So I have a function of the form:
$$f(x) = \frac{1}{1-x p_{2m}(x)}$$
where $p_{2m}(x)$ is a polynomial of degree $2m$. I can expand the function around $x=0$ as: $$f(x) =\sum_{n=0}^{\infty}c_n x^n$$ for some coefficients $c_n$. I would like to prove that the $c_n>0$. One attempt I made was to write: $$f(x) =\sum_{n=0}^{\infty} x^n (p_{2m}(x))^n$$ since I know that $p_{2m}(x)>0$ for all $x\in\mathbb{R}$ and $m\in\mathbb{N}$ with $m\geq1$: $$p_{2m}(x) =\sum_{j=1}^{2m+1} \,(-1)^{j-1}\, a_j\, x^{j-1}$$ with $a_j>0$. So I thought that I could evaluate the function at $x=1$ and compare the two expansions. Obvioulsy this doesn't work because to evaluate the series term to term one would first need to reorganize the expansion to have the same power of $x$. So, I first have to evaluate:
$$(p_{2m}(x))^n = \sum_{k_1,\ldots,k_{2m+1}\\k_1+\ldots+k_{2m+1}=n}\frac{n!}{k_1!\ldots k_{2m+1}!}a(1)^{k_1}\ldots a({2m+1})^{k_{2m+1}}\,\,(-x)^{\sum_{i=1}^{2m+1} i \,k_i-n}$$ which means: $$f(x) =\sum_{n=0}^{\infty}\sum_{k_1,\ldots,k_{2m+1}\\k_1+\ldots+k_{2m+1}=n}\frac{n!}{k_1!\ldots k_{2m+1}!}a(1)^{k_1}\ldots a({2m+1})^{k_{2m+1}}\,\,(-1)^{\sum_{i=1}^{2m+1} i \,k_i-n}x^{\sum_{i=1}^{2m+1} i \,k_i}$$ unfortunately I am stuck. How do I compare the two expansions? Moreover, knowing that $p(x)>0$ is it sufficent to prove that $c_n>0$?
Edit:
There is another important property that I found that I forgot to mention, namely: $$\frac{a(j)}{a(j+1)}>j+1$$ so: $$a(1)>a(2)>a(3)>\ldots>a(2m+1)$$ The example in the comments clearly does not respect this property. With $0<a(j)<1$
Edit 2:
Actually I can give the coefficients:
$$a(j) ={2m+1\choose j}\frac{(2m+2-j)!}{(2m+2)!}$$
One way you might tackle something like this is to look at the partial fraction decomposition of $f(x)$. This depends on the roots of the denominator $1 - x p_{2m}(x)$. If those roots are distinct, $$ f(x) = \sum_{j=1}^{2m+1} \frac{\alpha_j}{x-r_j} = \sum_{n=0}^\infty \sum_{j=1}^{2m+1} \frac{-\alpha_j}{r_j^{n+1}} x^n$$ where $r_j$ are the roots. For large $n$ the coefficient is dominated by the terms for roots of the smallest absolute value. If those are positive, there are only finitely many other terms to check. Note that if $p_{2m}(x) > 0$ for all $x$, all real roots are positive, which will make the corresponding terms $-\alpha_j/r_j^{n+1}$ have constant sign. But there may be complex roots, which will complicate matters.