# 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

Archimedes encounters the series in his work Quadrature of the Parabola. He finds 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, $1+{\frac {1}{4}}+{\frac {1}{4^{2}}}+{\frac {1}{4^{3}}}+\cdots ={\frac {4}{3}}.$