May 5, 2021 - 15:04

An infinite series problem.

By Vamshi Jandhyala in mathematics

May 5, 2021

Problem

Solution

From Cassini’s identity, we have

$$ F_{2n+1}^2 - 1 = F_{2n}F_{2n+2} $$

Therefore,

$$ \begin{aligned} \sum_{n=1}^\infty \frac{1}{F_{2n+1} - 1}+ \sum_{n=1}^\infty \frac{1}{F_{2n+1} + 1} &= \sum_{n=1}^\infty \frac{2F_{2n+1}}{F_{2n+1}^2 - 1} \\
&= \sum_{n=1}^\infty \frac{2F_{2n+1}}{F_{2n}F_{2n+2}} \\
&= \sum_{n=1}^\infty \frac{2}{F_{2n}} - \frac{2}{F_{2n+2}}\\
&= \frac{2}{F_2} = 2 \end{aligned} $$

Using Binet’s formula, we have

$$ \begin{aligned} F_{2n+1} &= \frac{\phi^{2n+1}+\frac{1}{\phi^{2n+1}} }{\sqrt{5}} \\
\end{aligned} $$

where $\phi = \frac{1 + \sqrt{5}}{2}$.

Using the above, we get

$$ \begin{aligned} \sum_{n=1}^\infty \frac{1}{F_{2n+1} + 1} &= \sum_{n=1}^\infty \frac{\sqrt{5}\phi^{2n+1}}{\phi^{4n+2} + \sqrt{5}\phi^{2n+1} + 1} \\
&= \sum_{n=1}^\infty \frac{\sqrt{5}\phi^{2n+1}}{\phi^{4n+2} + \phi^{2n+2} + \phi^{2n} + 1} \\
&= \sum_{n=1}^\infty \frac{\sqrt{5}\phi^{2n}(\phi^2-1)}{(1+\phi^{2n+2})(1+\phi^{2n})} \\
&= \sum_{n=1}^\infty \sqrt{5} \left( \frac{1}{1+\phi^{2n}} - \frac{1}{1+\phi^{2n+2}} \right) \\
&= \frac{\sqrt{5}}{1 + \phi^2} = \frac{\sqrt{5}}{\sqrt{5}\phi} = \phi - 1 = \frac{\sqrt{5}-1}{2} \end{aligned} $$