The statement is not correct. The correct statement is:
$$\lim_{n\rightarrow\infty}\frac{1}{2n}\sum_{k=1}^{2n}\left\{ \frac{F(2n)}{F(k)}\right\}=\frac{3\log 2}{4}.$$
That is, the limit is $\approx 0.519860385$ instead of $\approx 0.518788175$. As a numerical verification, for $n=10^6$, the average under the limit is $\approx 0.519853615$ (according to SAGE).
More generally, we shall prove the following.
$$\sum_{k=1}^n\left\{\frac{F(n)}{F(k)}\right\}=
\begin{cases}
\frac{\pi}{8}n+O(\sqrt{n}),&\text{$n$ is odd};\\
\frac{3\log 2}{4}n+O(\sqrt{n}),&\text{$n$ is even}.
\end{cases}$$
The idea of the proof is that the fractional part of $F(n)/F(k)$ is typically very close to $0$ or $1$, and the two cases follow a simple pattern.
We can restrict to $k\nmid n$ in the sum, because otherwise the fractional part is zero. Accordingly, we shall consider
$$n=2mk\pm r\qquad\text{with}\qquad r\in\{1,2,\dotsc,k-1\}.$$
If $L(v)$ denotes the $v$-th Lucas number, then we have the identity
$$F(u)L(v)=\begin{cases}
F(u+v)+(-1)^v F(u-v),&u\geq v;\\
F(u+v)-(-1)^u F(v-u),&u\leq v.
\end{cases}$$
Plugging $(mk,mk\pm r)$ for $(u,v)$, we obtain the following congruences:
\begin{alignat*}{2}
F(2mk+r)&\equiv F(r)(-1)^{mk}&&\pmod{F(mk)},\\
F(2mk-r)&\equiv F(r)(-1)^{mk-r-1}&&\pmod{F(mk)}.
\end{alignat*}
These congruences are also valid modulo $F(k)$, because $F(k)$ divides $F(mk)$. We infer that:
$$\{F(n)/F(k)\}=
\begin{cases}
F(r)/F(k),&n=2mk+r,\quad mk\equiv 0\pmod{2};\\
1-F(r)/F(k),&n=2mk+r,\quad mk\not\equiv 0\pmod{2};\\
1-F(r)/F(k),&n=2mk-r,\quad mk\equiv n\pmod{2};\\
F(r)/F(k),&n=2mk-r,\quad mk\not\equiv n\pmod{2}.\\
\end{cases}$$
It is now convenient to introduce the notation
$$I(n,t):=\mathbb{N}\cap\left(\frac{n}{t+1},\frac{n}{t}\right),$$
because then the cases $n=2mk+r$ correspond to $k\in I(n,2m)$, while the cases $n=2mk-r$ correspond to $k\in I(n,2m-1)$. Then we see that:
- for odd $m$ the contribution of $k\in I(n,2m)$ is $\frac{1}{2}|I(n,2m)|+O(1)$;
- for odd $m$ the contribution of $k\in I(n,2m-1)$ is $\frac{1}{2}|I(n,2m-1)|+O(1)$;
- for even $m$ the contribution of $k\in I(n,2m)$ is $O(1)$;
- for even $m$ and odd $n$ the contribution of $k\in I(n,2m-1)$ is
$O(1)$;
- for even $m$ and even $n$ the contribution of $k\in I(n,2m-1)$ is
$|I(n,2m-1)|+O(1)$.
Now we choose a positive integer $M$, and sum up the contributions of $I(n,2m)$ and $I(n,2m-1)$ for $m\in\{1,2,\dotsc,M\}$. This way we see that
$$\sum_{k=\lceil n/(2M+1)\rceil}^{n}\left\{\frac{F(n)}{F(k)}\right\}=\sum_{m=1}^M\sum_{k\in I(n,2m)}\left\{\frac{F(n)}{F(k)}\right\}
+\sum_{m=1}^M\sum_{k\in I(n,2m-1)}\left\{\frac{F(n)}{F(k)}\right\}.\tag{$\ast$}$$
If $n$ is odd, then up to an error of $O(M)$, the right-hand side of $(\ast)$ equals
$$\frac{1}{2}\sum_{\substack{1\leq m\leq M\\\text{$m$ odd}}}\left(\frac{n}{2m-1}-\frac{n}{2m+1}\right)=\frac{\pi}{8}n+O\left(\frac{n}{M}\right).$$
The contribution of $k<n/(2M+1)$ is $O(n/M)$, hence in fact
$$\sum_{k=1}^n\left\{\frac{F(n)}{F(k)}\right\}=\frac{\pi}{8}n+O\left(M+\frac{n}{M}\right).$$
If $n$ is even, then up to an error of $O(M)$, the right-hand side of $(\ast)$ equals
$$\frac{1}{2}\sum_{\substack{1\leq m\leq M\\\text{$m$ odd}}}\left(\frac{n}{2m-1}-\frac{n}{2m+1}\right)+\sum_{\substack{1\leq m\leq M\\\text{$m$ even}}}\left(\frac{n}{2m-1}-\frac{n}{2m}\right)=\frac{3\log 2}{4}n+O\left(\frac{n}{M}\right).$$
The contribution of $k<n/(2M+1)$ is $O(n/M)$, hence in fact
$$\sum_{k=1}^n\left\{\frac{F(n)}{F(k)}\right\}=\frac{3\log 2}{4}n+O\left(M+\frac{n}{M}\right).$$
In both cases we choose $M=\lfloor\sqrt{n}\rfloor$, and we obtain the claimed asymptotic formula with error term $O(\sqrt{n})$.
Best Answer
The conjecture is true. For $t\in[0,1]$, let $N(t)$ be the number of $k\in\{1,2,\dotsc,n\}$ satisfying $\{n/k\}<t$. On the one hand, $$\sum_{k=1}^n f\left(\frac{n}{k}\right)=\int_0^{1/2}\bigl(N(t+1/2)-N(t)\bigr)\,dt.$$ On the other hand, for any positive integer $M$, $$N(t)=\sum_{m=1}^\infty\left(\biggl\lfloor\frac{n}{m}\biggr\rfloor-\biggl\lfloor\frac{n}{m+t}\biggr\rfloor\right)=\sum_{m=1}^M\left(\frac{n}{m}-\frac{n}{m+t}\right)+O\left(M+\frac{n}{M}\right).$$ It follows that \begin{align*} \sum_{k=1}^n f\left(\frac{n}{k}\right) &=\sum_{m=1}^M\int_0^{1/2}\left(\frac{n}{m+t}-\frac{n}{m+t+1/2}\right)\,dt+O\left(M+\frac{n}{M}\right)\\ &=\sum_{m=1}^M n\log\left(\frac{(2m+1)^2}{(2m)(2m+2)}\right)+O\left(M+\frac{n}{M}\right)\\ &=n\log\left(\prod_{m=1}^M\frac{(2m+1)^2}{(2m)(2m+2)}\right)+O\left(M+\frac{n}{M}\right). \end{align*} By Wallis's product, $$\prod_{m=1}^\infty\frac{(2m+1)^2}{(2m)(2m+2)}=\frac{4}{\pi},$$ hence also $$\prod_{m=1}^M\frac{(2m+1)^2}{(2m)(2m+2)}=\frac{4}{\pi}\prod_{m=M+1}^\infty\frac{(2m)(2m+2)}{(2m+1)^2}=\frac{4}{\pi}\left(1+O\left(\frac{1}{M}\right)\right).$$ Taking the logarithm of both sides, and going back to the $k$-sum, $$\sum_{k=1}^n f\left(\frac{n}{k}\right)=n\log\left(\frac{4}{\pi}\right)+O\left(M+\frac{n}{M}\right).$$ Finally, we choose $M=\lfloor\sqrt{n}\rfloor$ to conclude that $$\sum_{k=1}^n f\left(\frac{n}{k}\right)=n\log\left(\frac{4}{\pi}\right)+O\left(\sqrt{n}\right).$$