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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.

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

1 Answer 1

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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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