A geometric sequence in which the number of terms increases without bounds is called an infinite geometric series. Show
If the absolute value of the common ratio $$r$$ is less than $$1$$, $${S_n}$$, the sum of $$n$$ terms always approaches a definite limit as $$n$$ increases without bounds. As we have proved, the sum of a finite geometric series is We rewrite it as If now $$r$$ is numerically less than $$1$$, i.e., $$\left r \right < 1$$, the numerical value of $${r^n}$$ decreases as $$n$$ increases. By taking $$n$$ which is sufficiently large, we can make $${r^n}$$ as small as we want. Hence, by taking $$n$$ large enough, we can make $${S_n}$$ differ from $$\frac{{{a_1}}}{{1 – r}}$$ by as little as we want, i.e., we can make $${S_n}$$ approach $$\frac{{{a_1}}}{{1 – r}}$$ as a limit. Symbolically Here $${S_\infty }$$ is the sum of an infinite geometric progression with the first term as $${a_1}$$ and common ratio $$r$$. We can also call it an infinite series. Accordingly, the expression $${a_1} + {a_1}r + {a_1}{r^2} + \cdots $$ is called an infinite geometric series. If the terms continuously decrease as $${S_\infty }$$ approaches a limiting value as $$n$$ becomes infinitely large, it is said to be a convergent infinite series. Example: Find the sum of the infinite geometric sequence $$2,{\text{ }}\frac{4}{3},{\text{ }}\frac{8}{9},{\text{ }}\frac{{16}}{{27}}, \cdots $$ Solution: We have $${a_1} = 2$$, $$r = \frac{2}{3} < 1$$, then $${S_\infty } = \frac{{{a_1}}}{{1 – r}} = \frac{2}{{1 – \frac{2}{3}}} = 6$$ Recurring or Periodic Decimals An interesting application of a geometric progression with infinitely many terms is the evaluation of recurring or periodic decimals. When we attempt to express a common fraction such as $$\frac{3}{8}$$ or $$\frac{4}{{11}}$$ as a decimal fraction, the decimal always either terminates or ultimately repeats in blocks. Thus $$\frac{3}{8} = 0.375$$ (decimal terminates) $$\frac{4}{{11}} = 0.363636…$$ (decimal repeats) In the division process by which we express the fraction $$\frac{p}{q}$$ as a decimal fraction the remainders can only be the numbers $$0,1,2,3,4, \ldots ,q – 1$$. If at any stage in the division we obtain a remainder of $$0$$, the process terminates. Otherwise, after not more than $$q$$ divisors, one of the remainders $$0,1,2,3,4, \ldots ,q – 1$$ must recur and the decimal begins to repeat. Example: Express the recurring decimal fraction $$0.5378378378…$$ as a common fraction. Solution: The given decimal fraction can be written in the form Hence our number consists of the decimal $$0.5$$ plus the sum of an infinite geometric progression with the first term $${a_1} = 0.0378$$ and common ratio $$r = 0.001$$. The sum of the infinite progression is expressible as the fraction Hence $$0.5378378… = 0.5 + \frac{7}{{185}} = \frac{{199}}{{370}}$$ 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 unit square is partitioned into an infinite number of Lshaped 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 1 2 + 1 4 + 1 8 + 1 16 + ⋯ {\displaystyle {\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 {\displaystyle 1/2} . In general, a geometric series is written as a + a r + a r 2 + a r 3 + . . . {\displaystyle a+ar+ar^{2}+ar^{3}+...} , where a {\displaystyle a} is the coefficient of each term and r {\displaystyle r} is the common ratio between adjacent terms. The geometric series had an important role in the early development of calculus, is used throughout mathematics, and can serve as an introduction to frequently used mathematical tools such as the Taylor series, the complex Fourier series, and the matrix exponential.The name geometric series indicates each term is the geometric mean of its two neighboring terms, similar to how the name arithmetic series indicates each term is the arithmetic mean of its two neighboring terms. The sequence of geometric series terms (without any of the additions) is called a geometric sequence or "geometric progression". Coefficient aThe 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 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.[1] 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[1] of the geometric series written as ∑ k = 0 ∞ {\displaystyle \sum _{k=0}^{\infty }} arkand a closed form of the geometric series written as a 1 − r for  r  < 1. {\displaystyle {\frac {a}{1r}}{\text{ for }}r<1.}The derivation of the closed form from the expanded form is shown in this article's Sum section. 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. It is often a convenience in notation to set the series equal to the sum s and work with the geometric series s = a + ar + ar2 + ar3 + ar4 + ... in its normalized form s / a = 1 + r + r2 + r3 + r4 + ... or in its normalized vector form s / a = [1 1 1 1 1 ...][1 r r2 r3 r4 ...]T or in its normalized partial series form sn / a = 1 + r + r2 + r3 + r4 + ... + rn, where n is the power (or degree) of the last term included in the partial sum sn.Changing even one of the coefficients to something other than coefficient a would 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 rComplex geometric series (coefficient a = 1 and common ratio r = 0.5 eiω0t) 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 headtotail vector summation that converges to the circle. The circle intersects the real axis at 2 (= 1/(11/2) 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:
The convergence of the geometric series depends on the value of the common ratio r:
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 reiθ 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 reiθ, the expanded form of the geometric series is a + areiθ + ar2ei2θ + ar3ei3θ + ... 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 + areiω0t + ar2ei2ω0t + ar3ei3ω0t + ... , 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. SumClosedform formulaAs an alternative to the algebraic derivation of the geometric series closed form formula, there is also the following geometric derivation. (TOP) Represent the first n+1 terms of s/a as areas of overlapped similar triangles.[2] For example, the area of the biggest overlapped (red) triangle is bh/2 = (2)(1)/2 = 1, which is the value of the first term of the geometric series. The area of the second biggest overlapped (green) triangle is bh/2 = (2r1/2)(r1/2)/2 = r, which is the value of the second term of the geometric series. Each progressively smaller triangle has its base and height scaled down by another factor of r1/2, resulting in a sequence of triangle areas 1, r, r2, r3, ... which is equal to the sequence of terms in the normalized geometric series. (MIDDLE) In the order from largest to smallest, remove each triangle's overlapped area, which is always a fraction r of its area, and scale the remaining 1−r of the triangle's nonoverlapped area by 1/(1−r) so the area of the formerly overlapped triangle, now the area of a nonoverlapped trapezoid, remains the same. (BOTTOM) Aggregate the resulting n+1 nonoverlapped trapezoids into a single nonoverlapped trapezoid and calculate its area. The area of that aggregated trapezoid represents the value of the partial series. That area is equal to the outermost triangle minus the empty triangle tip: sn/a = (1−rn+1) / (1−r), which simplifies to s/a = 1/(1−r) when n approaches infinity and r < 1.The sum of the first n terms of a geometric series, up to and including the r n1 term, is s n = a r 0 + a r 1 + ⋯ + a r n − 1 = ∑ k = 0 n − 1 a r k = ∑ k = 1 n a r k − 1 = { a ( 1 − r n 1 − r ) , for r ≠ 1 a n , for r = 1 {\displaystyle {\begin{aligned}s_{n}&=ar^{0}+ar^{1}+\cdots +ar^{n1}\\&=\sum _{k=0}^{n1}ar^{k}=\sum _{k=1}^{n}ar^{k1}\\&={\begin{cases}a\left({\frac {1r^{n}}{1r}}\right),{\text{ for }}r\neq 1\\an,{\text{ for }}r=1\end{cases}}\end{aligned}}} where r is the common ratio. One can derive that closedform formula for the partial sum, sn, by subtracting out the many selfsimilar terms as follows:[3][4][5] s n = a r 0 + a r 1 + ⋯ + a r n − 1 , r s n = a r 1 + a r 2 + ⋯ + a r n , s n − r s n = a r 0 − a r n , s n ( 1 − r ) = a ( 1 − r n ) , s n = a ( 1 − r n 1 − r ) , for r ≠ 1. {\displaystyle {\begin{aligned}s_{n}&=ar^{0}+ar^{1}+\cdots +ar^{n1},\\rs_{n}&=ar^{1}+ar^{2}+\cdots +ar^{n},\\s_{n}rs_{n}&=ar^{0}ar^{n},\\s_{n}\left(1r\right)&=a\left(1r^{n}\right),\\s_{n}&=a\left({\frac {1r^{n}}{1r}}\right),{\text{ for }}r\neq 1.\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 s = a + a r + a r 2 + a r 3 + a r 4 + ⋯ = ∑ k = 0 ∞ a r k = ∑ k = 1 ∞ a r k − 1 = a 1 − r , for  r  < 1. {\displaystyle {\begin{aligned}s&=a+ar+ar^{2}+ar^{3}+ar^{4}+\cdots \\&=\sum _{k=0}^{\infty }ar^{k}=\sum _{k=1}^{\infty }ar^{k1}\\&={\frac {a}{1r}},{\text{ for }}r<1.\end{aligned}}} The formula also holds for complex r, with the corresponding restriction that 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 finite 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 the distance between two values. However, the padic 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 convergenceWe can prove that the geometric series converges using the sum formula for a geometric progression: 1 + r + r 2 + r 3 + ⋯ = lim n → ∞ ( 1 + r + r 2 + ⋯ + r n ) = lim n → ∞ 1 − r n + 1 1 − r . {\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 {1r^{n+1}}{1r}}.\end{aligned}}}The second equality is true because if  r  < 1 , {\displaystyle r<1,} then r n + 1 → 0 {\displaystyle r^{n+1}\to 0} as n → ∞ {\displaystyle n\to \infty } and ( 1 + r + r 2 + ⋯ + 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 . {\displaystyle {\begin{aligned}(1+r+r^{2}+\cdots +r^{n})(1r)&=((1r)+(rr^{2})+(r^{2}r^{3})+...+(r^{n}r^{n+1}))\\&=(1+(r+r)+(r^{2}+r^{2})+...+(r^{n}+r^{n})r^{n+1})\\&=1r^{n+1}.\end{aligned}}}Alternatively, a geometric interpretation of the convergence is shown in the adjacent diagram. The area of the white triangle is the series remainder = s  sn = arn+1 / (1  r). Each additional term in the partial series reduces the area of that white triangle remainder by the area of the trapezoid representing the added term. The trapezoid areas (i.e., the values of the terms) get progressively thinner and shorter and closer to the origin. In the limit, as the number of trapezoids approaches infinity, the white triangle remainder vanishes as it is filled by trapezoids and therefore sn converges to s, provided r<1. In contrast, if r>1, the trapezoid areas representing the terms of the series instead get progressively wider and taller and farther from the origin, not converging to the origin and not converging as a series. Rate of convergenceConverging alternating geometric series with common ratio r = 1/2 and coefficient a = 1. (TOP) Alternating positive and negative areas. (MIDDLE) Gaps caused by addition of adjacent areas. (BOTTOM) Gaps filled by broadening and decreasing the heights of the separated trapezoids.After knowing that a series converges, there are some applications in which it is also important to know how quickly the series converges. For the geometric series, one convenient measure of the convergence rate is how much the previous series remainder decreases due to the last term of the partial series. Given that the last term is arn and the previous series remainder is s  sn1 = arn / (1  r)), this measure of the convergence rate of the geometric series is arn / (arn / (1  r)) = 1  r, if 0 ≤ r < 1. If r < 0, adjacent terms in the geometric series alternate between being positive and negative. A geometric interpretation of a converging alternating geometric series is shown in the adjacent diagram in which the areas of the negative terms are shown below the x axis. Pairing and summing each positive area with its negative smaller area neighbor results in nonoverlapped trapezoids separated by gaps. To remove the gaps, broaden each trapezoid to cover the rightmost 1  r2 of the original triangle area instead of just the rightmost 1  r. However, to maintain the same trapezoid areas during this broadening transformation, scaling is needed: scale*(1  r2) = (1  r), or scale = (1  r) / (1  r2) = (1 + r) / (1  r2) = (1 + r) / ((1 + r)(1  r)) = 1 / (1  r) where 1 < r ≤ 0. Note that because r < 0 this scale decreases the amplitude of the separated trapezoids in order to fill in the separation gaps. In contrast, for the case r > 0 the same scale 1 / (1  r) increases the amplitude of the nonoverlapped trapezoids in order to account for the loss of the overlapped areas. With the gaps removed, pairs of terms in a converging alternating geometric series become a converging (nonalternating) geometric series with common ratio r2 to account for the pairing of terms, coefficient a = 1 / (1  r) to account for the gap filling, and the degree (i.e., highest powered term) of the partial series called m instead of n to emphasize that terms have been paired. Similar to the r > 0 case, the r < 0 convergence rate = ar2m / (s  sm1) = 1  r2, which is the same as the convergence rate of a nonalternating geometric series if its terms were similarly paired. Therefore, the convergence rate does not depend upon n or m and, perhaps more surprising, does not depend upon the sign of the common ratio. One perspective that helps explain the variable rate of convergence that is symmetric about r = 0 is that each added term of the partial series makes a finite contribution to the infinite sum at r = 1 and each added term of the partial series makes a finite contribution to the infinite slope at r = 1. Historic insightsZeno of Elea (c.495 – c.430 BC)2,500 years ago, Greek mathematicians had a problem when walking from one place to another: they thought[6] 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." A geometric interpretation for the same case of common ratio r>1. (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 nonoverlapped right area portion (11/r = (r1)/r) and scale that nonoverlapped trapezoid by r/(r1) so its area is the same as the area of the original overlapped triangle. (BOTTOM) Calculate the area of the aggregate trapezoid as the area of the large triangle less the area of the empty small triangle at the large triangle's left tip. The large triangle is the largest overlapped triangle scaled by r/(r1). The empty small triangle started as a but that area was transformed into a nonoverlapped scaled trapezoid leaving an empty left area portion (1/r). However, that empty triangle of area a/r must also be scaled by r/(r1) so its slope matches the slope of all the nonoverlapped scaled trapezoids. Therefore, Sn = area of large triangle  area of empty small triangle = arn+1/(r1)  a/(r1) = a(rn+11)/(r1).Euclid's Elements of Geometry[7] Book IX, Proposition 35, proof (of the proposition in adjacent diagram's caption):
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,
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:
Similarly, here is a sentence by sentence interpretation of the proof:
Archimedes of Syracuse (c.287 – c.212 BC)Archimedes' dissection of a parabolic segment into infinitely many trianglesArchimedes 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 ( 1 8 ) + 4 ( 1 8 ) 2 + 8 ( 1 8 ) 3 + ⋯ . {\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 1 + 1 4 + 1 16 + 1 64 + ⋯ . {\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 ∑ n = 0 ∞ 4 − n = 1 + 4 − 1 + 4 − 2 + 4 − 3 + ⋯ = 4 3 . {\displaystyle \sum _{n=0}^{\infty }4^{n}=1+4^{1}+4^{2}+4^{3}+\cdots ={4 \over 3}.}The sum is 1 1 − r = 1 1 − 1 4 = 4 3 . {\displaystyle {\frac {1}{1r}}\;=\;{\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[8] 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 / (1r) = (1/2) / (11/2) = 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 / (1r) = S / (1r) = a / (1r) / (1r) = (1/2) / (11/2) / (11/2) = 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 ddimensional geometric series converging to Sd / a = 1 / (1r)d within the range r<1. Pascal's triangle and long division reveals the coefficients of these multidimensional geometric series, where the closed form is valid only within the range r<1. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 {\displaystyle {\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}}} d S d / a (closed form) S d / a (expanded form) 1 1 / ( 1 − r ) 1 + r + r 2 + r 3 + r 4 + ⋯ 2 1 / ( 1 − r ) 2 1 + 2 r + 3 r 2 + 4 r 3 + 5 r 4 + ⋯ 3 1 / ( 1 − r ) 3 1 + 3 r + 6 r 2 + 10 r 3 + 15 r 4 + ⋯ 4 1 / ( 1 − r ) 4 1 + 4 r + 10 r 2 + 20 r 3 + 35 r 4 + ⋯ {\displaystyle {\begin{aligned}&d\quad S^{d}/a\ {\text{(closed form)}}\quad &&S^{d}/a\ {\text{(expanded form)}}\\&1\quad 1/(1r)\quad &&1+r+r^{2}+r^{3}+r^{4}+\cdots \\&2\quad 1/(1r)^{2}\quad &&1+2r+3r^{2}+4r^{3}+5r^{4}+\cdots \\&3\quad 1/(1r)^{3}\quad &&1+3r+6r^{2}+10r^{3}+15r^{4}+\cdots \\&4\quad 1/(1r)^{4}\quad &&1+4r+10r^{2}+20r^{3}+35r^{4}+\cdots \\\end{aligned}}}
ApplicationsEconomicsIn 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 {\displaystyle I} ), where I {\displaystyle I} is the yearly interest rate.Similarly, a payment of $100 two years in the future has a present value of $100 / (1 + I {\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 ∑ n = 1 ∞ $ 100 ( 1 + I ) n , {\displaystyle \sum _{n=1}^{\infty }{\frac {\$100}{(1+I)^{n}}},}which is the infinite series: $ 100 ( 1 + I ) + $ 100 ( 1 + I ) 2 + $ 100 ( 1 + I ) 3 + $ 100 ( 1 + I ) 4 + ⋯ . {\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 + I {\displaystyle I} ). The sum is the first term divided by (one minus the common ratio): $ 100 / ( 1 + I ) 1 − 1 / ( 1 + I ) = $ 100 I . {\displaystyle {\frac {\$100/(1+I)}{11/(1+I)}}\;=\;{\frac {\$100}{I}}.}For example, if the yearly interest rate is 10% ( I {\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 financial asset assuming a stable growth rate. Fractal geometryFor 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 ( 1 9 ) + 12 ( 1 9 ) 2 + 48 ( 1 9 ) 3 + ⋯ . {\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 1 + a 1 − r = 1 + 1 3 1 − 4 9 = 8 5 . {\displaystyle 1\,+\,{\frac {a}{1r}}\;=\;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. IntegrationThe derivative of f ( x ) = arctan ( u ( x ) ) is f ′ ( x ) = u ′ ( x ) / ( 1 + [ u ( x ) ] 2 ) {\displaystyle f(x)=\arctan(u(x)){\text{ is }}f'(x)=u'(x)/(1+[u(x)]^{2})} because,[10] letting y and u represent f ( x ) and u ( x ) , {\displaystyle y{\text{ and }}u{\text{ represent }}f(x){\text{ and }}u(x),} y = arctan ( u ) implies u = tan ( y ) in the range − π / 2 < y < π / 2 and u ′ = sec 2 y ⋅ y ′ by applying the quotient rule to tan ( y ) = sin ( y ) / cos ( y ) , y ′ = u ′ / sec 2 y by dividing both sides by sec 2 y , = u ′ / ( 1 + tan 2 y ) by using the trigonometric identity derived by dividing sin 2 y + cos 2 y = 1 by cos 2 y , = u ′ / ( 1 + u 2 ) by recalling u = tan ( y ) . {\displaystyle {\begin{aligned}y&=\arctan(u)&&\quad {\text{ implies }}\\u&=\tan(y)&&\quad {\text{ in the range }}\pi /2<y<\pi /2{\text{ and }}\\u'&=\sec ^{2}y\cdot y'&&\quad {\text{ by applying the quotient rule to }}\tan(y)=\sin(y)/\cos(y),\\y'&=u'/\sec ^{2}y&&\quad {\text{ by dividing both sides by }}\sec ^{2}y,\\&=u'/(1+\tan ^{2}y)&&\quad {\text{ by using the trigonometric identity derived by dividing }}\sin ^{2}y+\cos ^{2}y=1{\text{ by }}\cos ^{2}y,\\&=u'/(1+u^{2})&&\quad {\text{ by recalling }}u=\tan(y).\end{aligned}}}Therefore, letting u ( x ) = x , arctan ( x ) {\displaystyle u(x)=x,\arctan(x)} is the integral arctan ( x ) = ∫ d x 1 + x 2 in the range − π / 2 < arctan ( x ) < π / 2 , = ∫ d x 1 − ( − x 2 ) by writing integrand as closed form of geometric series with r = − x 2 , = ∫ ( 1 + ( − x 2 ) + ( − x 2 ) 2 + ( − x 2 ) 3 + ⋯ ) d x by writing geometric series in expanded form , = ∫ ( 1 − x 2 + x 4 − x 6 + ⋯ ) d x by calculating the sign and power of each term in integrand , = x − x 3 3 + x 5 5 − x 7 7 + ⋯ by integrating each term , = ∑ n = 0 ∞ ( − 1 ) n 2 n + 1 x 2 n + 1 by writing series in generator form , {\displaystyle {\begin{aligned}\arctan(x)&=\int {\frac {dx}{1+x^{2}}}\quad &&{\text{in the range }}\pi /2<\arctan(x)<\pi /2,\\&=\int {\frac {dx}{1(x^{2})}}\quad &&{\text{by writing integrand as closed form of geometric series with }}r=x^{2},\\&=\int \left(1+\left(x^{2}\right)+\left(x^{2}\right)^{2}+\left(x^{2}\right)^{3}+\cdots \right)dx\quad &&{\text{by writing geometric series in expanded form}},\\&=\int \left(1x^{2}+x^{4}x^{6}+\cdots \right)dx\quad &&{\text{by calculating the sign and power of each term in integrand}},\\&=x{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}{\frac {x^{7}}{7}}+\cdots \quad &&{\text{by integrating each term}},\\&=\sum _{n=0}^{\infty }{\frac {(1)^{n}}{2n+1}}x^{2n+1}\quad &&{\text{by writing series in generator form}},\end{aligned}}}which is called Gregory's series and is commonly attributed to Madhava of Sangamagrama (c. 1340 – c. 1425). Specific geometric series
Geometric seriesA compact illustration of different perspectives on the geometric series to make perspective switching easier. Quadrant A1 contains the why perspective briefly describing why geometric series matter. Quadrant A2 contains the how perspective listing a snippet of Julia code that computes the geometric series. Quadrant A3 contains the what (algebra) perspective with the algebraic proof of the geometric series closed form formula. Quadrant A4 contains the what (geometry) perspective showing three steps of the geometric proof of the geometric series closed form formula. And the right margin shows the timeline of some of the historical insights about the geometric series.The geometric series has two degrees of freedom: one for its coefficient a and another for its common ratio r. In the map of polynomials, the big red circle represents the set of all geometric series. Converging geometric seriesOnly a subset of all geometric series converge. Specifically, a geometric series converges if and only if its common ratio r < 1. In the map of polynomials, the red triangle represents the set of converging geometric series and being drawn inside the big red circle representing the set of all geometric series indicates the converging geometric series is a subset of the geometric series. Repeated decimalsOnly a subset of all converging geometric series converge to decimal fractions that have repeated patterns that continue forever (e.g., 0.7777... or 0.9999... or 0.123412341234...). In the map of polynomials, the little yellow triangle represents the set of geometric series that converge to infinitely repeated decimal patterns. It is drawn inside the red triangle to indicate it is a subset of the converging geometric series, which in turn is drawn inside the big red circle indicating both the converging geometric series and the geometric series that converge to infinitely repeated patterns are subsets of the geometric series. Although fractions with infinitely repeated decimal patterns can only be approximated when encoded as floating point numbers, they can always be defined exactly as the ratio of two integers and those two integers can be calculated using the geometric series. For example, the repeated decimal fraction 0.7777... can be written as the geometric series 0.7777 … = 7 10 + 7 10 1 10 + 7 10 1 10 2 + 7 10 1 10 3 + ⋯ {\displaystyle 0.7777\ldots \;=\;{\frac {7}{10}}\,+\,{\frac {7}{10}}{\frac {1}{10}}\,+\,{\frac {7}{10}}{\frac {1}{10^{2}}}\,+\,{\frac {7}{10}}{\frac {1}{10^{3}}}\,+\,\cdots }where coefficient a = 7/10 and common ratio r = 1/10. The geometric series closed form reveals the two integers that specify the repeated pattern: 0.7777 … = a 1 − r = 7 / 10 1 − 1 / 10 = 7 / 10 9 / 10 = 7 9 . {\displaystyle 0.7777\ldots \;=\;{\frac {a}{1r}}\;=\;{\frac {7/10}{11/10}}\;=\;{\frac {7/10}{9/10}}\;=\;{\frac {7}{9}}.}This approach extends beyond baseten numbers. In fact, any fraction that has an infinitely repeated pattern in baseten numbers also has an infinitely repeated pattern in numbers written in any other base. For example, looking at the floating point encoding for the number 0.7777...
julia> bitstring(Float32(0.77777777777777777777)) reveals the binary fraction 0.110001110001110001... where the binary pattern 0b110001 repeats indefinitely and can be written in mostly (except for the powers) binary numbers as 0.110001110001110001 … = 110001 1000000 + 110001 1000000 1 1000000 + 110001 1000000 1 1000000 2 + 110001 1000000 1 1000000 3 + ⋯ {\displaystyle 0.110001110001110001\ldots \;=\;{\frac {110001}{1000000}}\,+\,{\frac {110001}{1000000}}{\frac {1}{1000000}}\,+\,{\frac {110001}{1000000}}{\frac {1}{1000000^{2}}}\,+\,{\frac {110001}{1000000}}{\frac {1}{1000000^{3}}}\,+\,\cdots }where coefficient a = 0b110001 / 0b1000000 = 49 / 64 and common ratio r = 1 / 0b1000000 = 1 / 64. Using the geometric series closed form as before 0.7777 … = 0 b 0.110001110001110001 … = a 1 − r = 49 / 64 1 − 1 / 64 = 49 / 64 63 / 64 = 49 63 = 7 9 . {\displaystyle 0.7777\ldots \;=\;0b0.110001110001110001\ldots \;=\;{\frac {a}{1r}}\;=\;{\frac {49/64}{11/64}}\;=\;{\frac {49/64}{63/64}}\;=\;{\frac {49}{63}}\;=\;{\frac {7}{9}}.}You may have noticed that the floating point encoding does not capture the 0b110001 repeat pattern in the last couple (least significant) bits. This is because floating point encoding rounds the remainder instead of truncating it. Therefore, if the most significant bit of the remainder is 1, the least significant bit of the encoded fraction gets incremented and that will cause a carry if the least significant bit of the fraction is already 1, which can cause another carry if that bit of the fraction is already a 1, which can cause another carry, etc. This floating point rounding and the subsequent carry propagation explains why the floating point encoding for 0.99999... is exactly the same as the floating point encoding for 1.
julia> bitstring(Float32(0.99999999999999999999)) As an example that has four digits in the repeated pattern, 0.123412341234... can be written as the geometric series 0.123412341234 … = 1234 10000 + 1234 10000 1 10000 + 1234 10000 1 10000 2 + 1234 10000 1 10000 3 + ⋯ {\displaystyle 0.123412341234\ldots \;=\;{\frac {1234}{10000}}\,+\,{\frac {1234}{10000}}{\frac {1}{10000}}\,+\,{\frac {1234}{10000}}{\frac {1}{10000^{2}}}\,+\,{\frac {1234}{10000}}{\frac {1}{10000^{3}}}\,+\,\cdots }where coefficient a = 1234/10000 and common ratio r = 1/10000. The geometric series closed form reveals the two integers that specify the repeated pattern: 0.123412341234 … = a 1 − r = 1234 / 10000 1 − 1 / 10000 = 1234 / 10000 9999 / 10000 = 1234 9999 . {\displaystyle 0.123412341234\ldots \;=\;{\frac {a}{1r}}\;=\;{\frac {1234/10000}{11/10000}}\;=\;{\frac {1234/10000}{9999/10000}}\;=\;{\frac {1234}{9999}}.}Power seriesLike the geometric series, the power series has one degree of freedom for its common ratio r (along the xaxis) but has n+1 degrees of freedom for its coefficients (along the yaxis), where n represents the power of the last term in the partial series. In the map of polynomials, the big blue circle represents the set of all power series. Taylor seriesBinary encoded numbersZeno of Elea's geometric series with coefficient a=1/2 and common ratio r=1/2 is the foundation of binary encoded approximations of fractions in digital computers. Concretely, the geometric series written in its normalized vector form is s/a = [1 1 1 1 1 …][1 r r2 r3 r4 …]T. Keeping the column vector of basis functions [1 r r2 r3 r4 …]T the same but generalizing the row vector [1 1 1 1 1 …] so that each entry can be either a 0 or a 1 allows for an approximate encoding of any fraction. For example, the value v = 0.34375 is encoded as v/a = [0 1 0 1 1 0 …][1 r r2 r3 r4 …]T where coefficient a = 1/2 and common ratio r = 1/2. Typically, the row vector is written in the more compact binary form v = 0.010110 which is 0.34375 in decimal. Similarly, the geometric series with coefficient a=1 and common ratio r=2 is the foundation for binary encoded integers in digital computers. Again, the geometric series written in its normalized vector form is s/a = [1 1 1 1 1 …][1 r r2 r3 r4 …]T. Keeping the column vector of basis functions [1 r r2 r3 r4 …]T the same but generalizing the row vector [1 1 1 1 1 …] so that each entry can be either a 0 or a 1 allows for an encoding of any integer. For example, the value v = 151 is encoded as v/a = [1 1 1 0 1 0 0 1 0 …][1 r r2 r3 r4 r5 r6 r7 r8 …]T where coefficient a = 1 and common ratio r = 2. Typically, the row vector is written in reverse order (so that the most significant bit is first) in the more compact binary form v = …010010111 = 10010111 which is 151 in decimal. Bit fields for encoding a 32bit floating point number according to IEEE 754 standard.As shown in the adjacent figure, the standard binary encoding of a 32bit floating point number is a combination of a binary encoded integer and a binary encoded fraction, beginning at the most significant bit with
Building upon the previous example of 0.34375 having binary encoding of 0.010110, a floating point encoding (according to the IEEE 754 standard) of 0.34375 is
Although encoding floating point numbers by hand like this is possible, letting a computer do it is easier and less error prone. The following Julia code confirms the hand calculated floating point encoding of the number 0.34375:
julia> bitstring(Float32(0.34375)) Laurent seriesComplex Fourier seriesThe front and back planes show the sum of the first +/n terms of the complex Fourier series that has coefficients set to converge to a tracing of the letter 'e' for exponential. (The Julia source code that generates the frames and audio of this animation is here[11] in Appendix B.) As an example of the ability of the complex Fourier series to trace any 2D closed figure, in the adjacent animation a complex Fourier series traces the letter 'e' (for exponential). Given the intricate coordination of motions shown in the animation, a definition of the complex Fourier series can be surprisingly compact in just two equations: s ( t ) = ∑ n = − ∞ ∞ c n e 2 π i n t c n = ∫ 0 1 s ( t ) e − 2 π i n t d t {\displaystyle {\begin{aligned}s(t)&=\sum _{n=\infty }^{\infty }c_{n}e^{2\pi int}\\c_{n}&=\int _{0}^{1}s(t)e^{2\pi int}dt\\\end{aligned}}}where parameterized function s(t) traces some 2D closed figure in the complex plane as the parameter t progresses through the period from 0 to 1. To help make sense of these compact equations defining the complex Fourier series, note that the complex Fourier series summation looks similar to the complex geometric series except that the complex Fourier series is basically two complex geometric series (one set of terms rotating in the positive direction and another set of terms rotating in the negative direction), and the coefficients of the complex Fourier series are complex constants that can vary from term to term. By allowing terms to rotate in either direction, the series becomes capable of tracing any 2D closed figure. In contrast, the complex geometric series has all the terms rotating in the same direction and it can trace only circles. Allowing the coefficients of the complex geometric series to vary from term to term would expand upon the shapes it can trace but all the possible shapes would still be limited to being puffy and cloudlike, not able to trace the shape of a simple line segment, for example going back and forth between 1 + i0 and 1 + i0. However, Euler's formula shows that the addition of just two terms rotating in opposite directions can trace that line segment between 1 + i0 and 1 + i0: e i θ = cos θ + i sin θ e − i θ = cos θ − i sin θ cos θ = e i θ + e − i θ 2 . {\displaystyle {\begin{aligned}e^{i\theta }&=\cos \theta +i\sin \theta \\e^{i\theta }&=\cos \theta i\sin \theta \\\cos \theta &={\frac {e^{i\theta }+e^{i\theta }}{2}}.\\\end{aligned}}}Concerning the complex Fourier series second equation defining how to calculate the coefficients, the coefficient of the nonrotating term c0 can be calculated by integrating the complex Fourier series first equation over the range of one period from 0 to 1. Over that range, all the rotating terms integrate to zero, leaving just c0. Similarly, any of the terms in the complex Fourier series first equation can be made to be a nonrotating term by multiplying both sides of the equation by e − 2 π i n t {\displaystyle e^{2\pi int}} before integrating to calculate cn, and that is the complex Fourier series second equation.Matrix polynomialMatrix exponentialSee also
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