Dissection of the parabolic segment

The main idea of the proof is the dissection of the parabolic segment into infinitely many triangles, as shown in the figure to the right. Each of these triangles is inscribed in its own parabolic segment in the same way that the blue triangle is inscribed in the large segment.

Areas of the triangles

In propositions eighteen through twenty-one, Archimedes proves that the area of each green triangle is ${\displaystyle {\tfrac {1}{8}}}$ the area of the blue triangle, so that both green triangles together sum to ${\displaystyle {\tfrac {1}{4}}}$ the area of the blue triangle. From a modern point of view, this is because the green triangle has ${\displaystyle {\tfrac {1}{2}}}$ the width and ${\displaystyle {\tfrac {1}{4}}}$ the height of the blue triangle:^[7]

Following the same argument, each of the ${\displaystyle 4}$ yellow triangles has ${\displaystyle {\tfrac {1}{8}}}$ the area of a green triangle or ${\displaystyle {\tfrac {1}{64}}}$ the area of the blue triangle, summing to ${\displaystyle {\tfrac {4}{64}}={\tfrac {1}{16}}}$ the area of the blue triangle; each of the ${\displaystyle 2^{3}=8}$ red triangles has ${\displaystyle {\tfrac {1}{8}}}$ the area of a yellow triangle, summing to ${\displaystyle {\tfrac {2^{3}}{8^{3}}}={\tfrac {1}{64}}}$ the area of the blue triangle; etc. Using the method of exhaustion, it follows that the total area of the parabolic segment is given by

${\displaystyle {\text{Area}}\;=\;T\,+\,{\frac {1}{4}}T\,+\,{\frac {1}{4^{2}}}T\,+\,{\frac {1}{4^{3}}}T\,+\,\cdots .}$

Here T represents the area of the large blue triangle, the second term represents the total area of the two green triangles, the third term represents the total area of the four yellow triangles, and so forth. This simplifies to give

${\displaystyle {\text{Area}}\;=\;\left(1\,+\,{\frac {1}{4}}\,+\,{\frac {1}{16}}\,+\,{\frac {1}{64}}\,+\,\cdots \right)T.}$

Sum of the series

Archimedes' proof that 1/4 + 1/16 + 1/64 + ⋯ = 1/3

To complete the proof, Archimedes shows that

${\displaystyle 1\,+\,{\frac {1}{4}}\,+\,{\frac {1}{16}}\,+\,{\frac {1}{64}}\,+\,\cdots \;=\;{\frac {4}{3}}.}$

The formula above is a geometric series—each successive term is one fourth of the previous term. In modern mathematics, that formula is a special case of the sum formula for a geometric series.

Archimedes evaluates the sum using an entirely geometric method,^[8] illustrated in the adjacent picture. This picture shows a unit square which has been dissected into an infinity of smaller squares. Each successive purple square has one fourth the area of the previous square, with the total purple area being the sum

${\displaystyle {\frac {1}{4}}\,+\,{\frac {1}{16}}\,+\,{\frac {1}{64}}\,+\,\cdots .}$

However, the purple squares are congruent to either set of yellow squares, and so cover ${\displaystyle {\tfrac {1}{3}}}$ of the area of the unit square. It follows that the series above sums to ${\displaystyle {\tfrac {4}{3}}}$ (since ${\displaystyle 1+{\tfrac {1}{3}}={\tfrac {4}{3}}}$).