Finite Complex Exponential Geometric Series with Negative Exponents So I want to know if I am doing my manipulations correctly. I have the following expression:
$$ X(\omega) = A\frac{1 - e^{-j \frac{1}{2} \omega N}}{1 - e^{j 
\frac{1}{2} \omega}} $$
Using the geometric series formula:
$$ \displaystyle\sum\limits_{n=0}^{N-1} A r^{n} = \frac{1 - r^{N}}{1 - r}$$
Can I rewrite the first expression as:
$$ X(\omega) = A\frac{1 - \big(e^{j \frac{1}{2} \omega }\big)^{-N}}{1 - e^{j 
\frac{1}{2} \omega}} $$
$$ X(\omega) = A  \displaystyle\sum\limits_{n=0}^{-(N-1)} e^{j \frac{1}{2} \omega n} $$
$$ X(\omega) = A  \displaystyle\sum\limits_{n=0}^{N-1} e^{-j \frac{1}{2} \omega n} $$
Somehow I don't think this is right, because if we reuse the geometric series expansion, the denominator will be different. But, I can't see to convince myself of it. 
Or this is manipulation correct?

Addendum
Let me try and clear your doubts about summing definition and handling.
There are fundamentally three ways to express a sum.
a) over a set
e.g.
$$
\sum\limits_{k\, \in \,\left\{ {1,2,3} \right\}} k  = 1 + 2 + 3 = 2 + 1 + 3 =  \ldots 
$$
b) under a condition on the index
e.g.
$$
\sum\limits_{1\, \le \,k\; \le \,3} k  = 1 + 2 + 3 =  \ldots \quad  \Rightarrow \quad \sum\limits_{1\, \le \,k\; \le \, - 3} k  = \sum\limits_\emptyset  k  = 0
$$
or
$$
\sum\limits_{\,k\;\backslash \,4} k  = 1 + 2 + 4
$$
etc.
c) as antidelta
that is
$$
\eqalign{
  & f(x) = \Delta _{\,x} F(x) = F(x + 1) - F(x) = \quad  \Rightarrow   \cr 
  &  \Rightarrow \quad F(x) = \Delta _{\,x} ^{\,\left( { - 1} \right)} f(x) = \sum\nolimits_x {f(x)} \quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \sum\nolimits_a^b {f(k)}  = F(b) - F(a) \cr} 
$$
So, concerning the geometric sum
when the exponent is positive, e.g. $3$, then
$$
{{1 - r^{\,3} } \over {1 - r}} = \sum\limits_{k\, \in \,\left\{ {0,1,2} \right\}} {r^{\,k} }  = \sum\limits_{0\, \le \,k\; \le \,2} {r^{\,k} }  = \sum\limits_{0\, \le \,k\; < \,3} {r^{\,k} }  = \sum\nolimits_0^3 {r^{\,k} } 
$$
the three definitions coincide.
But when it is negative
$$
\eqalign{
  & {{1 - r^{\, - 3} } \over {1 - r}}\quad  \ne \quad \sum\limits_{k\, \in \,\left\{ {0, - 1, - 2} \right\}} {r^{\,k} }  = {{1 - \left( {1/r} \right)^{\,3} } \over {1 - \left( {1/r} \right)}}  \cr 
  & \quad \quad \quad \;\; \ne \quad \sum\limits_{0\, \le \,k\; \le \, - 2} {r^{\,k} }  = 0 \cr} 
$$
while the antidelta gives
$$
\eqalign{
  & {{1 - r^{\, - 3} } \over {1 - r}} = \quad \sum\nolimits_0^{ - 3} {r^{\,k} }  =  - {{r^{\, - 3} } \over {1 - r}} + {{r^{\,0} } \over {1 - r}} =   \cr 
  &  =  - \sum\nolimits_{ - 3}^0 {r^{\,k} }  = \sum\limits_{ - 3\, \le \,k\; < \,0} {r^{\,k} }  = \sum\limits_{ - 3\, \le \,k\; \le \, - 1} {r^{\,k} }  =   \cr 
  &  = \sum\limits_{1\, \le \,k\; \le \,3} {\left( {1/r} \right)^{\,k} }  = \sum\limits_{0\, \le \,k\; \le \,3} {\left( {1/r} \right)^{\,k} }  - 1 = {{1 - \left( {1/r} \right)^{\,4} } \over {1 - \left( {1/r} \right)}} - 1 =   \cr 
  &  = {{r\left( {r^{\,4}  - 1} \right)} \over {r^{\,4} \left( {r - 1} \right)}} - 1 = {{r^{\,3}  - 1} \over {r^{\,3} \left( {r - 1} \right)}} \cr} 
$$
