TITLE: Functional inverse of $z=1+w+\cdots+w^{n-1}$
QUESTION [14 upvotes]: Migrated to MO.

I am interested in the functional inverse of
$$
z=1+w+\cdots+w^{n-1},\quad w\geq0,\ n>1.
$$
This function is strictly increasing on $w\geq0$ and thus admits an inverse.

My attempt:
By Lagrange's theorem we may write the inverse:
$$
w(z)=a+\sum _{k=1}^\infty g_k\frac{(z-f(a))^k}{k!},
$$
with $f(w)=1+w+\cdots+w^{n-1}$ and
$$
g_k=\lim_{w\to a}\partial_w^{k-1}\left({\frac {w-a}{f(w)-f(a)}}\right)^k.
$$
Expanding $f(w)$ around $w=1$ gives $f(w)=nF(1,1-n;2;1-w)$, which is a hypergeometric function. Choosing $a=1$ we write
$$
\frac{f(w)-f(1)}{w-1}=-n\frac{F(1,1-n,2,1-w)-1}{1-w}=\frac{n(n-1)}{2}F(1,2-n;3;1-w).
$$
It follows
$$
w(z)=1+\sum _{k=1}^\infty a_k\frac{2^k}{n^k(n-1)^k}\frac{(z-n)^k}{k!},
$$
with
$$
a_k=\lim_{w\to 1}\partial_w^{k-1}\left(F(1,2-n;3;1-w)\right)^{-k}.
$$
Evaluating the limit for $a_k$ is certainly non-trivial. One thought was to use the Faà di Bruno formula but I am a little unclear on the details of this calculation and am curious if $a_k$ can be written in a "nice" form that does not involve Bell polynomials. Could someone please fill me in on the details of computing the $a_k$'s? Also, for which values of $z$ does this series converge?
I was able to write a quick one line code in Mathematica to compute the $a_k$'s, which may lend itself to finding a pattern:
a[k_] := Limit[D[Hypergeometric2F1[1, 2 - n, c, 1 - w]^-k, {w, k - 1}], w -> 1] /. c -> 3

$a_1$, $a_2$, and $a_3$ seem to factor nicely while $a_4$ does not.

Edit:
Using Faà di Bruno's formula I was able to write down an explicit form for the $a_k$'s giving a final solution of
$$
\bbox[5px,border:2px solid #C0A000]{%
w(z)=1+\frac{2(z-n)}{n(n-1)}+\sum_{k=2}^\infty\sum_{\ell=1}^{k-1}(-k)^{(\ell)}B_{k-1,\ell}\left(\left\{(-1)^m\tfrac{m!(2-n)_m}{(3)_m}\right\}_{m=1}^{k-\ell}\right)\frac{\left(\frac{2(z-n)}{n(n-1)}\right)^k}{k!},%
}
$$
where $(s)^{(n)}=\Gamma(s+1)/\Gamma(s-n+1)$ is the falling factorial, $(s)_n=\Gamma(s+n)/\Gamma(s)$ is the Pochhammer symbol, and $B_{n,k}$ is the partial Bell polynomial. It is interesting to note that this result also works for the more general case $\{n\in\Bbb R:n>1\}$.
All that I am still curious about is the radius of convergence for this series which I believe is $|z-n|<n-1$.
Here is Mathematica code to compare the exact function $w(z)$ to approximation obtained by truncating its series expansion:
a[k_, n_] := 
 Sum[FactorialPower[-k, l] BellY[k - 1, l, 
    Table[(-1)^m (m! Pochhammer[2 - n, m])/Pochhammer[3, m], {m, 1, 
      k - l}]], {l, 1, k - 1}]
g[z_, n_, K_] := 
 1 + (2 (z - n))/(n (n - 1)) + 
  Sum[a[k, n] ((2 (z - n))/(n (n - 1)))^k/k!, {k, 2, K}]
gAprx[z_, n_] := 
 Quiet[N[Solve[(1 - w^n)/(1 - w) - z == 0, w, PositiveReals][[1, 1, 
     2]]]]
P[n_] := Manipulate[
  Plot[{gAprx[z, n], g[z, n, m]}, {z, 1, 2 n - 1}], {m, 2, 20, 1}]
P[5]

REPLY [2 votes]: Extended Comment:
For the case $n=2$, the radius of convergence for the inverse series $w(z)$ is infinite, which can be deduced from the fact that $a_k=0$ for all $k\geq 2$.
For $n=3,4,\dots$, the approach given by @Gary provides a means for finding the radius of convergence of the inverse series either exactly or numerically. Let $w_k$ represent any one of the $n-2$ solutions of $f^\prime(w)=0$. Evaluating the derivative of $f(w)=(1-w^n)/(1-w)$ and equating with zero we find
$$
\begin{align}
f^\prime(w_k)=0
&\implies(n-1)w_k^n-nw_k^{n-1}+1=0,\quad w_k\neq 1\\
&\implies w_k^n=\frac{w_k}{n-(n-1)w_k},\quad w\neq 1\\
\end{align}
$$
Substituting $w_k^n$ into $f$ then gives
$$
z_k:=f(w_k)=\frac{n}{n-(n-1)w_k}.
$$
Now, after a bit of algebra
$$
|z_k-n|=(n-1)\biggl|\left(1+\tfrac{w_k}{n(1-w_k)}\right)^{-1}\biggr|;
$$
thus, the radius of convergence for any $n$ can be found by computing the minimum $|z_k-n|$, that is
$$
r(n)=\min_{k\in\{1,\dots,n-2\}}(n-1)\biggl|\left(1+\tfrac{w_k}{n(1-w_k)}\right)^{-1}\biggr|.
$$
This can be implemented in Mathematica as
fprime[w_, n_] := (1 - w^n)/(1 - w)^2 - ((n w^(-1 + n))/(1 - w));
wk[n_] := Solve[fprime[w, n] == 0, w][[All, 1, 2]];
ROC[n_] := 
  ToRadicals[
   Min[FullSimplify[(n - 1) Abs[(1 + wk[n]/(n (1 - wk[n])))^-1]]]];

For $n=3,4,5$ we obtain exact solutions, namely,
$$
\begin{align}
n=3: &|z-3|<\frac{9}{4}\\
n=4: &|z-4|<4\sqrt{\frac{2}{3}}\\
n=5: &|z-5|<\frac{5}{8 \sqrt{\frac{3}{140-5\ 2^{2/3} \sqrt[3]{\frac{5}{3 \sqrt{6}-2}}+5^{2/3} \sqrt[3]{2
   \left(3 \sqrt{6}-2\right)}}}}\\
\end{align}
$$
Mathematica is not able to find closed forms for larger values of $n$; however, we can compute the radius of convergence numerically with the modified code
fprime[w_, n_] := (1 - w^n)/(1 - w)^2 - ((n w^(-1 + n))/(1 - w));
wk[n_] := NSolve[fprime[w, n] == 0, w][[All, 1, 2]];
ROC[n_] := 
  Min[FullSimplify[(n - 1) Abs[(1 + wk[n]/(n (1 - wk[n])))^-1]]];

This code also works for noninteger values of $n$, e.g.
ROC[10.7]=9.75823