Home » Uncategorized » Fibonacci series

Fibonacci series

Let \{F_n\}_{n \in \mathbb{N}} denote the Fibonacci sequence such that F_1=F_2=1 and F_3=2. Prove that

    \[\sum_{n=2}^{\infty} {\rm arctanh} \;  \frac{1}{F_{2n}}  =\frac{\log 3}{2}\]

Solution

It holds that

    \[{\rm arctanh}\; x - {\rm arctanh} \; y= {\rm arctanh} \left ( \frac{x-y}{1-xy} \right )\]

It follows from Cassini’s identity that

(1)   \begin{equation*} {F_{n-1} F_{n+1} - F_n^2 = (-1)^n \end{equation*}

Setting n \mapsto 2n back at (1) we get

(2)   \begin{equation*} F^2_{2n}=F_{2n-1} F_{2n+1} -1 \end{equation*}

Since F_n =F_{n-1} + F_{n-2} we get

(3)   \begin{equation*} F_{2n+1} =F_{2n} + F_{2n-1} \Leftrightarrow F_{2n} = F_{2n+1} - F_{2n-1} \end{equation*}

Hence,

    \begin{align*} \sum_{n=2}^{\infty} {\rm arctanh}\;   \frac{1}{F_{2n}}  &= \sum_{n=2}^{\infty} {\rm arctanh} \; \frac{F_{2n}}{F_{2n}^2}  \\ &=\sum_{n=2}^{\infty} {\rm arctanh} \frac{F_{2n-1} - F_{2n+1}}{1-F_{2n-1}F_{2n+1}} \\ &= \sum_{n=2}^{\infty} {\rm arctanh} \;  \frac{\frac{1}{F_{2n-1}} - \frac{1}{F_{2n+1}}}{1- \frac{1}{F_{2n-1} F_{2n+1}}} \\ &=\lim_{m \rightarrow +\infty} \sum_{n=2}^{m} \bigg [ {\rm arctanh} \left ( \frac{1}{F_{2n-1}} \right ) - \\ &\quad \quad \quad - {\rm arctanh} \left ( \frac{1}{F_{2n+1}} \right ) \bigg ] \\ &= \lim_{m \rightarrow +\infty} \left [ {\rm arctanh} \left ( \frac{1}{F_3} \right ) - {\rm arctanh} \left ( \frac{1}{F_{2m+1}} \right )\right ]\\ &= {\rm arctanh} \left ( \frac{1}{2} \right ) \\ &= \frac{\log 3}{2} \end{align*}

Read more

Leave a comment

Donate to Tolaso Network