A rational fraction with positive Taylor series Consider the polynomial
$$P_{a, m}(x)= \sum_{l=0}^m (-1)^l \binom{a}{l} x^l$$
It is a truncation of the binomial series $(1-x)^a$.
Assume that  $m$ even, $n$ odd, and $a\ge m$, $b\ge n$, and $a \le b$. One is to show that the rational fraction
$$\frac{P_{a, m}(x)}{P_{b,n}(x)}$$
has a Taylor series at $x=0$ with positive coefficients.
Example:
$$\frac{P_{6,4}(x)}{P_{7,3}(x)} = \frac{1 - 6 x + 15 x^2 - 20 x^3 + 15 x^4}{1 - 7 x + 21 x^2 - 35 x^3}$$
has Taylor series
$$1 + x + x^2 + x^3 + 36 x^4 + 266 x^5 + 1141 x^6 + 3661 x^7 + 10976 x^8 + 39886 x^9 + 176841 x^{10} + 784441 x^{11} + \cdots$$
Comments: The intuition behind the fact: for $x$ positive we have
$P_{a,m}(x)\ge (1-x)^a$ for even $m\le a$, and $P_{b,n}(x)\le (1-x)^b$ for odd $n\le b$. Therefore $P_{a,m}(x)\ge (1-x)^a \ge (1-x)^b\ge P_{b,n}(x)$. This indicates that the statement is plausible.
Also, it can be shown that $P_{a,m}> 0$ on $\mathbb{R}$ if $m$ even, and $a> m$ ( $P_{a, m}$ satisfies certain a differential equation of order $1$  of the form $ - \frac{(1-x)}{a} \cdot   f'(x)  + c\cdot  x^m = f(x)$ for some $a> 0$, approximately what $(1-x)^a$  does), while $P_{b,n}$ has a unique real root if $n$ odd and $b> n$.
Potential approach : One shows that
$$P_{a,m}(x) \cdot (1-x)^{-a} = 1 + \textrm{positive terms} $$
while
$$P_{b,n}(x) \cdot (1-x)^{-b} = 1 - (\textrm{positive terms}) $$
From here, we get some expressions for $P_{a,m}$, $P_{b,n}$.
Note that $a$, $b$ do not have to be integers.
Any feedback would be appreciated!
 A: First, notice that $P_{a,\infty} = (1-x)^a$ and
$$\frac{P_{a,m}(x)}{P_{b,n}(x)} =
\frac{P_{a,m}(x)}{P_{a,\infty}(x)} \frac{P_{a,\infty}(x)}{P_{b,\infty}(x)} \frac{P_{b,\infty}(x)}{P_{b,n}(x)}\\
= \left(P_{a,m}(x)(1-x)^{-a}\right)(1-x)^{a-b}\left(P_{b,n}(x)(1-x)^{-b}\right)^{-1}.\tag{1}$$
Thus, the original statement
$$
  \forall a \geq 0\;
  \forall b \geq a\;
  \forall m: m \leq a, \text{$m$ is even}\;
  \forall n: n \leq b, \text{$n$ is odd}\;\\
  \frac{P_{a,m}(x)}{P_{b,n}(x)}\text{ has a Taylor series with non-negative coefficients} \tag{2}
$$
is equivalent to the claim that all of the following 3 statements are true.

*

*$\forall a \geq 0\;\forall m: m \leq a, \text{$m$ is even}\;{}$
$P_{a,m}(x)(1-x)^{-a}$ has a Taylor series with non-negative coefficients.

*$\forall b \geq 0\;{}$
$(1-x)^{-b}$ has a Taylor series with non-negative coefficients.

*$\forall b \geq 0\; \forall n: n \leq b, \text{$n$ is odd}\;{}$
$\left(P_{b,n}(x)(1-x)^{-b}\right)^{-1}$ has a Taylor series with non-negative coefficients.

Statement #2 easily follows from the Taylor expansion formula. Thus, we will focus on #1 and #3. As you've noticed, it is enough to show
that for $a\geq0,n\leq a,n\in\mathbb{N}$ we have
$$P_{a,n}(x)(1-x)^{-a} = 1 + (-1)^n \cdot (\text{terms with non-negative coefficients}). \tag{3}$$
To show that, write
$$P_{a,n}(x)(1-x)^{-a} = 1 + \sum_{k=1}^{\infty}(-1)^kx^k \sum_{l=0}^{n} {a \choose l}{-a \choose k-l} \tag{4}$$
$$= 1 + \sum_{k=1}^{\infty}(-1)^{k+1}x^k \sum_{l=n+1}^{\infty} {a \choose l}{-a \choose k-l}.\tag{5}$$
Here the last some actually goes up to $l=k$ (or is zero when $k \leq n$).
Consider pairs of non-zero neighboring terms for $l$ and $l+1$. Since they are non-zero, $0\leq l \leq k-1$. The ratio between them is
$$\frac{{a \choose l}{-a \choose k-l}}{{a \choose l+1}{-a \choose k-l-1}}=-\frac{(k-l)(a-l)}{(l+1)(a+k-l-1)}.\tag{6}$$
This quantity is negative. It's absolute value is greater than 1 iff $(k-l)(a-l) > (l+1)(a+k-l-1)$, or, equivalently, when $l^2-l(k+a-1)+(k-1)(a-1)/2 > 0$, i.e. when $l < l_{\text{critical}} - \frac12$, where $l_{\text{critical}} = \frac{k+a}{2} - \frac12 \sqrt{k^2+a^2}$. In other words, the terms ${a \choose l}{-a \choose k-l}$ increase with $l$ till $l=l_{\text{critical}}$, and then decrease afterwards. If $l_{\text{critical}} < n + 1/2$ we use (5) to evaluate the coefficient, otherwise we use (4). In both cases the coefficient near $x^k$ takes the form
$$(-1)^{n} \sum_{m=0}^{\infty} (-1)^m c_m. \tag{7}$$
Here $m=l-n-1$ for $l_{\text{critical}} < n + 1/2$ and $m=n-l$ otherwise,
$c_m = (-1)^{k-l} {a \choose l}{-a \choose k-l}$.
Notice, that $c_m$ are non-negative, monotonically decreasing, and tending to $0$ as $m\to \infty$ (they are actually $0$ for large enough $m$). Thus, the sum in (7) is non-negative, hence (3) holds.
