Lets say I want to find the series representation for $f(x)$ which is defined as the product of an exponential function with a confluent hypergeometric function,
\begin{equation} f(x) = e^{-x}\, {_{1}}F_{1}(-a,-b,x) \end{equation}
where, $a > 0$, and $b> 0$.
Since both terms of the products have power series repsentations, I could simply take the Cauchy product such that,
\begin{equation} \begin{aligned} f(x) &= \left(\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}x^{n}\right)\, \left(\sum_{n=0}^{\infty}\frac{(-a)_{n}}{(-b)_{n}\,n!}x^{n}\right)\\ &=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}\frac{(-1)^{k}}{k!}\frac{(-a)_{n-k}}{(-b)_{n-k}(n-k)!}\right) \, x^{n} \end{aligned} \end{equation}
However, when the first arguement of the confluent hypergeometric function is a negative integer, the series is truncated to $a+1$ terms, i.e.
\begin{equation} {_{1}}F_{1}(-a,-b,x) = \sum_{n=0}^{a}\frac{(-a)_{n}}{(-b)_{n}\,n!}x^{n} \quad \text{when} \ a \in \mathbb{Z}^{+} \end{equation}
Gven the situation at hand (i.e. $-a\in\mathbb{Z}^{-}$) is the product above still valid? Can I just treat the hypergeometric function as infinite series knowing that all terms for $n=a \to\infty$ are zero?
Maybe a better way of framing this question is this: Is there a general solution for this product without prior knowledge of what value $-a$ is?