Partial sum of the harmonic series between two consecutive fibonacci numbers
In terms of the harmonic numbers $H_n$, your sequence is
$$ s_n = H_{F_{n+1}} - H_{F_n-1} $$
As $n \to \infty$ it's known that $H_n = \log n + \gamma + o(1)$, so
$$ \begin{align} s_n &= \log F_{n+1} + \gamma + o(1) - \log(F_n-1) - \gamma - o(1) \\ &= \log F_{n+1} - \log(F_n-1) + o(1). \end{align} $$
Now $F_m \sim \varphi^m/\sqrt{5}$, where $\varphi$ is the golden ratio, so using the fact that $a \sim b \implies \log a = \log b + o(1)$ we have
$$ \begin{align} s_n &= \log(\varphi^{n+1}/\sqrt{5}) - \log(\varphi^{n}/\sqrt{5}) + o(1) \\ &= \log \varphi + o(1). \end{align} $$
In other words,
$$ \lim_{n \to \infty} \sum_{k=F_n}^{F_{n+1}} \frac{1}{k} = \log \varphi. $$
The Fibonacci numbers increase as $\phi^n$ (where $\phi$ is the golden mean $\frac{1+\sqrt{5}}{2}$), and harmonic numbers increase as $\log n$ (i.e., the natural log). Therefore, the difference between the harmonic numbers for successive Fibonacci numbers will approach $\log\phi \approx 0.481211825...$
To expand a bit, the Fibonacci numbers can be expressed as $\frac{\phi^n - (1-\phi)^n}{\sqrt{5}}$. (Try it! The fact that the equation $f(x+2) - f(x+1) - f(x) = 0$ requires a sum of powers of $\phi$ and $1-\phi$ follows from the fact that these are the solutions to the equation $x^2 - x - 1 = 0$, and the coefficients come from f(1) = f(2) = 1.) The second term vanishes, so large Fibonacci numbers can be approximated quite well as $\frac{\phi^n}{\sqrt{5}}$.
Since one definition of the natural logarithm is the integral from 1 to the parameter of the function $t^{-1}$, the harmonic numbers can be approximated as the natural logarithm, and in fact the difference approaches a constant (called $\gamma$, about 0.577). If you're not familiar with integrals, the fact that the harmonic numbers increase as a logarithm is suggested by Oresme's proof that the harmonic series diverges...
$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \cdots > 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{16} + \cdots$$
...and it just so happens that that logarithm is the natural logarithm.
So if you accept that for very large n, the harmonic numbers approach $\log n$, and that the Fibonacci numbers approach $\frac{\phi^n}{\sqrt{5}}$, then you get for two successive...
$$\log\left(\frac{\phi^{n+1}}{\sqrt{5}}\right) - \log\left(\frac{\phi^n}{\sqrt{5}}\right) = \log\left(\frac{\phi^{n+1}}{\phi^n}\right) = \log\phi$$
($\log x - \log y = \log \frac{x}{y}$ is a natural inverse of $\frac{e^x}{e^y} = e^{x-y}$.)