General expression for coefficients of power expansion to a power $\left(\sum_{j=0}^\infty (ax)^j\right)^n$ I am working with a set of equations where I need to expand the sum of a geometric series to a power, something of the form:
\begin{equation}
\left[1+(ax)+(ax)^2+(ax)^3+\ldots\right]^n \quad \mbox{or} \quad \left(\sum_{j=0}^\infty (ax)^j\right)^n \quad \mbox{where} \quad |ax|<1
\end{equation}
This operation is necessary as I need to equate the resulting coefficients of powers of $x$. I have straightforwardly determined that the $i$^th coefficient, $C_i$, of the expansion is given by:
for $n=1$:
\begin{equation}
1+(ax)+(ax)^2+(ax)^3+\ldots \quad \mbox{so} \quad C_i = 1
\end{equation}
for $n=2$:
\begin{equation}
1+2(ax)+3(ax)^2+4(ax)^3+\ldots \quad \mbox{so} \quad C_i = i
\end{equation}
for $n=3$
\begin{equation}
1+3(ax)+6(ax)^2+10(ax)^3+\ldots\quad \mbox{so} \quad C_i = i\frac{\left(i+1\right)}{2}
\end{equation}
and for $n=4$:
\begin{equation}
1+4(ax)+10(ax)^2+15(ax)^3+\ldots \quad \mbox{so} \quad C_i = i\frac{\left(i+1\right)}{2}\frac{\left(i+2\right)}{3}
\end{equation}
and so on... Is there a general way to write this as a series in the form:
\begin{equation}
\sum_{i=1}^\infty f\left(C_i\right) (ax)^{i-1}
\end{equation}
I can see that this requires some form of factorial expression, but my factorial manipulation is very rusty. Any assistance gratefully received.
 A: I think you want a general form for
$$C_{i,{n+1}}=\frac{i(i+1)(i+2)\cdots (i+n-1)}{n!}$$ as you've defined it, with $C_{i,1}=1.$ This is given by
$$C_{i,{n+1}}=\frac{(i+n-1)!}{n!(i-1)!}$$ because the factorial descends from $i+n-1$, but needs a product underneath to cancel the other terms past $i$ in the numerator such as $i-1$ and so on.
A: The $n$-th power of a geometric series is a binomial series. We can perform a binomial series expansion and obtain

\begin{align*}
\color{blue}{\left(\sum_{j=0}^{\infty}\left(ax\right)^j\right)^n}&=\left(\frac{1}{1-ax}\right)^n\tag{1}\\
&=\frac{1}{(1-ax)^n}\\
&=\sum_{j=0}^{\infty}\binom{-n}{j}(-ax)^j\tag{2}\\
&\,\,\color{blue}{=\sum_{j=0}^{\infty}\binom{n+j-1}{j}(ax)^j}\tag{3}
\end{align*}

We see the wanted coefficient $f(C_{i+1})=\binom{n+j-1}{j}=\frac{(n+j-1)!}{j!(n-1)!}$.
Comment:

*

*In (1) we use the geometric series expansion.


*In (2) we use the binomial series expansion.


*In (3) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
