View source for Fibonacci zeta in terms of a sum of binomial coefficients

==Theorem==
The following formula holds:
$$(\sqrt{5})^{-z} F(z)=-\displaystyle\sum_{j=0}^{\infty} (-1)^j {{-z} \choose j} \dfrac{1}{\phi^{z+2j}-(-1)^j} ,$$
where $F(z)$ denotes the [[Fibonacci zeta function]], $\displaystyle{{-z} \choose j}$ denotes a [[binomial coefficient]], and $\phi$ denotes the [[golden ratio]].

==Proof==
By [[Binet's formula]],
$$F_n=\dfrac{\phi^n - (-\phi)^{-n}}{\sqrt{5}}.$$
Using the [[binomial series]], we see that
$$\dfrac{1}{F_k^z} = \left( \sqrt{5} \right)^z (\phi^k - (-\phi)^{-k})^{-z}=(\sqrt{5})^z \phi^{-zk} \left( 1 - \dfrac{(-\phi)^{-k}}{\phi^k} \right)^{-z}=(\sqrt{5})^z \phi^{-zk} \displaystyle\sum_{j=0}^{\infty} {{-z} \choose j} (-1)^j \left( \dfrac{(-\phi)^{-kj}}{\phi^{kj}} \right).$$
Therefore compute
$$\begin{array}{ll}
(\sqrt{5})^{-z} F(z) &= (\sqrt{5})^{-z} \displaystyle\sum_{k=1}^{\infty} \dfrac{1}{F_k^z} \\
&= (\sqrt{5})^{-z} \displaystyle\sum_{k=1}^{\infty} \displaystyle\sum_{j=0}^{\infty} (\sqrt{5})^z \phi^{-zk} {{-z} \choose j} (-1)^j \left( \dfrac{(-\phi)^{-kj}}{\phi^{kj}} \right) \\
&= \displaystyle\sum_{k=1}^{\infty} \displaystyle\sum_{j=0}^{\infty} \phi^{-zk} {{-z} \choose j} (-1)^j \left( \dfrac{(-\phi)^{-kj}}{\phi^{kj}} \right).
\end{array}$$
Now we interchange the summations (justify this) and apply the [[geometric series]] (justify why it applies)  to get
$$\begin{array}{ll}
(\sqrt{5})^{-z} F(z) &= \displaystyle\sum_{j=0}^{\infty} {{-z} \choose j} (-1)^j \displaystyle\sum_{k=1}^{\infty} \phi^{-zk} \left( \dfrac{(-\phi)^{-kj}}{\phi^{kj}} \right) \\
&= \displaystyle\sum_{j=0}^{\infty} {{-z} \choose j} (-1)^j \displaystyle\sum_{k=1}^{\infty} \phi^{-zk} \left( \dfrac{(-\phi)^{-kj}}{\phi^{kj}} \right) \\
&= \displaystyle\sum_{j=0}^{\infty} {{-z} \choose j} (-1)^j \displaystyle\sum_{k=1}^{\infty} \left((-1)^j \phi^{-z-2j} \right)^k \\
&= \displaystyle\sum_{j=0}^{\infty} {{-z} \choose j} (-1)^j \dfrac{(-1)^j \phi^{-z-2j}}{1-(-1)^j \phi^{-z-2j}} \\
&= \displaystyle\sum_{j=0}^{\infty} {{-z} \choose j} \dfrac{1}{\phi^{z+2j}-(-1)^j},
\end{array}$$
as was to be shown. 

==References==
* {{PaperReference|The Fibonacci Zeta Function|1976|Maruti Ram Murty|prev=Binet's formula|next=findme}} 

[[Category:Theorem]]
[[Category:Proven]]
[[Category:justify]]