TITLE: $f$ real-rooted forbid truncated $\frac1f$ to be so?
QUESTION [13 upvotes]: Let $f(x)$ be a polynomial in the ring $\mathbb{R}[x]$, the roots are all real and $f(0)=1$. Write the Taylor series of $1/f(x)$ around the origin as
$$\frac1{f(x)}=\sum_{k=0}^{\infty}a_kx^k,$$
and denote $P_n(x)=\sum_{k=0}^na_kx^k$.

Question. Is it true that the polynomial $P_{2n}(x)$ has no real roots at all?

A prototypical example: If we choose $f(x)=1-x$ then $P_{2n}(x)=\sum_{k=0}^{2n}x^k=\frac{1-x^{2n+1}}{1-x}$ has all roots $r$ on the unit circle and $r\neq\pm1$.

REPLY [8 votes]: Here is an alternative resolution motivated by Fedor's construction in his first line of argument.
Based on $f(0)=1$, it is clear that $\frac1{f(x)}=1+a_1x+a_2x^2+\cdots$ and now write
$$f(x)P_{2n}(x)=f(x)(1+a_1x+a_2x^2+\cdots+a_{2n}x^{2n})=
1-a_{2n+1}x^{2n+1}+\cdots.$$
The RHS, call it $Q(x)$, is a non-constant polynomial. 
More notably, observe that $Q(x)$ exhibits $2n$ (an even number of) vanishing consecutive coefficients. An application of Descartes' Rule of signs reveals $Q(x)$ has at least $2n$ non-real roots. This result is known as de Gua's Rule, which follows from Descartes' or Fourier-Budan Theorem. See the reference here, on page 28, Corollary 2.
Since $f(x)$ has only real roots, these non-real roots must come from the factor $P_{2n}(x)$ of $Q(x)$.