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How can we prove this identity: $$\sum _{n=1}^\infty\arctan\frac{(-1)^n}{\phi^{2n+1}} = \sum _{n=1}^\infty \frac{(-1)^n}{\sqrt5 \, \phi ^{4 n-2} \, (2n-1) \, F_{2n-1}},\tag{$\diamond$}$$ where $F_n$ are the Fibonacci numbers, and $\phi=\frac{1+\sqrt5}2$ is the golden ratio?

Numerically, both sides look the same: $$-0.1668065238140974961621200570365353119855145764209454554444463...$$

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By using the power series expansion of $\arctan(x)$ at $x=0$ (note that $|(-1)^n/\phi^{2n+1}|<1$), we have that \begin{align*} \sum _{n=1}^\infty\arctan\frac{(-1)^n}{\phi^{2n+1}}&= \sum _{n=1}^\infty\sum_{k=0}^{\infty}(-1)^{n+k}\frac{\phi^{-(2k+1)(2n+1)}}{2k+1}\\ &=\sum_{k=0}^{\infty}\frac{(-1)^k\phi^{-(2k+1)}}{2k+1}\sum _{n=1}^\infty (-\phi^{-(4k+2)})^n\\ &=\sum_{k=0}^{\infty}\frac{(-1)^{k}\phi^{-(2k+1)}}{2k+1}\cdot \frac{-\phi^{-(4k+2)}}{1+\phi^{-(4k+2)}}\\ &=\sum_{k=0}^{\infty}\frac{(-1)^{k+1}\phi^{-(6k+3)}}{2k+1}\cdot \frac{\phi^{2k+1}}{\phi^{(2k+1)}-(-\phi^{-1})^{2k+1}}\\ &=\sum_{k=0}^{\infty}\frac{(-1)^{k+1}\phi^{-(4k+2)}}{2k+1}\cdot \frac{1}{\sqrt{5}F_{2k+1}}\\ &=\frac{1}{\sqrt{5}}\sum_{n=1}^{\infty}\frac{(-1)^{n}\phi^{-(4n-2)}}{(2n-1)F_{2n-1}}. \end{align*}

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