Sum of Inverse Powers of Four

$\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \ldots = \frac{1}{3} $


The full area of the square is $\square$. The sum of the black squares is thus $\sum_n \blacksquare = \sum_n \frac{1}{4^n} \square$.
We can give an alternative description in terms of the orange highlighted area $\definecolor{colorYellow}{RGB}{255,165,0}\textcolor{colorYellow}{\boxplus}$ (mouse over the above picture). Then $\sum_n \blacksquare = \sum_n \frac{1}{3} \textcolor{colorYellow}{\boxplus}$, and combining these equations, we get $\sum_n \frac{1}{4^n} \square = \sum_n \frac{1}{3} \textcolor{colorYellow}{\boxplus} = \frac{1}{3} \sum_n \textcolor{colorYellow}{\boxplus}$, but $\sum_n \textcolor{colorYellow}{\boxplus} = \square$, which concludes the proof $(\sum_n \frac{1}{4^n} \square = \frac{1}{3} \ \square)$.


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