Suppose $f(x)=\prod\_{k=1}^n(x-a\_k)$ where all $a\_k>0$. 


Expand the function $\frac1f$ at $\infty$ so that
$$\frac1{f(x)}=\frac{b\_n}{x^n}+\frac{b\_{n+1}}{x^{n+1}}+\cdots.$$
**Does it follow that each $b\_m$ is positive, for $m\geq n$?**



A polynomial and its reciprocal expansion
Yes, it's pretty easy.
$$
\frac{1}{f(x)}
= \prod\_{k=1}^n \frac{1}{x-a\_k}
= x^n \prod\_{k=1}^n \frac{1}{1-a\_k/x}
= x^n \prod\_{k=1}^n \sum\_{i=0}^\infty \left(\frac{a\_k}{x}\right)^i.
$$
From this is it clear that your $b\_n$ coefficients are positive.


