Infinite Fibonacci sums $\sum_{n=1}^{\infty} \frac{1}{f_nf_{n+2}}$ - diverge or converge

Notice $$\frac{1}{f_n f_{n+2}} = \frac{f_{n+1}}{f_nf_{n+1}f_{n+2}} =\frac{f_{n+2}-f_{n}}{f_n f_{n+1} f_{n+2}} = \frac{1}{f_nf_{n+1}}-\frac{1}{f_{n+1}f_{n+2}}$$ We are dealing with a telescoping sum and

$$\sum_{n=1}^\infty \frac{1}{f_nf_{n+2}} = \lim_{p\to\infty} \sum_{n=1}^p \frac{1}{f_nf_{n+2}} = \lim_{p\to\infty}\left(\frac{1}{f_1f_2} - \frac{1}{f_{p+1}f_{p+2}}\right) = \frac{1}{f_1 f_2} = 1$$

Another approach. The $n$th Fibonacci number is about $\varphi^n$, where $\varphi = (1+\sqrt{5})/2$ is the golden mean. Then your sum behaves like $\Sigma (1/\varphi^{2n})$. It's easy to show that converges. Of course you don't get the value of what it converges to, as in @achillehui 's nice answer.

Since it is easy to see that $f_n\geq n$ for $n\geq5$ you can bound the (positive) summand $\frac1{f_nf_{n+2}}<\frac1{n^2}$ for $n\geq5$; therefore since $\sum_{n\geq1}\frac1{n^2}$ converges, so does $\sum_{n\geq1}\frac1{f_nf_{n+2}}$.