Fibonacci numbers, arctangents, and pi

Here’s an unusual formula for π. Let Fn 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

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

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