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 $m\geq n$?
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$\begingroup$ Why the down vote? Is the problem too easy or ...? $\endgroup$– Lewi_SolCommented Jun 1, 2017 at 17:18
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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.
answered Jun 1, 2017 at 17:21
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$\begingroup$ You're welcome. One further tip. When moving $\infty$ to $0$, I often find it's easier to move it in both the domain and the range. In other words, conjugate by $x\to x^{-1}$ and look at $1/f(1/x)=b_nx^n+b_{n+1}x^{n+1}+\cdots$. This does not, intrinsically, change much (unless you're planning to iterate $f$), but somehow power series in $x$ seem more appealing than power series in $1/x$. $\endgroup$ Commented Jun 1, 2017 at 18:33