As the title states, I'm not sure how to prove $$\sum_{n=1}^{\infty} \arctan\left(\frac{1}{F_n}\right) \arctan\left(\frac{1}{F_{n+1}}\right)=\frac{\pi^2}{8}$$ where $F_n$ representes the $n$-th fibonacci number ($F_1=1, F_2=1, F_3=2$, etc).
This question comes from an Instagram post and WolframAlpha numerically verifies the series converges to $\frac{\pi^2}{8}$ for at least $60$ decimal points. I have seen several infinite series involving arctangent and Fibonacci numbers that end up in a telescoping sum through arctangent angle addition/subtraction identities, but I'm not sure how to approach this series with the product of two arctangent functions. I'm looking for a solution that doesn't rely on knowing the series converges to $\frac{\pi^2}{8}$.
Best Answer
Let $a_n=\arctan(1/F_n)$, $b_n=a_n^2$ for even $n$, and $b_n=a_{n-1}a_{n+1}$ for odd $n$ (here we assume $a_0=\pi/2$, so that $b_1=\pi^2/8$). Now I claim that $\color{blue}{a_n a_{n+1}=b_n-b_{n+1}}$. We have $$a_{n-1}-a_{n+1}=\arctan\frac{F_{n+1}-F_{n-1}}{F_{n-1}F_{n+1}+1}=\arctan\frac{F_n}{F_n^2+1+(-1)^n}$$ (with $n=1$ checked separately), hence $a_{n-1}-a_{n+1}=a_n$ for odd $n$, and \begin{align*} n\text{ is odd }&\implies b_n-b_{n+1}=a_{n-1}a_{n+1}-a_{n+1}^2=(a_{n-1}-a_{n+1})a_{n+1},\\n\text{ is even }&\implies b_n-b_{n+1}=a_n^2-a_n a_{n+2}=a_n(a_n-a_{n+2}), \end{align*} giving $a_n a_{n+1}$ in both cases. Thus $\sum_{n=1}^\infty a_n a_{n+1}=\sum_{n=1}^\infty(b_n-b_{n+1})=b_1$.