And if so, what is the radius of convergence of $x$?
I am inclined to think it converges absolutely for all $x$ but I can't prove it.
I have tried using an adaptation of the ratio test:
$\rho=\lim\frac{\frac{a^{n+2}}{\frac{n+2}{2}!}x^{n+2}}{\frac{a^n}{\frac{n}{2}!}x^n}=\lim\frac{a^{n+2}x^{n+2}}{\frac{n+2}{2}!}.\frac{\frac{n}{2}!}{a^nx^n}=\lim\frac{a^2x^2}{\frac{n+2}{2}}=\lim\frac{2a^2x^2}{n+2}$
and then this tends to $0$ and so, converges. But I'm sure that doesn't work. Any ideas?
The series converges absolutely for any real (or complex) $x$. One way to see this is to separate the series into even and odd terms. The even terms give $\sum_{k=0}^\infty (ax)^{2k}/k!$, which converges absolutely for any $x$ by the ratio test. The ratio test will also work for the odd terms, no matter what reasonable definition of $\frac{2k+1}2!$ you use. And the sum of two absolutely convergent series is again absolutely convergent, no matter what order you combine the terms in.