How to evaluate : $\sum_{n=0}^{\infty}\left(\frac{F_{n+x}}{F_{n+x+1}^2}-\frac{F_{n+x+2}}{F_{n+x+3}^2}\right)$ Thank you to &robjohn for proving this question, I got motivated by it, so I went on to investigate this sum $(1)$,
$$S_{x}=\sum_{n=0}^{\infty}\left(\frac{F_{n+x}}{F_{n+x+1}^2}-\frac{F_{n+x+2}}{F_{n+x+3}^2}\right)\tag1$$
using wolfram alpha we got the first six terms of $(1)$
$S_{0}=1$, $S_{1}=\frac{5}{4}$, $S_{2}=\frac{17}{36}$, $S_3=\frac{77}{225}$, $S_4=\frac{317}{1600}$ $S_5=\frac{1357}{10816}$$\cdots$
by observing the pattern of the sequence, I believe that the so-called closed-form is $(2)$
$$\left(\sum_{m=0}^{x}F_{m+1}^2\right)^2\times\sum_{n=0}^{\infty}\left(\frac{F_{n+x}}{F_{n+x+1}^2}-\frac{F_{n+x+2}}{F_{n+x+3}^2}\right)=1+2\sum_{j=0}^{x}F_jF_{j+1}F_{j+2}\tag2$$
Where $F_n$ is the Fibonacci number and $x\ge 0$
obviously $(2)$ can be written as
$$\sum_{n=0}^{\infty}\left(\frac{F_{n+x}}{F_{n+x+1}^2}-\frac{F_{n+x+2}}{F_{n+x+3}^2}\right)=\frac{1}{(F_{x+1}F_{x+2})^2}\left(1+2\sum_{j=0}^{x}F_jF_{j+1}F_{j+2}\right)\tag3$$
How do we prove $(2)?$

This seems to be a telescoping series:
$$
\begin{align}
S_{x}
&=\sum_{n=0}^{\infty}\left(\frac{F_{n+x}}{F_{n+x+1}^2}-\frac{F_{n+x+2}}{F_{n+x+3}^2}\right)\\
&=\sum_{n=0}^{\infty}\frac{F_{n+x}}{F_{n+x+1}^2}-\sum_{n=2}^{\infty}\frac{F_{n+x}}{F_{n+x+1}^2}\\
&=\frac{F_x}{F_{x+1}^2}+\frac{F_{x+1}}{F_{x+2}^2}
\end{align}
$$
