Can you find an analytical expression for the following series?
$$\sum_{n=m}^\infty (-1)^n \frac{(a)_n}{(n-m)!}x^n$$
where $m$ is a nonnegative integer, $x\in (0,1)$, $a > 0$, and $(a)_n$ is the Pochhammer symbol denoting a falling factorial:
$$(a)_n = a(a-1)(a-2)\ldots(a-n+1).$$
Using $(a)_n = (a)_m\cdot (a-m)_{n-m}$ and the binomial series, we obtain
$$\begin{align} \sum_{n=m}^\infty (-1)^n \frac{(a)_n}{(n-m)!}x^n &= (-1)^m(a)_mx^m\sum_{n=m}^\infty (-1)^{n-m} \frac{(a-m)_{n-m}}{(n-m)!}x^{n-m}\\ &= (-1)^m (a)_m x^m \sum_{k=0}^\infty (-1)^k \frac{(a-m)_k}{k!}x^k\\ &= (-1)^m (a)_mx^m \sum_{k=0}^\infty (-1)^k\binom{a-m}{k}x^k\\ &= (-1)^m (a)_mx^m \left(1-x\right)^{a-m}. \end{align}$$
