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.

Leave a Reply

Your email address will not be published. Required fields are marked *