<s>[INST] 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?
 [/INST] 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
$$
\eqalign{
  & {{1 - r^{\, - 3} } \over {1 - r}} = \quad \sum\nolimits_0^{ - 3} {r^{\,k} }  =  - {{r^{\, - 3} } \over {1 - r}} + {{r^{\,0} } \over {1 - r}} =   \cr 
  &  =  - \sum\nolimits_{ - 3}^0 {r^{\,k} }  = \sum\limits_{ - 3\, \le \,k\; < \,0} {r^{\,k} }  = \sum\limits_{ - 3\, \le \,k\; \le \, - 1} {r^{\,k} }  =   \cr 
  &  = \sum\limits_{1\, \le \,k\; \le \,3} {\left( {1/r} \right)^{\,k} }  = \sum\limits_{0\, \le \,k\; \le \,3} {\left( {1/r} \right)^{\,k} }  - 1 = {{1 - \left( {1/r} \right)^{\,4} } \over {1 - \left( {1/r} \right)}} - 1 =   \cr 
  &  = {{r\left( {r^{\,4}  - 1} \right)} \over {r^{\,4} \left( {r - 1} \right)}} - 1 = {{r^{\,3}  - 1} \over {r^{\,3} \left( {r - 1} \right)}} \cr} 
$$
</s>