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.

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

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? The cases $r=2$ and $3$ are easier to manage.

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

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.