1/4 + 1/16 + 1/64 + 1/256 + ⋯

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

Archimedes proves the proposition by first calculating $${\displaystyle \begin{array}{rcl}{\displaystyle B+C+\cdots +Z+\frac{B}{3}+\frac{C}{3}+\cdots +\frac{Z}{3}}& =& {\displaystyle \frac{4B}{3}+\frac{4C}{3}+\cdots +\frac{4Z}{3}}\\ & =& {\displaystyle \frac{1}{3}(A+B+\cdots +Y).}\end{array}}$$ On the other hand, $${\displaystyle \frac{B}{3}+\frac{C}{3}+\cdots +\frac{Y}{3}=\frac{1}{3}(B+C+\cdots +Y).}$$

Subtracting this equation from the previous equation yields $${\displaystyle B+C+\cdots +Z+\frac{Z}{3}=\frac{1}{3}A}$$ 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: $${\displaystyle 1+\frac{1}{4}+\frac{1}{4^2}+\cdots +\frac{1}{4^n}=\frac{1-{\left(\frac{1}{4}\right)}^{n+1}}{1-\frac{1}{4}}.}$$

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

Since the sum of an infinite series is defined as the limit of its partial sums, $${\displaystyle 1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\cdots =\frac{4}{3}.}$$