A Fibonacci series

Let $\mathcal{F}_n$ denote the $n$ – th Fibonacci number. Prove that

$\sum_{n=2}^{\infty}\arctan \left(\frac{\mathcal{F}_{n-1}}{\mathcal{F}_n \mathcal{F}_{n+1}+1}\right)\arctan \left(\frac{\mathcal{F}_{n+2}}{\mathcal{F}_n \mathcal{F}_{n+1}-1}\right)=\frac{\pi^2}{16}$

Solution

Let $S$ denote the given sum. Then,

$\begin{align*} \mathcal{S} & =\sum_{n=2}^{\infty}\arctan \left(\frac{\mathcal{F}_{n+1}-\mathcal{F}_n}{\mathcal{F}_n\mathcal{F}_{n+1}+1}\right)\arctan \left(\frac{\mathcal{F}_{n+1}+\mathcal{F}_n}{\mathcal{F}_n\mathcal{F}_{n+1}-1}\right) \\ &=\sum_{n=2}^{\infty}(\arctan \mathcal{F}_{n+1}-\arctan \mathcal{F}_n)(\pi-\arctan \mathcal{F}_{n+1}-\arctan \mathcal{F}_n) \\ &= \pi \sum_{n=2}^{\infty}(\arctan \mathcal{F}_{n+1}-\arctan \mathcal{F}_n)- \sum_{n=2}^{\infty}\left(\arctan^2 \mathcal{F}_{n+1} -\arctan^2 \mathcal{F}_n\right) \\ &= \pi \lim_{n \rightarrow +\infty}(\arctan \mathcal{F}_n-\arctan \mathcal{F}_2)- \lim_{n \rightarrow + \infty} \left( \arctan^2 \mathcal{F}_n -\arctan^2 \mathcal{F}_2 \right) \\ &= \pi \left ( \frac{\pi}{2} - \frac{\pi}{4} \right ) - \left ( \frac{\pi^2}{4} - \frac{\pi^2}{16} \right ) \\ &= \frac{\pi^2}{16} \end{align*}$