-x^{6}+\cdots \right)dx\\&=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+\cdots \\&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}x^{2n+1}.\end{aligned}}}$

By differentiating the geometric series, one obtains the variant[7]

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

Similarly obtained are:

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

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

• Grandi's series: 1 − 1 + 1 − 1 + ⋯
• 1 + 2 + 4 + 8 + ⋯
• 1 − 2 + 4 − 8 + ⋯
• 1/2 + 1/4 + 1/8 + 1/16 + ⋯
• 1/2 − 1/4 + 1/8 − 1/16 + ⋯
• 1/4 + 1/16 + 1/64 + 1/256 + ⋯
• A geometric series is a unit series (the series sum converges to one) if and only if |r| < 1 and a + r = 1 (equivalent to the more familiar form S = a / (1 - r) = 1 when |r| < 1). Therefore, an alternating series is also a unit series when -1 < r < 0 and a + r = 1 (for example, coefficient a = 1.7 and common ratio r = -0.7).
• The terms of a geometric series are also the terms of a generalized Fibonacci sequence (F_n = F_n-1 + F_n-2 but without requiring F₀ = 0 and F₁ = 1) when a geometric series common ratio r satisfies the constraint 1 + r = r², which according to the quadratic formula is when the common ratio r equals the golden ratio (i.e., common ratio r = (1 ± √5)/2).
• The only geometric series that is a unit series and also has terms of a generalized Fibonacci sequence has the golden ratio as its coefficient a and the conjugate golden ratio as its common ratio r (i.e., a = (1 + √5)/2 and r = (1 - √5)/2). It is a unit series because a + r = 1 and |r| < 1, it is a generalized Fibonacci sequence because 1 + r = r², and it is an alternating series because r < 0.