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.
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 $$ \frac{1}{\phi} - \frac{1}{\phi^2} = \frac{\phi - 1}{\phi^2} = \frac{1}{\phi^3} $$