<s>[INST] Prove $\sum_{n=1}^{\infty} \arctan\left(\frac{1}{F_n}\right) \arctan\left(\frac{1}{F_{n+1}}\right)=\frac{\pi^2}{8}$ 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 $\frac{\pi^2}{8}$.
 [/INST] COMMENT: May be this helps:
$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$
$\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy}$
Squaring both sides of relations and subtract we get:
$$4\tan^{-1}x \tan^{-1}y=\left(\tan^{-1}\frac{x+y}{1-xy}\right)^2-\left(\tan^{-1}\frac{x-y}{1+xy}\right)^2$$
$x=\frac 1{F_n}$, and, $y=\frac 1{F_{n+1}}$
Now we use some relations between Fibonacci numbers, golden ratio $c_n$ is:
$c_n=\frac {F_{n+1}}{F_n}=\frac {F_nF_{n+1}}{F_n^2}\rightarrow F_nF_{n+1}=c_nF_n^2$
$$x+y=\frac1{F_n}+\frac1{F_{n+1}}=\frac{F_{n+2}}{c_n F_n^2}\rightarrow \frac{x+y}{1-xy}=\frac{F_nF_{n+1}F_{n+2}}{c_nF_n^2(F_nF_{n+1}-1)}$$
Similarly:
$$x-y=\frac1{F_n}-\frac1{F_{n+1}}=\frac{F_{n-1}}{c_n F_n^2}\rightarrow \frac{x-y}{1+xy}=\frac{F_{n-1}F_nF_{n+1}}{c_nF_n^2(F_nF_{n+1}+1)}$$
Now puting $F_1, F_2, F_3 \cdots$ gives us a numerical sum which must result in $\frac{\pi^2}8$.
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