Sum One

Algebra Level 2

What does this picture best illustrate?

∏ n = 0 ∞ 1 2 n = 1 \prod_{n = 0}^{\infty}\frac{1}{2^n} = 1 ∑ n = 1 ∞ 1 2 n = 1 \sum_{n = 1}^{\infty}\frac{1}{2^n} = 1 ∑ n = 1 ∞ 1 n 2 = 1 \sum_{n = 1}^{\infty}\frac{1}{n^2} = 1 ∑ n = 0 ∞ 1 2 n = 1 \sum_{n = 0}^{\infty}\frac{1}{2^n} = 1 ∑ n = 0 ∞ 1 n 2 = 1 \sum_{n = 0}^{\infty}\frac{1}{n^2} = 1 ∏ n = 1 ∞ 1 2 n = 1 \prod_{n = 1}^{\infty}\frac{1}{2^n} = 1 ∏ n = 1 ∞ 1 n 2 = 1 \prod_{n = 1}^{\infty}\frac{1}{n^2} = 1 ∏ n = 0 ∞ 1 n 2 = 1 \prod_{n = 0}^{\infty}\frac{1}{n^2} = 1

Geoff Pilling by Geoff Pilling , 53, USA

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1 solution

Geoff Pilling
Mar 16, 2017

The picture shows that if you keep dividing the remaining area in half, and adding it to the previous total you eventually (at ∞ \infty ) get unity represented by the complete square, or ∑ n = 1 ∞ 1 2 n = 1 \boxed{\sum_{n = 1}^{\infty}\frac{1}{2^n} = 1}

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