$f$ real-rooted forbid truncated $\frac1f$ to be so? 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 $r\neq\pm1$. 
 A: 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 $Q(x)$.
A: We use a standard notation $[x^n]h(x)$ for a coefficient of $x^n$ in the series of $h(x)$. 
Assume that $t$ is a real root of $P_n$. First of all, note that $$0=a_0+a_1t+\dots+a_nt^n=t^n[x^n]\frac1{f(x)(1-x/t)},$$
thus for $g(x)=f(x)(1-x/t)$ we just need to prove that all coefficients $b_{2n}=[x^{2n}]1/g(x)$ are non-zero. We may write $g(x)=\prod_{i=1}^m(1-\alpha_ix)$, suppose for a moment that $\alpha$'s are different. Then $1/g=\sum c_i/(1-\alpha_ix)$, where $c_i=\alpha_i^{m-1}/\prod_{j\ne i}(\alpha_i-\alpha_j)$. Therefore $$b_{2n}=\sum_{i=1}^m\frac{\alpha_i^{m-1+2n}}{\prod_{j\ne i}(\alpha_i-\alpha_j)}.$$
It is a Schur's function corresponding to a partition $(2n,0,\dots,0)$, and this fact holds true even when some $\alpha$'s coincide (by continuity). I believe it should be well known that the value of this Schur function is always positive (when not all $\alpha$'s are equal to 0). Below is some proof.
We induct on $m$, base $m=1$ is clear. Again assume that $\alpha$'s are different. Note that $b_{2n}$ is a coefficient of $t^{m-1}$ in the polynomial $h(t)$ of degree at most $m-1$ satisfying $h(\alpha_i)=H(\alpha_i)$, where $H(\alpha)=\alpha^{m-1+2n}$. By Rolle's theorem for the function $h-H$ there exist $\beta_1,\dots,\beta_{m-1}$ between consecutive $\alpha$'s for which $h'(\beta_i)=H'(\beta_i)$. Therefore for $m$ distinct $\alpha$'s we may find $m-1$ distinct $\beta$'s, and this procedure allows a limit case in which some $\alpha's$ coincide (in this case some $\beta$'s also may coincide, but not all of them are equal to 0.) Thus we reduce $m$ to $m-1$ and may induct.
