generating function of summation equation

Question

Find a formula for the generating function $(1)_{(k)} + (2)_{(k)}x + (3)_{(k)}x^2 + (4)_{(k)}x^3 + (5)_{(k)}x^4 + ... = \sum_{n\geq 0}(n+1)_{(k)}x^n$. The subscript $k$ is falling factorial.

Answer 1

Denoting with $n^{\underline{k}}=n(n-1)\cdots(n-k+1)$ the falling factorials we observe the first $k-1$ terms of the series vanish.

We obtain \begin{align*} \color{blue}{\sum_{n\geq k}n^{\underline{k}}x^{n-1}}&=\sum_{n\geq k}\frac{n!}{(n-k)!}x^{n-1}\\ &=\frac{k!}{x}\sum_{n\geq k}\binom{n}{k}x^n\tag{1}\\ &=\frac{k!}{x}\sum_{n\geq 0}\binom{n+k}{n}x^{n+k}\tag{2}\\ &=k!x^{k-1}\sum_{n\geq 0}\binom{-k-1}{n}(-x)^n\tag{3}\\ &\color{blue}{=\frac{k!x^{k-1}}{(1-x)^{k+1}}}\tag{4} \end{align*}

Comment:

• In (1) we factor out $x$ and use binomial coefficients $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ for convenience.

• In (2) we shift the index $n$ by $k$ to start with $n=0$.

• In (3) we factor out $x^k$ and use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (4) we use a binomial series expansion.