Prove that $\sum_{n=1}^\infty \left(\phi-\frac{F_{n+1}}{F_{n}}\right)=\frac{1}{\pi}$

Notice:

• Golden Ratio: $$\phi=\frac{1+\sqrt{5}}{2}$$
• Limit of the Fibonacci numbers: $$\lim_{n\to\infty}\frac{\text{F}_{n+1}}{\text{F}_n}=\phi$$
• The $n^{\text{th}}$ Fibonacci number: $$\text{F}_n=\frac{\phi^n-\left(-\phi\right)^{-n}}{\sqrt{5}}=\frac{\phi^n-\frac{1}{\left(-\phi\right)^n}}{\sqrt{5}}$$

So, we get that:

$$\sum_{n=1}^{\infty}\left[\phi-\frac{\text{F}_{n+1}}{\text{F}_n}\right]=\sum_{n=1}^{\infty}\left[\phi-\frac{\frac{\phi^{n+1}-\frac{1}{\left(-\phi\right)^{n+1}}}{\sqrt{5}}}{\frac{\phi^n-\frac{1}{\left(-\phi\right)^n}}{\sqrt{5}}}\right]=\sum_{n=1}^{\infty}\left[\phi-\frac{1+(-\phi)^n\phi^{2+n}}{\phi\left(\left(-\phi^2\right)^n-1\right)}\right]=$$ $$\sum_{n=1}^{\infty}\frac{1+\phi^2}{\phi-\phi\left(-\phi^2\right)^n}=\frac{\sqrt{5}\left(5\pi-i\left(\ln(5)+2\psi_{\frac{1}{2}\left(-3-\sqrt{5}\right)}^{(0)}\left[1-\frac{2\pi}{\pi-i\text{arccosh}\left(\frac{3}{2}\right)}\right]\right)\right)}{2\left(\pi-i\text{arccosh}\left(\frac{3}{2}\right)\right)}$$

The Fibonacci numbers are given by $$F_n = \frac{\phi^n -(-\phi)^{-n}}{\sqrt{5}} $$ with $\phi=(1+\sqrt{5})/2$ the golden ratio.

So we need to calculate $$ \sum_{n=1}^\infty\left(\phi -\frac{F_{n+1}}{F_n}\right) = \sum_{n=1}^\infty \frac{1+\phi^2}{\phi(1+(-1)^{n+1}\phi^{2n})} =\sum_{n=1}^\infty \frac{\sqrt{5}\, 4^n}{4^n+(-1)^{n+1} \left(1+\sqrt{5}\right)^{2 n}}. $$

I am not sure if this last sum has any nice explicit expression.

But definitely, we can see that the sum converges absolutely, as $$ \left|\frac{\sqrt{5}\, 4^n}{4^n+(-1)^{n+1} \left(1+\sqrt{5}\right)^{2 n}}\right| = O((4/5)^n).$$