markush35q

2022-03-24

How can we prove this identity:
$\sum _{n=1}^{\mathrm{\infty }}\mathrm{arctan}\frac{{\left(-1\right)}^{n}}{{\varphi }^{2n+1}}=\sum _{n=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n}}{\sqrt{5},{\varphi }^{4n-2},\left(2n-1\right),{F}_{2n-1}}$
where ${F}_{n}$ are the Fibonacci numbers, and $\varphi =\frac{1+\sqrt{5}}{2}$ is the golden ratio?

Nunnaxf18

By using the power series expansion of $\mathrm{arctan}\left(x\right)$ at $x=0$ (note that $|\frac{{\left(-1\right)}^{n}}{{\varphi }^{2n+1}}|<1$), we have that
$\sum _{n=1}^{\mathrm{\infty }}\mathrm{arctan}\frac{{\left(-1\right)}^{n}}{{\varphi }^{2n+1}}=\sum _{n=1}^{\mathrm{\infty }}\sum _{k=0}^{\mathrm{\infty }}{\left(-1\right)}^{n+k}\frac{{\varphi }^{-\left(2k+1\right)\left(2n+1\right)}}{2k+1}$
$=\sum _{k=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k}{\varphi }^{-\left(2k+1\right)}}{2k+1}\sum _{n=1}^{\mathrm{\infty }}{\left(-{\varphi }^{-\left(4k+2\right)}\right)}^{n}$
$=\sum _{k=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k}{\varphi }^{-\left(2k+1\right)}}{2k+1}\cdot \frac{-{\varphi }^{-\left(4k+2\right)}}{1+{\varphi }^{-\left(4k+2\right)}}$
$=\sum _{k=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k+1}{\varphi }^{-\left(6k+3\right)}}{2k+1}\cdot \frac{{\varphi }^{2k+1}}{{\varphi }^{\left(2k+1\right)}-{\left(-{\varphi }^{-1}\right)}^{2k+1}}$
$=\sum _{k=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k+1}{\varphi }^{-\left(4k+2\right)}}{2k+1}\cdot \frac{1}{\sqrt{5}{F}_{2k+1}}$
$=\frac{1}{\sqrt{5}}\sum _{n=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n}{\varphi }^{-\left(4n-2\right)}}{\left(2n-1\right){F}_{2n-1}}$

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