So let
$$S_n=\sum_{a=1}^{n} \left(\frac{1}{2}\right)^a$$
then could rewrite it to
$$S_n=\frac{1}{2}\cdot\frac{1-\left(\frac{1}{2}\right)^n}{1-\frac{1}{2}}=1-\left(\frac{1}{2}\right)^n$$
But the new form $S_n=1-\left(\frac{1}{2}\right)^n$ could now also work even if $n$ isn't a whole number. If you plot $y=f(x)=1-\left(\frac{1}{2}\right)^n$ you'll get a continuous function. So now what does it mean to have non-whole $n$, as you can't sum to a non-whole term of a series.
In thoses cases in math we want to reach an explicit form and then just change the nature of the number in question examples :
Factorial
$$ n \in \mathbb{N}, \ n!\triangleq \prod_{k=1}^nk$$
But here product is cannot be transform into continous form directly so let's write it like follow :
$$n\in\mathbb{N}, \ n! \triangleq \int_0^\infty t^ne^tdt$$
In those terms we can define factorial for real and even for complex number : this is the so called Gamma function defined as follow :
$$ \forall z \in\mathbb{C}, \Re(z)>0,\\ z!\triangleq\Gamma(z+1)\triangleq\int_0^\infty t^{z-1}e^t$$
Note that this notation is rather informal.
Geometric sum
We can saw thing in two ways :
The first one is to define for $a$ complex number according that the integral can be complex.
$$ x \in \mathbb{R}, S_x(a)\triangleq \int_0^x(a)^tdt $$
The second is to use geometric sum formula and decide it is valid for real or even for complex number $a \neq 1$
-For a sum beginning to $0$
$$ x \in \mathbb{R}, S_x(a)=\dfrac{1-a^{x+1}}{1-a}$$
-For a sum beginning to $1$
$$ x \in \mathbb{R}, S_x(a)=a\dfrac{1-a^{x}}{1-a}$$
Note that you can do this with any object valid for whole numbers that you want to extend. But in thoses cases you have always to find something convenient (perhaps restrictive to manage to generalize)
Also see Fractional derivative.