In this question, the OP says that
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{F_kF_{k+1}}=\frac{1}{\phi}$$ where $F_n$ is the $n$th Fibonacci number, defined by $F_n=F_{n-1}+F_{n-2}$, with $F_1=F_2=1$.
I've been trying to prove this, but I haven't had much luck. Below is the work I have done.
I'll firstly write out a bunch of terms of the series. $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{F_kF_{k+1}}=\frac{1}{F_{1}F_{2}}-\frac{1}{F_{2}F_{3}}+\frac{1}{F_{3}F_{4}}-\frac{1}{F_{4}F_{5}}+\frac{1}{F_{5}F_{6}}-\frac{1}{F_{6}F_{7}}+\cdots$$ It therefore follows that $$\begin{align}\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{F_kF_{k+1}}&=\frac{1}{F_2}\left(\frac{1}{F_1}-\frac{1}{F_3}\right)+\frac{1}{F_4}\left(\frac{1}{F_3}-\frac{1}{F_5}\right)+\cdots\\ &=\frac{1}{F_1F_3}+\frac{1}{F_3F_5}+\cdots\\ &=\sum_{k=1}^{\infty}\frac{1}{F_{2k-1}F_{2k+1}}\tag{1}\end{align}$$ Similarly, $$\begin{align}\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{F_kF_{k+1}}&=\frac{1}{F_1F_2}-\frac{1}{F_3}\left(\frac{1}{F_2}-\frac{1}{F_4}\right)-\frac{1}{F_5}\left(\frac{1}{F_4}-\frac{1}{F_6}\right)-\cdots\\ &=1-\frac{1}{F_2F_4}-\frac{1}{F_4F_6}-\cdots\\ &=1-\sum_{k=1}^{\infty}\frac{1}{F_{2k}F_{2k+2}}\tag{2}\end{align}$$
Unfortunately I haven't had any luck in evaluating these. I did think of using Cassini's identity, which yielded the following when applied to $(1)$ and $(2)$ respectively: $$\begin{align}\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{F_kF_{k+1}}&=\sum_{k=1}^{\infty}\frac{1}{1+F_{2k}^2}\tag{3}\\ &=1-\sum_{k=1}^{\infty}\frac{1}{F_{2k+1}^2-1}\tag{4}\end{align}$$
But once more I could not see a way of evaluating these series either. Usually with summations involving Fibonacci numbers forming a telescoping series is effective, but I couldn't succeed in making this one telescope.
Thank you for your help, it is really appreciated. Any possible methods will be a great help.
