Harmonic Fibonacci Series

We know that a Generating Function of the Harmonic Numbers is given by:-

$\sum_{n=1}^{\infty}H_{n}z^{n} = -\frac{\ln(1-x)}{1-x}$

Also we know by Binet's Formula that the nth Fibonacci Number is given by $\displaystyle F_{n} = \frac{{\varphi}^{n}-(1-\varphi)^{n}}{\sqrt{5}}$

Where $\varphi = \frac{1+\sqrt{5}}{2}$ is the Golden Ratio .

So we have the following summation:-

$\displaystyle \frac{1}{\sqrt{5}}\sum_{n=1}^{\infty}H_{n}\left(\left(\frac{\varphi}{2}\right)^{n} -\left(\frac{1-\varphi}{2}\right)^{n}\right)$

So using the generating function we have our summation equal to:-

$\displaystyle \frac{1}{\sqrt{5}}\left(\frac{\ln(1-\frac{1-\varphi}{2})}{1-\frac{1-\varphi}{2}} - \frac{\ln(1-\frac{\varphi}{2})}{1-\frac{\varphi}{2}}\right)$ .

Now it is just a matter of Rearrangements. We know $\displaystyle \coth^{-1}(x) = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$ .

Using all of these we arrive at the answer:-

$\displaystyle \ln(4) +\dfrac{6}{\sqrt{5}}\coth^{-1}(\dfrac{3}{\sqrt{5}})$