How to prove $\sum_{n\ge 0} a_n x^n = \sum_{n\ge 0} \left(\sum_{k=0}^{n}(-1)^{k+1} \binom{n}{k}a_k\right) \frac{x^n}{(x-1)^{n+1}}$?

Question

While reading the Wikipedia article on the Binomial transform, which is defined for a given sequence $\{a_n\}$ as $$ s_n = \sum_{k=0}^{n}(-1)^k \binom{n}{k} a_k $$ I found the following relation between generating functions: If $f(x) = \sum_{n \ge 0} a_n x^n$ and $g(x) = \sum_{n \ge 0} s_n x^n$ then $$ (1-x) g(x) = f\left( \frac{x}{x-1}\right) $$ which recalling the original definitions for $g(x),f(x)$ and $s_n$ can be rewritten as $$ \sum_{n\ge 0} a_n x^n = \sum_{n\ge 0} \left(\sum_{k=0}^{n}(-1)^{k+1} \binom{n}{k}a_k\right) \frac{x^n}{(x-1)^{n+1}} $$

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The Wikipedia article doesn't mention how to prove this relationship between the generating functions, and I wondered how to prove it. My first idea was to expand $(x-1)^{-n-1}$ as a Taylor series such that we have several sums, but where only terms of the form $x^{\text{Something}}$ is present which resembles the original form we want to get at. This didn't seem to lead anywhere. Does anyone have any resources or suggestions on how to prove this? Thanks!

Answer 1

\begin{align} &\quad\sum_{n\ge 0} \left(\sum_{k=0}^{n}(-1)^{k+1} \binom{n}{k}a_k\right) \frac{x^n}{(x-1)^{n+1}} \\ &= \sum_{k\ge 0}(-1)^{k+1} a_k \sum_{n\ge k} \binom{n}{k}\frac{x^n}{(x-1)^{n+1}} \\ &= \frac{1}{x-1} \sum_{k\ge 0}(-1)^{k+1} a_k \sum_{n\ge k} \binom{n}{k}\left(\frac{x}{x-1}\right)^n \\ &= \frac{1}{x-1} \sum_{k\ge 0}(-1)^{k+1} a_k \left(\frac{x}{x-1}\right)^k \sum_{n\ge 0} \binom{n+k}{k}\left(\frac{x}{x-1}\right)^n \\ &= \frac{1}{x-1} \sum_{k\ge 0}(-1)^{k+1} a_k \left(\frac{x}{x-1}\right)^k \frac{1}{(1-x/(x-1))^{k+1}} \\ &= \frac{1}{x-1} \sum_{k\ge 0}(-1)^{k+1} a_k \left(\frac{x}{x-1}\right)^k (1-x)^{k+1} \\ &= \sum_{k\ge 0}a_k x^k \\ \end{align}