Geometric series

Rhind Papyrus, problem 79
Rhind Papyrus, problem 79
Geometric series are commonly attributed to, philosopher and mathematician, Pythagoras of Samos. However, they already appeared in one of the oldest Egyptian mathematical documents, the Rhind Papyrus around 1550 BC.\(\)

Summation of geometric sequence


A sequence is a list of numbers or terms. In a geometric sequence, each term is found by multiplying the previous term by a constant non-zero number. For example the geometric sequence \(\{2, 6, 18, 54, \ldots\}\). The general form of a geometric sequence is

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

The sum of all the terms, is called the summation of the sequence. The summation of an infinite sequence of values is called a series.

Historian Moritz Cantor translated problem 79 from the Rhind Papyrus as

An estate consisted of seven houses; each house had seven cats; each cat ate seven mice; each mouse ate seven heads of wheat; and each heat of wheat was capable of yielding seven hekat measures of grain. Houses, cats, mice, heads of wheat, and hekat measures of grain, how many of these in all were in the estate?

scribe Ahmes, Rhind Papyrus problem #79, translated by Moritz Cantor

The solution in the left column of the Papyrus suggests scribe Ahmes’ understanding of geometric sequences.

Problem 79 as a power series
object count count
Houses \(7\) \(7^1\)
Cats \(49\) \(7^2\)
Mice \(343\) \(7^3\)
Heads of wheat \(2,301\) \(7^4\)
Hekat measures \(16,807\) \(7^5\)
sum \(19,607\) \(19,607\)

To find the sum of the sequence \(7, 49, 343, 2401, 16807\), Ahmes approached it as \(7(1+7+49+343+2401)\). Since the sum of the terms inside the parentheses is \(2801\), he only had to multiply this number by \(7\), thinking of \(7\) as \(1+2+4\) so he could use repeated addition to do the multiplication

Since the first term of the geometric sequence \(7\) is equal to the common ratio of multiplication, the finite geometric series can be reduced to multiplications involving the finite series having one less term. In modern notation:

$$ \sum_{k=1}^n7^k = 7\left(1+\sum_{k=1}^{n-1}7^k\right) $$

Leonardo Fibonacci (1170-1250 AD) described a similarly amusing problem:

There are seven old women on the road to Rome. Each woman has seven mules; each mule carries seven sacks; each sack contains seven loaves; with each loaf are seven knives; and each knife is in seven sheaths. Women, mules, sacks, loaves, knives and sheaths, how many are there in all on the road to Rome.” Leonardo Fibonacci, Liber Abaci, 1202 AD

Even more recently:

As I was going to St. Ives, I met a man with seven wives; each wife had seven sacks, each sack had seven cats, each cat had seven kits. Kits, cats, sacks, and wives. How many were there going to St. Ives? Traditional nursery rhyme, 1730 AD

In general we write a geometric sequence, where \(a\) is the first term, \(r\) is the common ratio and \(m\) is the total number of terms.

$$ \{a,ar,ar^2,ar^3,\ldots, ar^{m-2},ar^{m-1}\} $$

The summation of that geometric sequence is

$$ S\triangleq a+ar+ar^2+ar^3+\ldots + ar^{m-2}+ar^{m-1}=\sum_{n=0}^{m-1} ar^n,\ \ \forall_{|r|>0}\label{eq:finite1} $$

To find the summation, multiply \(\eqref{eq:finite1}\) by \(r\)

$$ Sr=ar+ar^2+ar^3+ar^4+\ldots + ar^{m-1}+ar^{m} \label{eq:finite2} $$

subtract \(\eqref{eq:finite2}\) from \(\eqref{eq:finite1}\), so that all middle terms cancel out

$$ \begin{align} S-Sr &= a-ar^{m}\nonumber\\ S(1-r) &=a(1-r^m)\nonumber\\ S &=a\left(\frac{1-r^{m}}{1-r}\right) \end{align} $$

The summation of the geometric sequence follows as

$$ \shaded{ \sum_{n=0}^{m-1}ar^n=a\left(\frac{1-r^{m}}{1-r}\right) } \label{eq:finitegreometricseries} $$

Power series

A power series is the sum of an infinite sequence of the form

$$ \sum_{n=0}^\infty a_n(r-c)^n=a_0+a_1(r-c)^1+a_2(r-c)^2+\ldots $$

where \(a\) are coefficients independent on \(r\), \(c\) is a constant.

In many situations \(c=0\) and the coefficients are the same (\(a_n=a\)), so that power series takes the form

$$ \sum_{n=0}^\infty a r^n=a+ar^1+ar^2+\ldots \label{eq:power0} $$

Equation \(\eqref{eq:power0}\) resembles \(\eqref{eq:finitegreometricseries}\) for \(m\to\infty\)

$$ \begin{align} \lim_{m\to\infty}\sum_{n=0}^{m-1}ar^n=\lim_{m\to\infty}a\left(\frac{1-r^{m}}{1-r}\right)\nonumber\\ a\sum_{n=0}^{\infty}r^n=a\underbrace{\lim_{m\to\infty}\left(\frac{1-r^{m}}{1-r}\right)} \end{align} $$

The value of \(r\) in the right term determines when the series converges

$$ \lim_{m\to\infty}\left(\frac{1-r^{m}}{1-r}\right)= \begin{cases} \text{doesn’t exist}&r\leq-1\\ \frac{1}{1-r} & -1\lt r\lt 1\\ \infty & r\geq1 \end{cases} $$

The series converges for \(|r|\lt1\), and the formula for the finite geometric series follows

$$ \begin{align} \shaded{\frac{1}{1-r}=\sum_{n=0}^{\infty}r^n},&&|r|\lt1 \label{eq:geoseries} \end{align} $$

What if \(r\gt 1\)?

$$ \begin{align} \frac{1}{1-r} &=\frac{r^{-1}}{r^{-1}-1},&|r|\gt1\nonumber\\ &=-r^{-1}\frac{1}{1-r^{-1}},&|r|\gt1 \label{eq:gt0} \end{align} $$

Substitute \(r=a^{-1}\) in \(\eqref{eq:geoseries}\)

$$ \begin{align} \sum_{n=0}^{\infty}a^{-n}&=\frac{1}{1-a^{-1}},&|r|\gt1\label{eq:gt1} \end{align} $$

Apply \(\eqref{eq:gt1}\) to \(\eqref{eq:gt0}\) gives the converging series for \(|r|\gt1\), the so called modified finite geometric series

$$ \begin{align} \shaded{ \frac{1}{1-r}=-r^{-1}\sum_{n=0}^{\infty}r^{-n}=-\sum_{n=1}^{\infty}r^{-n} },&&|r|\gt1\\ \end{align} $$

Exponential function

Power series are often the result of a Taylor series expansion. A Taylor series represents a function as an infinite sum of terms that are calculated from the function’s derivatives at one point.

$$ f(x)=\frac{f(a)}{0!}+\frac{f^\prime(a)}{1!}(x-a)+\frac{f^{\prime\prime}(a)}{2!}(x-a)^2+\ldots=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n $$

To do Taylor’s expansion of the function \(\mathrm{e}^x\), we start with its definition

$$ \left(\mathrm{e}^x\right)^\prime=\frac{\text{d}}{\text{d}y}\mathrm{e}^x=\mathrm{e}^x $$

Taylor expansion of \(\mathrm{e}^x\) at \(a=0\) using \(\mathrm{e}^x=\left(\mathrm{e}^x\right)^{\prime}=\left(\mathrm{e}^x\right)^{\prime\prime}=\ldots=\left(\mathrm{e}^x\right)^{(n)}\) and \(\mathrm{e}^0=1\)

$$ \begin{align} \mathrm{e}^x(x)&=\frac{\mathrm{e}^0}{0!}+\frac{\mathrm{e}^0}{1!}x+\frac{\mathrm{e}^0}{2!}x^2+\frac{\mathrm{e}^0}{3!}x^3+\ldots\nonumber\\ &=\frac{1}{0!}+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots \end{align} $$


$$ \mathrm{e}^x = \sum_{n=0}^{\infty}\frac{x^n}{n!} $$

Therefore, the constant \(\mathrm{e}=\mathrm{e}^1\) is

$$ \shaded{ \mathrm{e}=\sum_{n=0}^{\infty}\frac{1}{n!} } \approx 2.71828 $$


Another common power series arises from the Taylor expansion of \(\sin(x)\) at \(x=0\). To expand this, we need to examine the nth derivative of \(sin(x)\) at \(x=0\)

$$ sin^{(n)}(0) = \begin{cases} 0 & n\bmod4=0\\ 1 & n\bmod4=1\\ 0 & n\bmod4=2\\ -1 & n\bmod4=-1 \end{cases} $$


$$ \sin(x) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots $$