Prove positivity of rational functions

Question

We say a rational function $F(z)$ is positive if the coefficients of its Maclaurin expansion, in the variable $z$, are non-negative.

In this context, let $$F_r(z):=\frac{1 - 2z + z^r - (1 - z)^r}{(1 - z)^{r - 1}(1 - 2z)}.$$

Is the following true? Note: $F_2(z)=0$ and $F_3(z)$ is easier to manage.

QUESTION. For $r\geq4$, each of the rational functions $F_r(z)$ is positive.

Example. After simplifications, $F_4(z)=\frac{2z}{(1-z)^2}$.

Answer 1

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}

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ADDED. The above bound implies a stronger statement: for $r\geq 2$ the function $$F_r(z) + 1 - \frac{1}{(1-z)^{r-2}}$$ is non-negative.