# 1/4 + 1/16 + 1/64 + 1/256 + ⋯

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. As it is a geometric series with first term 1/4 and common ratio 1/4, its sum is

$\sum _{n=1}^{\infty }{\frac {1}{4^{n}}}={\frac {\frac {1}{4}}{1-{\frac {1}{4}}}}={\frac {1}{3}}.$ ## Visual demonstrations

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, 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

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

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

$\sum _{n=1}^{\infty }{\frac {3}{4^{n}}}={\frac {3}{4}}+{\frac {3}{4^{2}}}+{\frac {3}{4^{3}}}+{\frac {3}{4^{4}}}+\cdots =1.$

See below for details on Archimedes' interpretation.

The same geometric strategy also works for triangles, as in the figure on the right: 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.

## Proof by Archimedes

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 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 areas A, B, C, D, … , Z, of which A is the greatest, and each is equal to four times the next in order, then

$A+B+C+D+\cdots +Z+{\frac {1}{3}}Z={\frac {4}{3}}A.$

Archimedes proves the proposition by first calculating

${\begin{array}{rcl}\displaystyle B+C+\cdots +Z+{\frac {B}{3}}+{\frac {C}{3}}+\cdots +{\frac {Z}{3}}&=&\displaystyle {\frac {4B}{3}}+{\frac {4C}{3}}+\cdots +{\frac {4Z}{3}}\\[1em]&=&\displaystyle {\frac {1}{3}}(A+B+\cdots +Y).\end{array}}$

On the other hand,

${\frac {B}{3}}+{\frac {C}{3}}+\cdots +{\frac {Y}{3}}={\frac {1}{3}}(B+C+\cdots +Y).$

Subtracting this equation from the previous equation yields

$B+C+\cdots +Z+{\frac {Z}{3}}={\frac {1}{3}}A$

and adding A to both sides gives the desired result.

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

$1+{\frac {1}{4}}+{\frac {1}{4^{2}}}+\cdots +{\frac {1}{4^{n}}}={\frac {1-\left({\frac {1}{4}}\right)^{n+1}}{1-{\frac {1}{4}}}}.$

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

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 take the limit of the above partial sums, but in modern calculus this step is easy enough:

$\lim _{n\to \infty }{\frac {1-\left({\frac {1}{4}}\right)^{n+1}}{1-{\frac {1}{4}}}}={\frac {1}{1-{\frac {1}{4}}}}={\frac {4}{3}}.$

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

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