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Recently, while spending time with Fibonacci numbers, I found that $$S=\sum_{k=1}^{n} \frac{(-1)^{k+1}}{F_k F_{k+1}}=\frac{F_{n}}{F_{n+1}}.~~~~(1)$$ Proof: Using a property of these numbers: $F_{n+1} F_{n+2}-F_n F_{n+3}=(-1)^n$, we can write: $$\frac{(-1)^{n+1}}{F_n F_{n+1}}=\frac{F_{n+3}}{F_{n+1}}-\frac{F_{n+2}}{F_n}=G_{n+1}-G_n.$$ By telescopic summation, we have $$S=\sum_{k=0}^{n} \frac{(-1)^{k+1}}{F_k F_{k+1}}=[(G_2-G_1)+(G_3-G_2)+(G_4-G_3)+....+(G_{n-1}-G_{n-2})+...+(G_n-G_{n-1})].$$ $$\implies S=[G_n-G_1]=\frac{F_{n+3}}{F_{n+1}}-\frac{F_3}{F_1}=\frac{F_{n+3}}{F_{n+1}}-2= \frac{F_n}{F_{n+1}}$$ Using another property of that $F_k f_{k+1}=F_1^2+F_2^2+F_3^2+F_4^2+...+F_k^2,$ we can write (1) as $$S=\sum_{k=1}^{n} \frac{(-1)^{k+1}}{\sum_{j=1}^{k} F^2_j}=\frac{F_{n}}{F_{n+1}}~~~~(2)$$

The question is: How (1,2) can be verified and proved otherwise?

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Z Ahmed
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  • What particular properties of the Fibonacci numbers are you avoiding using? I can only see two ways to show this sum: the "nice way" that you described above using properties of the Fibonacci numbers, and the "ugly way" by writing everything in terms of the Binet formula and using geometric series a bunch of times. What characteristics are you looking for in an alternate solution? – Carl Schildkraut Aug 26 '20 at 06:19
  • May be one can do in terms of $a,b$ by using $F_n=(a^n-b^n)/\sqrt{5}$, with $ab=-1, a+b=1.$. Induction could be another way. We don't know there may be some other surprising way! – Z Ahmed Aug 26 '20 at 06:50
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    You can simplify $\frac{F_{n+3}}{F_{n+1}}-2$ to $\frac{F_n}{F_{n+1}}$. – J.G. Aug 26 '20 at 11:09
  • @J.G Oh! yes, thanks. – Z Ahmed Aug 26 '20 at 11:21

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