Perturbation series for $x^5+\varepsilon x-1=0$ I want to find a closed form for the perturbation coefficients $a_n$ defined by the perturbative solution 
$$
x(\varepsilon)=1+\sum_{n=1}^\infty a_n \varepsilon^n
$$
to the quintic equation
$$
x^5+\varepsilon x-1=0.
$$
By computing the determinant, as was suggested for this related question regarding the cubic case, we can argue that the radius of convergence of the above series must be
$$
\rho=\frac{5}{4^{4/5}}=1.64938\dots
$$
Furthermore, the Lagrange-Bürmann theorem allows us to formally write down a closed form for the coefficients, namely 
$$
a_n=\frac{(-1)^n}{n!}\frac{d^{n-1}}{dx^{n-1}}\left(\frac{x}{1+x+x^2+x^3+x^4}\right)^n_{x=1}
$$ 
but this doesn't look very illuminating. (For the case of the cubic equation a slightly more explicit but still cumbersone rewriting was made possible by the simpler form of $a_n$). 
What I would like to achieve is obtaining $\rho$ from the explicit closed expression for the $a_n$. Therefore I set out to compute some of them. Here they are:
$$
a_1=-\frac{1}{5},\
a_2=-\frac{1}{5^2},\
a_3=-\frac{1}{5^3},\
a_4=0,\\
a_5=\frac{21}{5^6},\
a_6=\frac{78}{5^7},\
a_7=\frac{187}{5^8},\
a_8=\frac{286}{5^9},\
a_{9}=0,\\
a_{10}=-\frac{9367}{5^{12}},\
a_{11}=-\frac{39767}{5^{13}},\
a_{12}=-\frac{105672}{5^{14}},\
a_{13}=-\frac{175398}{5^{15}},\
a_{14}=0.
$$
The behavior of the $a_n$ for $n=1,\ldots,30$ supports the following conjecture:
$$\boxed{
a_n = -(-1)^{\lfloor n/5\rfloor}\frac{c_n}{5^{\alpha_n}}
}
$$
where
$$
\alpha_n=\sum_{k=0}^\infty \left\lfloor \frac{n}{5^k} \right\rfloor
$$
and the $c_n$ are nonnegative integer coefficients which are not divisible by $5$ and vanish for $n=5m-1$.
I think that $c_n$ should be something of the form
$$
\frac{1}{4n+1}\binom{5n}{n}
$$
which for $n=4$ goes very near to reproducing $c_8=286$ and has a scaling similar to that exhibited by the $c_n$. This problem is also motivated by this video where it is suggested that the answer can be guessed with some effort by staring at the coefficients hard enough (27.08).
I think I need some help, however!
 A: For convenience let $\varepsilon = 5s$.  Maple gives me the following:
$$\sum _{k=0}^\infty \frac{(-16 s^5)^k}{k!} \left(\frac{\left(\frac 2 5\right)_{2k} \left(-\frac 1 {10}\right)_{2k} }{\left(\frac 4 5\right)_k \left(\frac 3 5\right)_k \left(\frac 2 5\right)_k} 
- \frac{\left(\frac 4 5\right)_{2k} \left(\frac 3 {10}\right)_{2k} s }{ \left(\frac 6 5\right)_k \left(\frac 4 5\right)_k \left(
\frac 3 5\right)_k }
- \frac{\left(\frac{6}{5}\right)_{2k} \left(\frac{7}{10}\right)_{2k} s^2 }{ \left(\frac{7}{5}\right)_k \left(\frac{6}{5}\right)_k \left(
\frac{4}{5}\right)_k }  
- \frac{\left(\frac{8}{5}\right)_{2k} \left(\frac{11}{10}\right)_{2k} s^3 }{\left(\frac{8}{5}\right)_{k} \left(\frac{7}{5}\right)_k \left(
\frac{6}{5}\right)_k} \right) 
$$
using the Pochhammer symbols $$(a)_m = \frac{\Gamma(a+m)}{\Gamma(a)}$$
A: Making Robert Israel's formula a bit more compact we can write
$$\boxed{
x(\varepsilon)= 
\sum_{k=0}^\infty (-1)^{k+1}(2)^{4k}\sum_{j=0}^3(-1)^{\delta_{j0}}\frac{\left(\frac{2(j+1)}{5}\right)_{2k}\left(\frac{2(j+1)}{5}-\frac{1}{2}\right)_{2k}}{\left(\frac{j+5}{5}\right)_{k}\left(\frac{j+4}{5}\right)_{k}\left(\frac{j+3}{5}\right)_{k}\left(\frac{j+2}{5}\right)_{k}}\frac{\varepsilon^{5k+j}}{5^{5k+j}}
}
$$
where $(a)_m$ is the Pochhammer symbol and
$$
(a)_m = \frac{\Gamma(a+m)}{\Gamma(a)}=(a+m-1)(a+m-2)\cdots(a+1)a
$$
for positive integer $m$.
Applying Stirling's asymptotic expansion, 
$$\begin{aligned}
&\lim_{k\to\infty}\left[\frac{2^{4k}}{5^{5k+j}}\frac{\left(\frac{2(j+1)}{5}\right)_{2k}\left|\left(\frac{2(j+1)}{5}-\frac{1}{2}\right)_{2k}\right|}{\left(\frac{j+5}{5}\right)_{k}\left(\frac{j+4}{5}\right)_{k}\left(\frac{j+3}{5}\right)_{k}\left(\frac{j+2}{5}\right)_{k}}\right]^{1/{5k}}\\
=&\ \lim_{k\to\infty}\frac{2^{4/5}}{5}\frac{\left(2+\frac{2(j+1)}{5k}\right)^{2/5}\left(2+\frac{2(j+1)}{5k}-\frac{1}{2k}\right)^{2/5}}{\left(1+\frac{j+5}{k}\right)^{1/5}\left(1+\frac{j+4}{k}\right)^{1/5}\left(1+\frac{j+3}{k}\right)^{1/5}\left(1+\frac{j+2}{k}\right)^{1/5}}\\
=&\ \frac{2^{4/5}}{5}\cdot 2^{2/5}\cdot 2^{2/5}\\
=&\ \frac{4^{4/5}}{5}
\end{aligned}$$
independently of $j=0,1,2,3$. Therefore by Hadamard's formula, the radius of convergence is 
$$\boxed{
\rho=\frac{5}{4^{4/5}}.
}
$$
