<s>[INST] Sum of Sequence with Squares of Fibonacci Numbers in Denominator Find the sum of this sequence:
$$\frac{1}{1^2+1}-\frac{1}{2^2-1}+\frac{1}{3^2+1}-\frac{1}{5^2-1}+\frac{1}{8^2+1}-...$$
So, alternating series, but I've got nothing. I tried regrouping by pairs and got $$\frac{1}{6}+\frac{7}{120}+\frac{103}{10920}$$
which, helped me none.
 [/INST] Note: most of this is the identity
$$  F_{n+1} F_{n-1} - F_n^2 = (-1)^n $$ which is how I how I saw the two parts telescope.
The $+$ part is
$$ \frac{1}{1 \cdot 2} +  \frac{1}{2 \cdot 5} +  \frac{1}{5 \cdot 13} +  \frac{1}{13 \cdot 34} +  $$
Do the $\pm$ parts separately. The partial sums for the $+$ part are
$$  \frac{1}{2}, \frac{3}{5}, \frac{8}{13}, \frac{21}{34}, \cdots $$
which are ratios of Fibonacci numbers. 
The $-$ part is
$$ \frac{1}{1 \cdot 3} +  \frac{1}{3 \cdot 8} +  \frac{1}{8 \cdot 21} +  \frac{1}{21 \cdot 55} +  $$
Try this for the $-$ parts. 
$$  \frac{1}{3}, \frac{3}{8}, \frac{8}{21}, \frac{21}{55}, \cdots $$
Needs a bit of precision to get the limit of the difference.
I see
$$ \frac{1}{\phi} - \frac{1}{\phi^2} = \frac{\phi - 1}{\phi^2} = \frac{1}{\phi^3}  $$
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