In mathematics, the infinite series **1/4 + 1/16 + 1/64 + 1/256 + ⋯** is an example of one of the first infinite series to be summed in the history of mathematics; it was used by Archimedes circa 250–200 BC.^{[1]} As it is a geometric series with first term 1/4 and common ratio 1/4, its sum is

- <math>\sum_{n=1}^\infty \frac{1}{4^n}=\frac {\frac 1 4} {1 - \frac 1 4}=\frac 1 3.</math>

## Visual demonstrationsEdit

The series 1/4 + 1/16 + 1/64 + 1/256 + ⋯ lends itself to some particularly simple visual demonstrations because a square and a triangle both divide into four similar pieces, each of which contains 1/4 the area of the original.

In the figure on the left,^{[2]}^{[3]} if the large square is taken to have area 1, then the largest black square has area 1/2 × 1/2 = 1/4. Likewise, the second largest black square has area 1/16, and the third largest black square has area 1/64. The area taken up by all of the black squares together is therefore 1/4 + 1/16 + 1/64 + ⋯, and this is also the area taken up by the gray squares and the white squares. Since these three areas cover the unit square, the figure demonstrates that

- <math>3\left(\frac14+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+\cdots\right) = 1.</math>

Archimedes' own illustration, adapted at top,^{[4]} was slightly different, being closer to the equation

- <math>\sum_{n=1}^\infty \frac{3}{4^n}=\frac34+\frac{3}{4^2}+\frac{3}{4^3}+\frac{3}{4^4}+\cdots = 1.</math>

See below for details on Archimedes' interpretation.

The same geometric strategy also works for triangles, as in the figure on the right:^{[2]}^{[5]}^{[6]} if the large triangle has area 1, then the largest black triangle has area 1/4, and so on. The figure as a whole has a self-similarity between the large triangle and its upper sub-triangle. A related construction making the figure similar to all three of its corner pieces produces the Sierpiński triangle.^{[7]}

## Proof by ArchimedesEdit

Archimedes encounters the series in his work *Quadrature of the Parabola*. He is finding the area inside a parabola by the method of exhaustion, and he gets a series of triangles; each stage of the construction adds an area 1/4 times the area of the previous stage. His desired result is that the total area is 4/3 times the area of the first stage. To get there, he takes a break from parabolas to introduce an algebraic lemma:

Proposition 23.Given a series of areasA,B,C,D, … ,Z, of whichAis the greatest, and each is equal to four times the next in order, then^{[8]}

- <math>A + B + C + D + \cdots + Z + \frac13 Z = \frac43 A.</math>

Archimedes proves the proposition by first calculating

- <math>\begin{array}{rcl}

\displaystyle B+C+\cdots+Z+\frac{B}{3}+\frac{C}{3}+\cdots+\frac{Z}{3} & = &\displaystyle \frac{4B}{3}+\frac{4C}{3}+\cdots+\frac{4Z}{3} \\[1em] & = &\displaystyle \frac13(A+B+\cdots+Y). \end{array}</math> On the other hand,

- <math>\frac{B}{3}+\frac{C}{3}+\cdots+\frac{Y}{3} = \frac13(B+C+\cdots+Y).</math>

Subtracting this equation from the previous equation yields

- <math>B+C+\cdots+Z+\frac{Z}{3} = \frac13 A</math>

and adding *A* to both sides gives the desired result.^{[9]}

Today, a more standard phrasing of Archimedes' proposition is that the partial sums of the series 1 + 1/4 + 1/16 + ⋯ are:

- <math>1+\frac{1}{4}+\frac{1}{4^2}+\cdots+\frac{1}{4^n}=\frac{1-\left(\frac14\right)^{n+1}}{1-\frac14}.</math>

This form can be proved by multiplying both sides by 1 − 1/4 and observing that all but the first and the last of the terms on the left-hand side of the equation cancel in pairs. The same strategy works for any finite geometric series.

## The limitEdit

Archimedes' Proposition 24 applies the finite (but indeterminate) sum in Proposition 23 to the area inside a parabola by a double *reductio ad absurdum*. He does not *quite*^{[10]} take the limit of the above partial sums, but in modern calculus this step is easy enough:

- <math>\lim_{n\to\infty} \frac{1-\left(\frac14\right)^{n+1}}{1-\frac14} = \frac{1}{1-\frac14} = \frac43.</math>

Since the sum of an infinite series is defined as the limit of its partial sums,

- <math>1+\frac14+\frac{1}{4^2}+\frac{1}{4^3}+\cdots = \frac43.</math>

## NotesEdit

- ↑ Shawyer and Watson p. 3.
- ↑
^{2.0}^{2.1}Nelsen and Alsina p. 74. - ↑ Ajose and Nelson. p. 230
- ↑ Heath p. 250
- ↑ Stein p. 46.
- ↑ Mabry. p. 63
- ↑ Nelson and Alsina p. 56
- ↑ This is a quotation from Heath's English translation (p. 249).
- ↑ This presentation is a shortened version of Heath p. 250.
- ↑ Modern authors differ on how appropriate it is to say that Archimedes summed the infinite series. For example, Shawyer and Watson (p. 3) simply say he did; Swain and Dence say that "Archimedes applied an indirect limiting process"; and Stein (p. 45) stops short with the finite sums.

## ReferencesEdit

- Ajose, Sunday and Roger Nelsen (June 1994). "Proof without Words: Geometric Series".
*Mathematics Magazine*.**67**(3): 230. doi:10.2307/2690617. JSTOR 2690617. - Heath, T. L. (1953) [1897].
*The Works of Archimedes*. Cambridge UP. Page images at Casselman, Bill. "Archimedes' quadrature of the parabola". Retrieved 2007-03-22. HTML with figures and commentary at Otero, Daniel E. (2002). "Archimedes of Syracuse". Archived from the original on 7 March 2007. Retrieved 2007-03-22. - Mabry, Rick (February 1999). "Proof without Words: <math>\frac{1}{4}</math> + <math>(\frac{1}{4})^2</math> + <math>(\frac{1}{4})^3</math> + ⋯ = <math>\frac{1}{3}</math>".
*Mathematics Magazine*.**72**(1): 63. doi:10.1080/0025570X.1999.11996702. JSTOR 2691318. - Nelsen, Roger B.; Alsina, Claudi (2006).
*Math Made Visual: Creating Images for Understanding Mathematics*. MAA. ISBN 0-88385-746-4. - Shawyer, Bruce; Watson, Bruce (1994).
*Borel's Methods of Summability: Theory and Applications*. Oxford UP. ISBN 0-19-853585-6. - Stein, Sherman K. (1999).
*Archimedes: What Did He Do Besides Cry Eureka?*. MAA. ISBN 0-88385-718-9. - Swain, Gordon; Dence, Thomas (April 1998). "Archimedes' Quadrature of the Parabola Revisited".
*Mathematics Magazine*.**71**(2): 123–30. doi:10.2307/2691014. JSTOR 2691014.