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, ${{-z} \choose j}$ 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 works) 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.
