The Bring-Jerrard quintic,
$$x^5+x+t=0$$
can be solved as,
$$x = -\sum_{k=0}^\infty(-1)^k\frac{(5k)!}{k!(4k+1)!}\;t^{4k+1}\tag1$$
when,
$$|t|<\frac{4}{5^{5/4}}\approx 0.53\dots$$
This paper mentions how Passare and Tsikh found that a version of the principal quintic,
$$Bx^5+Ax^2+x+1 = 0$$
can be solved as,
$$x = -\sum_{j=0}^\infty \sum_{k=0}^\infty (-1)^k \frac{(2j+5k)!}{j!k!(j+4k+1)!}\;A^j B^k\tag2$$
The domain of convergence is, let $|A|,|B| = a,b$, then,
$$5^5b^2 - 4^4b + 108a^5 - 27a^4 + 1600a b - 2250a^2b<0$$
For example, let $A=173/5000,\;B=1/16$ and we find the series indeed yields $x=-0.9773\dots$
Question: Is there a double-sum solution like $(2)$ for the Bring-Jerrard sextic,
$$Bx^6+Ax^2+x+1=0$$
and if there is, what is its domain of convergence?
If you write the equation as $x + 1 + t (c x^2 + d x^6) = 0$, then expand a root of this in powers of $t$, you get a series $$\eqalign{ x &= \sum_{n=0}^\infty C_n(c,d) t^n \cr &= -1 + (-c-d) t + (-2 c^2-8 c d-6 d^2) t^2 + \ldots }\tag{1}$$ where each $C_n(c,d)$ is a homogeneous polynomial of degree $n$ in $c,d$. For any $c,d$ this should converge for $t$ in some neighbourhood of $0$. The Lagrange inversion formula says $$\eqalign{C_n(c,d) &= \left.\dfrac{1}{n!} \left(\dfrac{d^{n-1}}{dx^{n-1}}\right) (-cx^2-dx^6)^n \right|_{x=-1}\cr &= - \sum_{j=0}^n \dfrac{(6n-4j)!}{ j! (n-j)! (5n-4j+1)!} c^j d^{n-j}}$$ That is, your double-sum formula is $$ x = - \sum_{j=0}^\infty \sum_{k=0}^\infty \dfrac{(2j+6k)!}{j! k! (j + 5 k + 1)!} A^j B^k \tag{2}$$
EDIT: The discriminant of $x+1 + t(cx^2 + d x^6)$ with respect to $x$ is $$-{d}^{3}{t}^{4} \left( 1024\,{c}^{6}{t}^{5}-256\,{c}^{5}{t}^{4}+13824 \,{c}^{3}d{t}^{3}-43200\,{c}^{2}d{t}^{2}+22500\,cdt+46656\,{d}^{2}t- 3125\,d \right) $$ Neglecting the factor $-d^3 t^4$ (which comes from the fact that when $t=0$ or $d=0$ the degree is less than $6$), the radius of convergence of the series (1) should be at least the smallest of the absolute values of roots of $1024\,{c}^{6}{t}^{5}-256\,{c}^{5}{t}^{4}+13824 \,{c}^{3}d{t}^{3}-43200\,{c}^{2}d{t}^{2}+22500\,cdt+46656\,{d}^{2}t- 3125\,d$ with respect to $t$. In general I'm not sure this corresponds exactly to convergence of series (2) because of potential cancellations, but at least if $A,B > 0$ there are no cancellations to worry about. Thus we should have convergence of (2) at least in the shaded region of the first quadrant shown below, where the boundary curve is $-1024 A^6+256 A^5-13824 A^3 B+43200A^2 B-22500 A B-46656 B^2+3125 B = 0$.