# 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.

• Numerical estimates suggest that the answer is $1/\varphi^3$. Commented Apr 6, 2017 at 1:05
• That's a cute problem! Where did you see this? Commented Apr 6, 2017 at 2:10
• High school math competition in Texas. Helping some students I know. Commented Apr 6, 2017 at 3:24

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