Proving the Fibonacci sum $\sum_{n=1}^{\infty}\left(\frac{F_{n+2}}{F_{n+1}}-\frac{F_{n+3}}{F_{n+2}}\right) = \frac{1}{\phi^2}$ and its friends In this article, (eq.92) has,
$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{F_{n+1}F_{n+2}} = \frac{1}{\phi^2}\tag1$$
and I wondered if this could be generalized to the tribonacci numbers. It seems it can be. Given the Fibonacci, tribonacci, tetranacci (in general, the Fibonacci k-step numbers) starting with $n=1$,
$$F_n = 1,1,2,3,5,8\dots$$
$$T_n = 1, 1, 2, 4, 7, 13, 24,\dots$$
$$U_n = 1, 1, 2, 4, 8, 15, 29, \dots$$
and their limiting ratios, $x_k$, the root $x_k \to 2$ of,
$$(2-x)x^k = 1$$
with Fibonacci constant $x_2$, tribonacci constant $x_3$, etc, it can be empirically observed that,
$$\sum_{n=1}^{\infty}\left(\frac{F_{n+2}}{F_{n+1}}-\frac{F_{n+3}}{F_{n+2}}\right) = \frac{1}{x_2^2}\tag2$$
$$\sum_{n=1}^{\infty}\left(\frac{T_{n+2}}{T_{n+1}}-\frac{T_{n+3}}{T_{n+2}}\right) = \frac{1}{x_3^3}$$
$$\sum_{n=1}^{\infty}\left(\frac{U_{n+2}}{U_{n+1}}-\frac{U_{n+3}}{U_{n+2}}\right) = \frac{1}{x_4^4}$$
and so on. Q: How do we rigorously prove the observation indeed holds for all integer $k\geq2$?
Edit:
To address a comment that disappeared, to transform $(1)$ to $(2)$, we use a special case of Catalan's identity,
$$F_{n+2}^2-F_{n+1}F_{n+3} = (-1)^{n+1}$$
so,
$$\begin{aligned}
\frac{(-1)^{n+1}}{F_{n+1}F_{n+2}} 
&= \frac{F_{n+2}^2-F_{n+1}F_{n+3}}{F_{n+1}F_{n+2}}\\
&= \frac{F_{n+2}}{F_{n+1}} - \frac{F_{n+3}}{F_{n+2}}
\end{aligned}$$
hence the alternating series $(1)$ is equal to $(2)$. 

Let's start off with the case of the Fibonacci sequence. We have
$$\sum_{n = 1}^k \left( \frac{F_{n+2}}{F_{n+1}} -\frac{F_{n+3}}{F_{n+2}} \right) = \left(\frac{F_3}{F_2} - \frac{F_4}{F_3}\right) + \left(\frac{F_4}{F_3} - \frac{F_5}{F_4}\right) + \cdots + \left(\frac{F_{k+2}}{F_{k+1}}- \frac{F_{k+3}}{F_{k+2}}\right)\\
= \frac{F_3}{F_2} - \frac{F_{k+3}}{F_{k+2}} = 2 - \frac{F_{k+3}}{F_{k+2}}.$$
Setting the limit as $k$ approaches infinity, we get
$$2 - \lim_{k\to\infty} \frac{F_{k+3}}{F_{k+2}} = 2- x = \frac{1}{x^2}$$
If you want more rigor, you can easily transform this into an induction argument.
Since $(2 - x)(x^2) = 1 \implies 2 - x = \frac{1}{x^2}$, the result follows. The proof generalizes really easily.
