I want to represent the function:
\begin{equation} f(x)=e^{-a(x-b)^{2}} \end{equation}
where, $0<a<1$, $x\in\mathbb{R}$, and $b\in\mathbb{R}$.
As a power series for an integral I am working on. First, I re-wrote $f(x)$ using the power series definition of the exponential to get,
\begin{equation} f(x) = \sum_{n=0}^{\infty} \frac{(-a)^{n}}{n!}(x-b)^{2n} \end{equation}
Using binomial expansion on the $(x-b)^{2n}$ yields
\begin{equation} f(x) = \sum_{n=0}^{\infty} \frac{(-a)^{n}}{n!}\sum_{k=0}^{2n}\binom{2n}{k}(-b)^{2n-k}\,x^{k} \end{equation}
I want to then simplify this expression by switching the order of summation. Here is what I did:
\begin{equation} \begin{aligned} f(x) &= \sum_{k=0}^{\infty}x^{k} \sum_{n=k/2}^{\infty}\binom{2n}{k}\frac{(-a)^{n}}{n!}(-b)^{2n-k}\\ &=\sum_{k=0}^{\infty}\frac{(-\sqrt{-a}x)^{k}}{\frac{k}{2}!}\,{_{1}}F_{1}\left[\frac{k+1}{2},\frac{1}{2},-ab^{2}\right] \end{aligned} \end{equation}
where, ${_{1}}F_{1}(a,b,z)$ is the confluent hypergeometric function of the 1st kind.
At this point the form of $f(x)$ makes the solution to my integral much simpler. However, I am not sure if switching the order of summation like this is allowed under these circumstances. Looks like the sum has an imaginary component that one would not expect. Any thoughts?
The rearrangement appears to be okay. Assuming that $b$ is positive, if we start by switching variables to $z=-x$, then we're looking at $$ f(z) = e^{-a(b+z)^2} $$ which is entire, so the series expansion $$ f(z) = \sum_{n=0}^\infty \frac{(-a)^n}{n!}(b+z)^{2n} $$ converges absolutely for all $z$. In particular for positive $z$, binomial expansion on $(b+z)^{2n}$ will simply break each term of the absolutely convergent series into finitely many terms that all have the same sign, so the broken-up sequence is still absolutely convergent and can be rearranged freely.
When, by rearranging, you get everything down to a power series in $z$, you know that this power series will converge correctly for every positive $z$, which means that it converges (absolutely) for every $z$ where $|z|$ is smaller than a positive number -- that is, for every $z$. And by the identity theorem, then, it converges towards the right thing everywhere in the complex plane.
Finally, changing the sign of some of the coefficients to get a series in $x$ instead of $z$ is of course not going to change the convergence properties.
I lack the prerequisites to verify your final ${}_1F_1$ form, though.