How to prove that $$\pi=2\sum_{n=0}^{\infty} \arctan \frac{1}{F_{2n+1}}$$ Where $F_{n}$ is the Fibonacci Number.
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Interesting, where did you get this from? – AnotherPerson Jan 03 '16 at 12:22
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2Maybe useful: $\arctan(x)+\arctan(y) = \arctan((x+y) / (1-xy))$. – Martín-Blas Pérez Pinilla Jan 03 '16 at 12:29
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You might find this interesting:http://math.stackexchange.com/questions/552727/infinite-series-sum-n-0-infty-arctan-frac1f-2n1?rq=1 – Olivier Oloa Jan 03 '16 at 12:30
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The goal is to write $\arctan\left(\dfrac1{F_{2n+1}}\right)$ as $\arctan(a_{n+1}) - \arctan(a_{n})$. This means we need $$\dfrac{a_{n+1}-a_n}{1+a_na_{n+1}} = \dfrac1{F_{2n+1}}$$ Recall that from Cassini/Catalan identity we have $$F_{2n+1}^2 = 1+F_{2n+2}F_{2n}$$ Hence, let $a_n = F_{2n}$. We then have $$\dfrac{a_{n+1}-a_n}{1+a_na_{n+1}} = \dfrac{F_{2n+2}-F_{2n}}{1+F_{2n+2}F_{2n}} = \dfrac{F_{2n+1}}{F_{2n+1}^2} = \dfrac1{F_{2n+1}}$$ Hence, we have $$\arctan\left(\dfrac1{F_{2n+1}}\right) = \arctan(F_{2n+2})-\arctan(F_{2n})$$ I trust you can finish off from here.
Hence, we have $$\sum_{n=0}^m \arctan\left(\dfrac1{F_{2n+1}}\right) = \arctan(F_{2m+2}) \implies \sum_{n=0}^{\infty} \arctan\left(\dfrac1{F_{2n+1}}\right) = \lim_{m \to \infty}\arctan(F_{2m+2}) = \dfrac{\pi}2$$
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