Notice that
$$F_r(z) = \frac{1}{(1-z)^{r-1}} - \sum_{k=0}^{r-1} \left(\frac{z}{1-z}\right)^k$$
and therefore for $r\geq 4$ and $n\geq 1$, we have
\begin{split}
[x^n]\ F_r(z) &= \binom{n+r-2}{r-2} - \sum_{k=1}^{r-1} \binom{n-1}{k-1} \\
& = \binom{n+r-2}{r-2} - \binom{n-1}{r-2} - \binom{n-1}{r-3} - \sum_{k=1}^{r-3} \binom{n-1}{n-k} \\
&\geq \binom{n+r-2}{r-2} - \binom{n}{r-2} - \sum_{k=1}^{r-3} \binom{n-1+r-3-k}{n-k} \\
&= \binom{n+r-2}{r-2} - \binom{n}{r-2} - \binom{n+r-4}{r-3} \\
&\geq \binom{n+r-2}{r-2} - \binom{n+r-4}{r-2} - \binom{n+r-4}{r-3} \\
&= \binom{n+r-3}{r-3}\\
&> 0.
\end{split}