1/4 + 1/16 + 1/64 + 1/256 + ⋯

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Archimedes' figure with a = 3/4

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

\frac {\frac 1 4} {1 - \frac 1 4}=\frac 1 3.

Visual demonstrations[edit]

3s = 1.

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

3\left(\frac14+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+\cdots\right) = 1.

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

3s = 1 again
\frac34+\frac{3}{4^2}+\frac{3}{4^3}+\frac{3}{4^4}+\cdots = 1.

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 Sierpinski triangle.[7]


This curve is a parabola. The dots on the secant line AE are equally spaced. Archimedes showed that the sum of the areas of triangles ABC and CDE is 1/4 of the area of triangle ACE. He then constructs another layer of four triangles atop those, the sum of whose areas is 1/4 of the sum of the areas of ABC and CDE, and then another layer of eight triangles atop that, having 1/4 of that area, and so on. He concluded that the area between the secant line and the curve is 4/3 the area of triangle ACE.

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 in that the total area is 4/3 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 areas A, B, C, D, … , Z, of which A is the greatest, and each is equal to four times the next in order, then[8]

A + B + C + D + \cdots + Z + \frac13 Z = \frac43 A.

Archimedes proves the proposition by first calculating

\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).

On the other hand,

\frac{B}{3}+\frac{C}{3}+\cdots+\frac{Y}{3} = \frac13(B+C+\cdots+Y).

Subtracting this equation from the previous equation yields

B+C+\cdots+Z+\frac{Z}{3} = \frac13 A

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:


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 limit[edit]

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:

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

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

1+\frac14+\frac{1}{4^2}+\frac{1}{4^3}+\cdots = \frac43.


  1. ^ Shawyer and Watson p. 3.
  2. ^ a b Nelsen and Alsina p. 74.
  3. ^ Ajose and Nelson.
  4. ^ Heath p.250
  5. ^ Stein p. 46.
  6. ^ Mabry.
  7. ^ Nelson and Alsina p.56
  8. ^ This is a quotation from Heath's English translation (p.249).
  9. ^ This presentation is a shortened version of Heath p.250.
  10. ^ 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.


  • 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: 14 + (14)2 + (14)3 + · · · = 13". Mathematics Magazine 72 (1): 63. JSTOR 2691318. 
  • Nelsen, Roger B. and Claudi Alsina (2006). Math Made Visual: Creating Images for Understanding Mathematics. MAA. ISBN 0-88385-746-4. 
  • Shawyer, Bruce and Bruce Watson (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 and Thomas Dence (April 1998). "Archimedes' Quadrature of the Parabola Revisited". Mathematics Magazine 71 (2): 123–30. doi:10.2307/2691014. JSTOR 2691014.