Golden ratio II

Nice hypergeometric series involving the Fibonacci numbers $F_{2k+3}$

Today we show the proof of this nice series proposed by @diegorattaggi $$\sum_{n=1}^{\infty} \binom{n+k-1}{k} \frac{F_{n}}{2^n} = 2^k F_{2k+3}$$ For the proof we will use some properties of the rising factorial (Pochhammer polynomial), the Gaussian hypergeometric function and the Binet's formula for Fibonacci numbers.

Proof:

$$S = \sum_{n=1}^{\infty} \binom{n+k-1}{k} \frac{F_{n}}{2^n} = \sum_{n=1}^{\infty} \frac{(n+k-1)!}{(n-1)!k!} \frac{F_{n}}{2^n} = \sum_{n=1}^{\infty} \frac{\Gamma(n+k)}{(n-1)!k!} \frac{F_{n}}{2^n}$$ Recall that the pochhammer polynomial (rising factorial) can be expressed as the quotient of two gamma functions: $$\frac{\Gamma(n+x)}{\Gamma(x)} = (x)_{n}$$ where $$(x)_{n} = x(x+1)(x+2)\cdots (x+n-1)$$ Hence $$S = \sum_{n=1}^{\infty} \frac{\Gamma(n+k)}{(n-1)!k!} \frac{F_{n}}{2^n} = \frac{\Gamma(k)}{k!}\sum_{n=1}^{\infty} \frac{(k)_{n}}{(n-1)!} \frac{F_{n}}{2^n} = \frac{1}{k}\sum_{n=1}^{\infty} \frac{(k)_{n}}{(n-1)!} \frac{F_{n}}{2^n}$$ If we let the change of variable $n=j+1$ $$S = \frac{1}{2k}\sum_{j=0}^{\infty} \frac{(k)_{j+1}}{j!} \frac{F_{j+1}}{2^{j}}$$ From the recursion formula for the degree of the Pochhammer polynomial: $$(x)_{n+1} = x(x+1)_{n}$$ we have $$S = \frac{1}{2k}\sum_{j=0}^{\infty} \frac{(k)_{j+1}}{j!} \frac{F_{j+1}}{2^{j}} = \frac{1}{2}\sum_{j=0}^{\infty} \frac{(k+1)_{j}}{j!} \frac{F_{j+1}}{2^{j}}$$ From the Binet's formula $$F_{j+1} = \frac{\phi^{j+1}-\psi^{j+1}}{\sqrt{5}}$$ $$S = \frac{\phi}{2\sqrt{5}}\sum_{j=0}^{\infty} \frac{(k+1)_{j}}{j!} \frac{\phi^{j}}{2^{j}} - \frac{\psi}{2\sqrt{5}}\sum_{j=0}^{\infty} \frac{(k+1)_{j}}{j!} \frac{\psi^{j}}{2^{j}}$$ Multiplying and dividing by $(1)_{j}$ we have an hypergeometric representation: $$\begin{align*} S =& \frac{\phi}{2\sqrt{5}}\sum_{j=0}^{\infty} \frac{(k+1)_{j}}{j!} \frac{\phi^{j}}{2^{j}} - \frac{\psi}{2\sqrt{5}}\sum_{j=0}^{\infty} \frac{(k+1)_{j}}{j!} \frac{\psi^{j}}{2^{j}} \\ =& \frac{\phi}{2\sqrt{5}}\sum_{j=0}^{\infty} \frac{(k+1)_{j}(1)_{j}}{(1)_{j}} \frac{\left(\frac{\phi}{2}\right)^j}{j!} - \frac{\psi}{2\sqrt{5}}\sum_{j=0}^{\infty} \frac{(k+1)_{j}(1)_{j}}{(1)_{j}} \frac{\left(\frac{\psi}{2} \right)^j}{j!}\\ =& \frac{\phi}{2\sqrt{5}}F\left(1,k+1;1;\frac{\phi}{2}\right)- \frac{\psi}{2\sqrt{5}}F\left(1,k+1;1;\frac{\psi}{2}\right) \end{align*}$$ This is a special case of the Gaussian hypergeometric fucntion (generalized Binomial theorem): $$F(a,b;a;z ) = (1-z)^{-b}$$ If we put $a=1, b=k+1$ $$S = \frac{\phi}{2\sqrt{5}}F\left(1,k+1;1;\frac{\phi}{2}\right)- \frac{\psi}{2\sqrt{5}}F\left(1,k+1;1;\frac{\psi}{2}\right) = \frac{\phi}{2\sqrt{5}}\left(1-\frac{\phi}{2} \right)^{-k-1}-\frac{\psi}{2\sqrt{5}}\left(1-\frac{\psi}{2} \right)^{-k-1}$$ Note $$\begin{align*} \frac{\phi}{2\sqrt{5}}\left(1-\frac{\phi}{2} \right)^{-k-1} =& \frac{\phi}{2\sqrt{5}}\left(\frac{2}{2-\phi} \right)^{k+1}\\ =& \frac{\phi}{2\sqrt{5}}\left(3+\sqrt{5}\right)^{k+1}\\ =& \frac{\phi}{2\sqrt{5}}\left(2+2\phi\right)^{k+1}\\ =& \frac{\phi 2^{k+1}}{2\sqrt{5}}\left(1+\phi\right)^{k+1}\\ =& \frac{\phi 2^{k+1}}{2\sqrt{5}}(\phi^2)^{k+1} \quad (1+\phi = \phi^2)\\ =& \frac{2^k \phi^{2k+3}}{\sqrt{5}} \end{align*}$$ $$\begin{align*} \frac{\psi}{2\sqrt{5}}\left(1-\frac{\psi}{2} \right)^{-k-1} = & \frac{\psi}{2\sqrt{5}}\left(1-\frac{1-\phi}{2} \right)^{-k-1} \quad (\psi = 1-\phi)\\ =& \frac{\psi}{2\sqrt{5}}\left(\frac{2}{1+\phi} \right)^{k+1}\\ =& \frac{2^k\psi}{\sqrt{5}}\left(\frac{1}{\phi^2} \right)^{k+1} \quad (\phi+1 = \phi^2) \\ =& \frac{2^k\psi}{\sqrt{5}}\left(\psi^2\right)^{k+1}\\ =& \frac{2^k\psi^{2k+3}}{\sqrt{5}} \end{align*}$$ Hence $$S = \frac{2^k \phi^{2k+3}}{\sqrt{5}} -\frac{2^k}{\sqrt{5}}\psi^{2k+3} = 2^{k}\left[\frac{\phi^{2k+3}}{\sqrt{5}} -\frac{\psi^{2k+3}}{\sqrt{5}}\right] = 2^kF_{2k+3}$$ Therefore $$\boxed{ \sum_{n=1}^{\infty} \binom{n+k-1}{k} \frac{F_{n}}{2^n} = 2^k F_{2k+3}}$$