Proving $\pi=2\sum_{n=0}^{\infty} \arctan \frac{1}{F_{2n+1}}$ How to prove that $$\pi=2\sum_{n=0}^{\infty} \arctan \frac{1}{F_{2n+1}}$$
Where  $F_{n}$ is the Fibonacci Number.

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 $$\sum_{n=0}^m \arctan\left(\dfrac1{F_{2n+1}}\right) = \arctan(F_{2m+2}) \implies \sum_{n=0}^{\infty} \arctan\left(\dfrac1{F_{2n+1}}\right) = \lim_{m \to \infty}\arctan(F_{2m+2}) = \dfrac{\pi}2$$

