I'm trying to figure out a way to solve the value of this: $$\frac{1}{1\times 2}-\frac{1}{2\times 3}+\frac{1}{3\times 5}-\frac{1}{5\times 8}+\frac{1}{8\times 13}-\dots$$
The only thing I can come up with is a summation involving the $nth$ fibonacci term formula that uses $\phi$. Any other insights?
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You can do this using partial fractions.
Note that: $$\begin{align} \frac{1}{1 * 2} &= \frac{1}{1} + \frac{-1}{2} \\ \frac{1}{2 * 3} &= \frac{-1}{2} + \frac{2}{3} \\ \frac{1}{3 * 5} &= \frac{2}{3} + \frac{-3}{5} \\ \frac{1}{5 * 8} &= \frac{-3}{5} + \frac{5}{8} \\ \frac{1}{8 * 13} &= \frac{5}{8} + \frac{-8}{13} \end{align}$$ The pattern can be formulated so that this identity can be proven later on $$ \frac{1}{f_i * f_{i+1}} = (-1)^{i-1}\frac{f_{i-1}}{f_i} + (-1)^i\frac{f_i}{f_{i+1}}. $$ Then the partial sums of the series are equivalent to $$\begin{multline} \frac{1}{1} + \frac{-1}{2} + \frac{1}{2} + \frac{-2}{3} + \frac{2}{3} + \frac{-3}{5} + \frac{3}{5} + \frac{-5}{8} + \frac{5}{8} + \frac{-8}{13} + \cdots + \frac{-f_i}{f_{i+1}} = \\ = 1 - \frac{f_i}{f_{i+1}} = \frac{f_{i+1}}{f_{i+1}} - \frac{f_i}{f_{i+1}} = \frac{f_{i-1}}{f_{i+1}} \end{multline}$$
Note that $$\frac1{F_n}-\frac1{F_{n+1}}=\frac{F_{n+1}-F_n}{F_nF_{n+1}}=\frac{F_{n-1}}{F_nF_{n+1}}, $$ hence $$ \frac1{F_{n-1}F_n}-\frac1{F_nF_{n+1}}=\frac1{F_{n-1}F_{n+1}}$$ and your sum turns into $$\frac1{1\cdot 3}+\frac1{3\cdot 8}+\frac1{8\cdot 21}+\ldots$$ Playing with the first few partial sums, $\frac13,\frac 38, \frac8{21}$, it looks like the partial sum up to $\frac1{F_{n-1}F_{n+1}}$ equals $\frac{F_{n-1}}{F_{n+1}}$. Show by induction that this is inded the case and conclude that the limit is $\phi^2$.