# May 5, 2021 - 15:04

#### An infinite series problem.

By Vamshi Jandhyala in mathematics

May 5, 2021

## Problem

— 級数bot (@infseriesbot) May 5, 2021

## 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}
$$