Geometric Series with Geometric Proofs

Problem: \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots = 1

Solution:


Problem\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots = \frac{1}{2}

Solution:

Problem\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \dots = \frac{1}{3}

Solution:


Problem: 1 + r + r^2 + \dots = \frac{1}{1-r} if r < 1.

Solution:

This last one follows from similarity of the subsequent trapezoids: the right edge of the teal(ish) trapezoid has length r, and so the right edge of the neighboring trapezoid, x, is found by \frac{r}{1} = \frac{x}{r}, and we see that it has length r^2.

We may come up with infinitely many proofs of these geometric series! All we need is a figure which can be dissected into n self-similar parts, where the geometric series is a sum of powers of \frac{1}{n}. Awesome.

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