Root of a polynomial $p(x)$ and asymptotic behaviour of the Taylor series of $\frac{1}{p(x)}$ I noticed that for a larger number of polynomials $p(x)$, it is possible to obtain the smallest positive real root by computing the Taylor series (more precisely the Maclaurin series) for $\frac{1}{p(x)}=\sum_{k=0}^{\infty}a_kx^k$ as the limit of $\frac{a_k}{a_{k+1}}$ for $k$ going to infinity.
Example:
Take $p(x)=1-x-x^2-x^3-x^7$, then the Maclaurin series of $\frac{1}{p(x)}$ is given by
$$\frac{1}{p(x)}=1+x+2x^2+4x^3+\dots+227675342x^{32}+422234731x^{33}+\dots
$$
and
$$\frac{a_{32}}{a_{33}}=\frac{227675342}{422234731}=.5392150983... \, ,
$$
which is an excellent approximation to the smallest positive real root of $p(x)$.
I'm pretty sure that this fact is already known, but can't find any reference for it. Does somebody have any reference for it?
Thank you very much for any hint.
 A: Suppose that:


*

*$p(0)\neq0$;

*$p(x)$ has a real root $r$;

*for every root $\lambda$ of $p(x)$, $|\lambda|\geqslant|r|$.


Then a standard Complex Analysis theorem tells us that the radius of convergence of the Taylor series $\sum_{n=0}^\infty a_nx^n$ of $\frac1{p(x)}$ is $|r|$. This is equivalent to asserting that$$\limsup_{n\in\mathbb N}\sqrt[n]{|a_n|}=\frac1r.$$In a large amount of cases, this happens if and only if$$\lim_{n\in\mathbb N}\left|\frac{a_n}{a_{n+1}}\right|=r.$$
A: Yes, your conjecture is correct if the polynomial has a single root
of least absolute value (among all complex roots), i.e.


*

*$p(0) \ne 0$

*$z_1 \in \Bbb C$ is a root: $p(z_1) = 0$,

*All other roots $z_j \in \Bbb C$ satisfy $|z_j| > |z_1|$.


This is a special case of the following more general result:

Let $f$ be holomorphic in some disk $B_R(0)$ with a single zero $z_1 \ne 0$
  in $B_R(0)$, and let the Maclaurin series for $\frac 1f$ be
  $$
 \frac{1}{f(z)} = \sum_{k=0}^\infty a_k z^k \, .
$$
  Then $a_k \ne 0$ for all sufficiently large $k$,  $(\frac{a_k}{a_{k+1}})_k$ is convergent and
  $$
 \lim_{k \to \infty} \frac{a_k}{a_{k+1}} = z_1 \, .
$$

Remark: The assumption that there is a single root of least
absolute value cannot be dropped, as $f(z) = 1-z^2$ or any even
function shows.
Proof: We can write $f(z) = \left(1 - \frac{z}{z_1} \right) g(z)$
where $g$ is holomorphic and non-zero in $B_R(0)$. The Maclaurin
series for $\frac 1g$
$$
  \frac{1}{g(z)} = \sum_{k=0}^\infty b_k z^k
$$
converges for $|z| < R$. Then
$$
\frac{1}{f(z)} = \left( \sum_{k=0}^\infty \frac{z^k}{z_1^k} \right)
\left( \sum_{k=0}^\infty b_k z^k \right) 
= \sum_{k=0}^\infty \left( \sum_{l=0}^k \frac{b_l}{z_1^{k-l}}\right) z^k
$$
and therefore
$$
 a_k = \sum_{l=0}^k \frac{b_l}{z_1^{k-l}} = \frac{1}{z_1^k} \sum_{l=0}^k b_l z_1^l  
 = \frac{1}{z_1^k} \left( \frac{1}{g(z_1)} - \sum_{l=k+1}^\infty b_l z_1^l  \right) = \frac{1}{z_1^k} \left( \frac{1}{g(z_1)} - R_k \right)
$$
where
$$
 R_k = \sum_{l=k+1}^\infty b_l z_1^l 
$$
is the series remainder of the convergent series for $1/g(z_1)$,
and that converges to zero for $k \to \infty$.
It follows that $a_k \ne 0$ for all sufficiently large $k$, and
$$
 \frac{a_k}{a_{k+1}} = z_1 \frac{\frac{1}{g(z_1)} - R_k }{\frac{1}{g(z_1)} - R_{k+1} }
 \to z_1
$$
for $k \to \infty$.
A: I do not know if this could help you.
Considering that we want to solve $f(x)=0$, let us write $\frac 1{f(x)}$ as a $[1,n]$ Padé approximant around $x=0$
$$\frac 1{f(x)}=\frac{a^{(n)}_0+a^{(n)}_1 x}{1+\sum_{k=1}^n b_k x^k}$$ making the approximate solution to be $$x_{(n)}=-\frac{a^{(n)}_0 } {a^{(n)}_1 }=(n+1)\frac{ f^{(n)}(0)}{f^{(n+1)}(0)}$$ 
