The exponential function can be represented by the series
$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$
therefore we can represent $e^{x/2}$ by
$$e^{x/2}=\sum\limits_{n = 0}^\infty\dfrac{(\frac{x}{2})^n}{n!}$$
and it appears that the same logic holds when the indices outside of the summation are negative but represent the same value (example from Wolfram Alpha)
$$e^{x/2}=\sum\limits_{n = 0}^\infty\dfrac{(\frac{x}{2})^{n-1}}{(n-1)!}$$ $$e^{x/2}=\sum\limits_{n = 0}^\infty\dfrac{(\frac{x}{2})^{n-2}}{(n-2)!}$$ $$e^{x/2}=\sum\limits_{n = 0}^\infty\dfrac{(\frac{x}{2})^{n-3}}{(n-3)!}$$ $$\vdots$$ $$e^{x/2}=\sum\limits_{n = 0}^\infty\dfrac{(\frac{x}{2})^{n-m}}{(n-m)!}$$
where $m$ and $n$ are integers. I would assume that this is true because we can make an index substitution. Although, this seems strange to me since
$$\sum\limits_{n = 0}^\infty\dfrac{(\frac{x}{2})^{n-1}}{(n-1)!}=\sum\limits_{n = -1}^\infty\dfrac{(\frac{x}{2})^{n}}{n!}\neq e^{x/2}$$
Why can we change the indices so that
$$e^{x/2}=\sum\limits_{n = 0}^\infty\dfrac{(\frac{x}{2})^{\color{red}{n}}}{\color{red}{n}!}=\sum\limits_{n = 0}^\infty\dfrac{(\frac{x}{2})^{\color{blue}{(n-3)}}}{\color{blue}{(n-3)}!}?$$
WolframAlpha appears to treat factorials of negative integers (which are normally not defined) as being (complex) infinite, and it treats their reciprocal as zero. Therefore, the first few terms of your sum are all treated as zero, and the remainder of the sum converges to $e^{x/2}$ as expected.