Generating series for binomial coefficients Is there a simple expression for the generating series $$f_n(z)=\sum_{k\geq 1}\binom{kn}{n}z^k$$
for $n=1$, I know it's $\frac{z}{(1-z)^2}$, but what happens for other $n$? Thanks for any help.
 A: Let
$$ g_n(x) = \sum_{k \ge 0} \binom{k}{n}x^k = \frac{x^{n}}{(1 - x)^{n + 1}}. $$
Let $\zeta_n = \exp\left(2 \pi i/n \right)$. Then
$$ \frac{g_n(x) + g_n(\zeta_n x) + \cdots + g_n(\zeta_n^{n - 1} x)}{n} = \sum_{k \ge 0} \binom{kn}{n} x^{kn}. $$
So
$$ f_n(x) = \frac{g_n(x^{1/n}) + g_n(\zeta_n x^{1/n}) + \cdots + g_n(\zeta_n^{n - 1} x^{1/n})}{n}. $$
A: I think the general solution will be of the form $f_n(z)=\frac{P_n(z)}{(1-z)^{n+1}}$, where $P_n(z)$ is a polynomial. This is because $\binom{nk}{k}=\frac{nk(nk-1)..(nk-n)}{n!}$, so that you need to know how to evaluate $\sum_{k=0}^\infty nk(nk-1)...(nk-n)z^k$, which can be done by computing $n$ derivatives of $\frac{1}{(1-z^{n})}$. This is not fun.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\left.\vphantom{\Large A}\mrm{f}_{n}\pars{z}
\right\vert_{\ \verts{z}\ <\ 1} \equiv
\sum_{k \geq 1}{kn \choose n}z^{k}.\quad
\mbox{Note that}\quad \mrm{f}_{0}\pars{z} = {z \over 1 - z}}$.

Then,
\begin{align}
\left.\vphantom{\Large A}\mrm{f}_{n}\pars{z}
\right\vert_{\ n\ \in\ \mathbb{N}_{\large\ \geq\ 1}} & \equiv \sum_{k \geq 1}
{kn \choose n}z^{k} =
\sum_{k = 0}^{\infty}{kn \choose n}z^{kn/n} =
\sum_{k = n}^{\infty}{k \choose n}z^{k/n}
\,{\sum_{\ell = 0}^{n - 1}r_{\ell}^{k} \over n}
\end{align}

where $\ds{r_{\ell} = \exp\pars{{2\ell\pi \over n}\,\ic}.\ \ell = 0,1,2,\ldots,n - 1}$.

Then,
\begin{align}
\left.\vphantom{\Large A}\mrm{f}_{n}\pars{z}
\right\vert_{\ n\ \in\ \mathbb{N}_{\large\ \geq\ 1}} & =
{1 \over n}\sum_{\ell = 0}^{n - 1}\sum_{k = n}^{\infty}
{k \choose k - n}\pars{z^{1/n}r_{\ell}}^{k} =
{1 \over n}\sum_{\ell = 0}^{n - 1}\sum_{k = n}^{\infty}
{-n - 1 \choose k - n}\pars{-1}^{k - n}\pars{z^{1/n}r_{\ell}}^{k}
\\[5mm] & =
{1 \over n}\sum_{\ell = 0}^{n - 1}\sum_{k = 0}^{\infty}
{-n - 1 \choose k}\pars{-1}^{k}\pars{z^{1/n}r_{\ell}}^{k + n}
\\[5mm] & =
{1 \over n}\sum_{\ell = 0}^{n - 1}\pars{z^{1/n}r_{\ell}}^{n}
\sum_{k = 0}^{\infty}{-n - 1 \choose k}\pars{-z^{1/n}r_{\ell}}^{k} =
{z \over n}\sum_{\ell = 0}^{n - 1}\pars{1 - z^{1/n}r_{\ell}}^{-n - 1}
\end{align}

$$
\\
\bbox[15px,#ffe,border:1px dotted navy]{\ds{%
\mrm{f}_{n}\pars{z} =
\left\{\begin{array}{lcl}
\ds{z \over 1 - z} & \mbox{if} & \ds{n = 0}
\\[2mm]
\ds{{z \over n}\sum_{\ell = 0}^{n - 1}
\bracks{1 - z^{1/n}\exp\pars{-\,{2\pi\ell \over n}\,\ic}}^{-n - 1}} & \mbox{if} & \ds{n \geq 1}
\end{array}\right.}}\\
$$
