Fibonacci, arctangents, and pi

Here’s an unusual formula for π. Let F_n be the nth Fibonacci number. Then

$\pi = 4 \sum_{n=1}^\infty \arctan\left( \frac{1}{F_{2n+1}} \right)$

As mysterious as this equation may seem, it’s not hard to prove. The arctangent identity

$\arctan\left(\frac{1}{F_{2n+1}}\right) = \arctan\left(\frac{1}{F_{2n}}\right) - \arctan\left(\frac{1}{F_{2n+2}}\right)$

shows that the sum telescopes, leaving only the first term, arctan(1) = π/4. To prove the arctangent identity, take the tangent of both sides, use the addition law for tangents, and use the Fibonacci identity

$F_{n+1} F_{n-1} - F_n^2 = (-1)^n$

See this post for an even more remarkable formula relating Fibonacci numbers and π.

One thought on “Fibonacci numbers, arctangents, and pi”

Nat Kuhn

9 December 2017 at 13:52

This is lovely, thanks! In the second equation, I believe the first minus sign should be an equal sign.

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