<s>[INST] Prove positivity of rational functions 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, $F_4(z)=\frac{2z}{(1-z)^2}$.
 [/INST] 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}

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.
</s>