Mathematica is able to evaluate the summation $\sum_{i=1}^{n}\frac{F_i}{i}$ in terms of the Lerch transcendent. It is natural to consider whether or not this summation can be expressed in a more simple way.
It is conceivable that it is possible that $a_{n} = \sum_{i=1}^{n}\frac{F_i}{i}$ may be expressed in a more compact way, perhaps in terms of Fibonacci and Lucas numbers somehow, using the nearest integer function or the floor function.
Integer sequences such as $(\lfloor a_{n} \rfloor)_{n \in \mathbb{N}}$, $([a_{n}])_{n \in \mathbb{N}}$, $( n! a_{n} )_{n \in \mathbb{N}}$, and $( \text{den}(a_{n}) )_{n \in \mathbb{N}}$ are not currently in the On-Line Encyclopedia of Integer Sequences.
It appears that $\lim_{n \to \infty} \frac{\ln a_{n}}{n} \approx 0.481$ converges to a constant. What is this constant? Can this constant be used to somehow express the sequence $(a_{n})_{n \in \mathbb{N}}$ in closed form?
EXPANDED: Let's define : $$\tag{1}a_n(x):=\sum_{k=1}^{n}\frac{F_k}{k}x^k$$ and use $\,F_k=\dfrac {\varphi^{\,k}-\psi^{,k}}{\sqrt{5}}\;$ with $\;\varphi=\dfrac{1+\sqrt{5}}2,\ \psi=\dfrac{1-\sqrt{5}}2=-\dfrac 1{\varphi}\;$ (as suggested by Aravind and Wikipedia) then :
\begin{align} a_n(x)&=\frac 1{\sqrt{5}}\sum_{k=1}^{n}\left(\varphi^{\,k}-\psi^{,k}\right)\frac{x^k}k\tag{2}\\ &=\frac 1{\sqrt{5}} \int\sum_{k=1}^{n}\left(\varphi^{\,k}-\psi^{,k}\right) \frac {x^{k}}x\,dx\\ &=\frac 1{\sqrt{5}}\left(\int \frac{(\varphi\,x)^{\,n}-1}{\varphi\,x-1}\varphi\,dx-\int\frac{(\psi\,x)^{\,n}-1}{\psi\,x-1}\psi\,dx\right)\\ &=\frac 1{\sqrt{5}}\int_{\psi\,x}^{\varphi\,x} \frac{u^{\,n}-1}{u-1}\,du=\frac 1{\sqrt{5}}\int_{\psi\,x}^{\varphi\,x}\ \sum_{k=0}^{n-1}\;u^{\,k}\,du\tag{3}\\ &\quad\text{writing the integrand at the left as a sum of two terms gives :}\\\\ a_n:=a_n(1)&=\frac 1{\sqrt{5}}\left(-B_{\varphi}(n+1,0)+B_{\psi}(n+1,0)-\log(1-\varphi)+\log(1-\psi)\right)\tag{4}\\ \end{align} with $B_x(a,b)$ the incomplete beta function (the singularities in the integrals are removable and will not be considered).
All this is probably not very useful for practical evaluations but $(2)$ applied to $x=1$ allows to find a closed form for your limit $\;\displaystyle l:=\lim_{n \to \infty} \frac{\ln a_{n}}{n}$ using $\varphi>1$ and $|\psi|<1$ : \begin{align} a_n&=\frac 1{\sqrt{5}}\sum_{k=1}^{n}\frac{1}k\left(\varphi^{\,k}-\psi^{,k}\right)\\ a_n&\sim \frac 1{\sqrt{5}}\sum_{k=1}^{n}\frac{\varphi^{\,k}}k,\quad n\to\infty\\ &\sim \frac 1{\sqrt{5}}\frac{\varphi^{\,n}}{n}\sum_{k=1}^{n}\frac{\varphi^{\,k-n}\;n}k,\quad n\to\infty\\ &\quad\text{setting $\;j:=n-k\,$ gives}\\ &\sim \frac 1{\sqrt{5}}\frac{\varphi^{\,n}}{n}\sum_{j=0}^{n-1}\frac 1{\varphi^{\,j}\;\frac {n-j}n},\quad n\to\infty\\ &\sim \frac 1{\sqrt{5}}\frac{\varphi^{\,n}}{n}\sum_{j=0}^{n-1}\frac {1+\sum_{m>0} \left(\frac jn\right)^m}{\varphi^{\,j}},\quad n\to\infty\\ &\sim \frac 1{\sqrt{5}}\frac{\varphi^{\,n}}{n}\left(\frac 1{1-\frac 1{\varphi}}+\sum_{m>0}\frac 1{n^m}\sum_{j=0}^{n-1}\frac {j^m}{\varphi^{\,j}}\right) ,\quad n\to\infty\\ &\sim \frac 1{\sqrt{5}}\frac{\varphi^{\,n}}{n}\left(\frac {\varphi}{\varphi-1}+\sum_{m>0}\frac {\Phi\left(\frac 1{\varphi},-m,0\right)}{n^m}\right) ,\quad n\to\infty\tag{5}\\ a_n&\sim \frac{\varphi^{\,n+1}}{\sqrt{5}\;n\,(\varphi-1)},\quad n\to\infty\tag{6}\\ \end{align} With $\Phi$ the Lerch transcendent $\;\displaystyle\Phi(z,s,a):=\sum_{n\ge 0}\frac{z^n}{(n+a)^s}$.
The limit will simply be : $$l=\lim_{n \to \infty} \frac{\ln a_{n}}{n}=\log(\varphi)=\log\dfrac{1+\sqrt{5}}2\approx 0.4812118251$$ $$-$$ Let's use $(5)$ to get a more precise expansion defining $\;\displaystyle f_m(z):=\sum_{n\ge 0}n^m\,z^n=\Phi(z,-m,0)$.
For $m$ a nonnegative integer we have $\;\displaystyle f_{m+1}(z)=z\,f_{m}(z)'=z\sum_{n\ge 0}n^m\,n\,z^{n-1}=\sum_{n\ge 0}n^{m+1}\,z^{n}\ $
giving us the closed forms : \begin{align} f_0(z)&=\frac 1{(1-z)}\\ f_1(z)&=\frac z{(1-z)^2}\\ f_2(z)&=\frac {z^2+z}{(1-z)^3}\\ f_3(z)&=\frac {z^3+4z^2+z}{(1-z)^4}\\ \cdots \end{align} Using $\;z:=\dfrac 1{\varphi}\;$ so that $\,\dfrac 1{1-z}=\dfrac {\varphi}{\varphi-1}\;$ we may rewrite $(5)$ as :
\begin{align} a_n&\sim \frac 1{\sqrt{5}}\frac{\varphi^{\,n}}{n}\left(\frac {\varphi}{\varphi-1}+\frac {\varphi}{(\varphi-1)^2\,n}+\frac {\varphi+\varphi^2}{(\varphi-1)^3\,n^2}+\frac {\varphi+4\varphi^2+\varphi^3}{(\varphi-1)^4\,n^3}+\cdots\right) \\ a_n&\sim \frac 1{\sqrt{5}}\frac{\varphi^{\,n}}{n}\left(\varphi+1+\frac {2\varphi+1}{n}+\frac {8\varphi+5}{n^2}+\frac {50\varphi+31}{n^3}+\frac{416\varphi+257}{n^4}+\cdots\right) \\ \end{align}
The sequences of integers appearing there are known : OEIS A000557 and A000556 allowing us to write this exponential generating function : \begin{align} \frac{\varphi+e^{-x}}{1-2\sinh(x)}&=(\varphi+1)+(2\varphi+1)\frac x{1!}+(8\varphi+5)\frac {x^2}{2!}+(50\varphi+31)\frac {x^3}{3!}+(416\varphi+257)\frac {x^4}{4!}+\cdots\\ \end{align}
The OEIS links should help you to find further relations and interesting links.