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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.

hydrologist
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2 Answers2

1

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.

Alexander Conrad
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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 $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

Markus Scheuer
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