<s>[INST] Show that: $\sum_{n=0}^{\infty}(-1)^{n+1}\left(\frac{F_n}{F_{n+1}F_{n+2}}\right)^2=\frac{1}{\phi^3}$ How to show that?
$$S=\sum_{n=0}^{\infty}(-1)^{n+1}\left(\frac{F_n}{F_{n+1}F_{n+2}}\right)^2=\frac{1}{\phi^3}$$
Where $F_n$ Fibonacci  number
$F_n=\frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}}$
$$F_n^2=\frac{\phi^{2n}-2\phi^n(-\phi)^{-n}+(-\phi)^{-2n}}{5}$$
$$S=\sum_{n=1}^{\infty}(-1)^{n+1}2\left(\frac{\phi^n}{F_{n+1}F_{n+2}}\right)^2+\sum_{n=1}^{\infty}\left(\frac{1}{F_{n+1}F_{n+2}}\right)^2$$
 [/INST] Lemma: Let $F_n$ be Fibonacci Numbers and $L_n$ be Lucas Numbers, then
$$
F_nF_{n+1}-F_{n-k}F_{n+k+1}=\frac{(-1)^{n+1}+(-1)^{n-k}L_{2k+1}}5\tag1
$$
Proof: Using $F_n=\frac{\phi^n-(-1/\phi^n)}{\sqrt5}$ and $L_n=\phi^n+(-1/\phi)^n$, we get
$$
\begin{align}
&5(F_nF_{n+1}-F_{n-k}F_{n+k+1})\\
&=\left(\phi^n-(-1/\phi)^n\right)\left(\phi^{n+1}-(-1/\phi)^{n+1}\right)
-\left(\phi^{n-k}-(-1/\phi)^{n-k}\right)\left(\phi^{n+k+1}-(-1/\phi)^{n+k+1}\right)\\
&=(-1)^{n+1}(\phi-1/\phi)+(-1)^{n-k}\left(\phi^{2k+1}-1/\phi^{2k+1}\right)\\
&=(-1)^{n+1}+(-1)^{n-k}L_{2k+1}
\end{align}
$$
$\large\square$
Therefore,
$$
\begin{align}
\sum_{k=0}^\infty\frac1{F_{2k+1}F_{2k+3}}
&=\sum_{k=0}^\infty\frac{F_{2k+1}F_{2k+2}-F_{2k}F_{2k+3}}{F_{2k+1}F_{2k+3}}\tag{2a}\\
&=\sum_{k=0}^\infty\left(\frac{F_{2k+2}}{F_{2k+3}}-\frac{F_{2k}}{F_{2k+1}}\right)\tag{2b}\\
&=\frac1\phi\tag{2c}
\end{align}
$$
Explanation:
$\text{(2a)}$: apply the Lemma substituting $(n,k)\mapsto(2k+1,1)$
$\text{(2b)}$: algebra
$\text{(2c)}$: the partial sum to $n$ telescopes to $\frac{F_{2n+2}}{F_{2n+3}}$, which limits to $\frac1\phi$

$$
\begin{align}
\sum_{n=0}^\infty(-1)^{n+1}\left(\frac{F_n}{F_{n+1}F_{n+2}}\right)^2
&=\sum_{n=0}^\infty(-1)^{n+1}\left(\frac{F_{n+2}-F_{n+1}}{F_{n+1}F_{n+2}}\right)^2\tag{3a}\\
&=\sum_{n=0}^\infty(-1)^{n+1}\left(\frac1{F_{n+1}^2}+\frac1{F_{n+2}^2}-\frac2{F_{n+1}F_{n+2}}\right)\tag{3b}\\
&=\sum_{n=0}^\infty(-1)^{n+1}\frac1{F_{n+1}^2}-\sum_{n=1}^\infty(-1)^{n+1}\frac1{F_{n+1}^2}\\
&+2\sum_{n=0}^\infty\left(\frac1{F_{2n+1}F_{2n+2}}-\frac1{F_{2n+2}F_{2n+3}}\right)\tag{3c}\\
&=-1+2\sum_{n=0}^\infty\frac1{F_{2n+1}F_{2n+3}}\tag{3d}\\[3pt]
&=-1+\frac2\phi\tag{3e}\\[6pt]
&=\frac1{\phi^3}\tag{3f}
\end{align}
$$
Explanation:
$\text{(3a)}$: $F_n=F_{n+2}-F_{n+1}$
$\text{(3b)}$: algebra
$\text{(3c)}$: break the sum into three pieces
$\phantom{\text{(3c):}}$ substitute $n\mapsto n-1$ in the second piece
$\phantom{\text{(3c):}}$ break the third piece into two to remove $(-1)^{n+1}$
$\text{(3d)}$: cancel identical terms in the first two sums
$\phantom{\text{(3d):}}$ combine terms in the third sum using $F_{2k+3}-F_{2k+1}=F_{2k+2}$
$\phantom{\text{(3d):}}$ and cancel $F_{2k+2}$ in the numerator and denominator
$\text{(3e)}$: apply $(2)$
$\text{(3f)}$: $2-\phi=\frac1{\phi^2}$
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