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?
 A: The simpler way first
$$
\eqalign{
  & {{X(\omega )} \over A} = {{1 - e^{\, - jw/2N} } \over {1 - e^{\,jw/2} }} = {{1 - e^{\, - jw/2N} } \over {e^{\,jw/2} \left( {e^{\, - jw/2}  - 1} \right)}} =   \cr 
  &  =  - e^{\, - jw/2} {{1 - e^{ - jw/2N} } \over {\left( {1 - e^{\, - jw/2} } \right)}} =  - e^{\, - jw/2} \sum\limits_{0\, \le \,k\, \le \,N - 1} {e^{\, - jw/2\,k} }  =   \cr 
  &  =  - \sum\limits_{1\, \le \,k\, \le \,N} {e^{\, - jw/2\,k} }  \cr} 
$$
Following instead your way, to take the geometric series from $0$ to $-N$, you shall write
$$
\eqalign{
  & {{X(\omega )} \over A} = {{1 - e^{\, - jw/2N} } \over {1 - e^{\,jw/2} }} = {{1 - \left( {e^{\,jw/2} } \right)^{\, - N} } \over {1 - e^{\,jw/2} }} =   \cr 
  &  = \sum\nolimits_{k = 0}^{ - N} {e^{\,jw/2\;k} }  =  - \sum\nolimits_{k =  - N}^0 {e^{\,jw/2\;k} }  =  - \sum\limits_{ - N\, \le \,k\, \le \, - 1} {e^{\,jw/2\,k} }  =   \cr 
  &  =  - \sum\limits_{ - N\, \le \,k\, \le \, - 1} {e^{\,jw/2\,k} }  =  - \sum\limits_{1\, \le \, - k\, \le \,N} {e^{\,jw/2\,k} }  =  - \sum\limits_{1\, \le \,k\, \le \,N} {e^{\, - \,jw/2\,k} }  \cr} 
$$
where
$$
\sum\nolimits_{k = 0}^X {f(k)} 
$$
indicates the "Indefinite" Sum
computed within the indicated bounds. In summary we have
$$
\eqalign{
  & f(k) = \Delta _{\,k} \,F(k) = F(k + 1) - F(k)\quad  \Leftrightarrow   \cr 
  &  \Leftrightarrow \quad F(k) = \Delta _{\,k} ^{\left( { - 1} \right)} \,f(k) = \sum\nolimits_k {f(k)} \quad  \Leftrightarrow   \cr 
  &  \Leftrightarrow \quad \sum\nolimits_{k = 0}^X {f(k)}  = F(X) - F(0)\quad  \Leftrightarrow   \cr 
  &  \Leftrightarrow \quad \sum\nolimits_{k = 0}^X {f(k)} \quad \left| {\,X \in Z} \right. = \left\{ {\matrix{
   { - \sum\limits_{X\, \le \,k\, \le \, - 1} {f(k)} } & {X < 0}  \cr 
   {\sum\limits_{0\, \le \,k\, \le \,X - 1} {f(k)} } & {0 \le X}  \cr 
 } } \right. \cr} 
$$
---  in reply to your comment  ----
Among the basic properties of the Indefinite Sum, also called Antidelta, we have
$$
\left\{ \matrix{
  \sum\nolimits_a^a {f(k)}  = 0 \hfill \cr 
  \sum\nolimits_a^b {f(k)}  + \sum\nolimits_b^c {f(k)}  = \sum\nolimits_a^c {f(k)}  \hfill \cr}  \right.
$$
which implies
$$
\eqalign{
  & \sum\nolimits_a^b {f(k)}  + \sum\nolimits_b^a {f(k)}  = \sum\nolimits_a^a {f(k)}  = 0  \cr 
  & \sum\nolimits_b^a {f(k)}  =  - \sum\nolimits_a^b {f(k)}  \cr} 
$$
The Indefinite Sum is the discrete analog of the Integral and mimicks many (not all) of its properties.
Take, in our case
$$
F(x) =  - {{r^{\,x} } \over {1 - r}} =  - r^{\,x} \sum\limits_{0\, \le \,k} {r^{\,k} }  =  - \sum\limits_{0\, \le \,k} {r^{\,k + x} } 
$$
then
$$
f(x) = \Delta _{\,x} F(x) = F(x + 1) - F(x) = {{r^{\,x}  - r^{\,x + 1} } \over {1 - r}} = r^{\,x} 
$$
and
$$
\sum\nolimits_a^b {f(k)}  = \sum\nolimits_a^b {r^{\,k} }  =  - {{r^{\,b} } \over {1 - r}} + {{r^{\,a} } \over {1 - r}} = {{r^{\,a}  - r^{\,b} } \over {1 - r}}
$$
which allows to define the sum even for real bounds $a$ and $b$.
In particular, for a non-negative integer $N$
$$
\eqalign{
  & \sum\nolimits_0^{ - N} {r^{\,k} }  = {{1 - r^{\, - N} } \over {1 - r}} = \left( {1 - r^{\, - N} } \right)\sum\limits_{0\, \le \,k} {r^{\,k} }  
 = \sum\limits_{0\, \le \,k} {r^{\,k} }  - \sum\limits_{0\, \le \,k} {r^{\,k - N} }  =   \cr 
  &  = \sum\limits_{0\, \le \,k} {r^{\,k} }  - \sum\limits_{ - N\, \le \,k - N} {r^{\,k - N} } 
 =  - \sum\limits_{ - N\, \le \,j\; \le \, - 1} {r^{\,j} }  =  - \sum\nolimits_{ - N}^0 {r^{\,k} }  =   \cr 
  &  =  - \sum\limits_{1\, \le \,j\; \le \,N} {\left( {r^{\, - 1} } \right)^{\,j} }
  =  - {{1 - \left( {r^{\, - 1} } \right)^{\,N} } \over {1 - 1/r}} + 1 = {{ - r\left( {1 - r^{\, - N} } \right) + r - 1} \over {r - 1}} =   \cr 
  &  = {{1 - r^{\, - N} } \over {1 - r}} \cr} 
$$
that is to say 
$$
{{1 - r^{\, - N} } \over {1 - r}} =  - \sum\limits_{1\, \le \,j\; \le \,N} {\left( {r^{\, - 1} } \right)^{\,j} }  =  - \sum\limits_{1\, \le \,j\; \le \,N} {\left( {1/r} \right)^{\,j} }  = 1 - \sum\limits_{0\, \le \,j\; \le \,N} {\left( {1/r} \right)^{\,j} } 
$$
A: The manipulation is not correct in general.

Starting with OP's last line and applying the finite geometric series formula we obtain
  \begin{align*}
\color{blue}{A\sum_{n=0}^{N-1}e^{-j\frac{1}{2}\omega n}}&=A\sum_{n=0}^{N-1}e^{\left(-\frac{j\omega}{2}\right)^n}\\
&=A\frac{1-e^{\left(-\frac{j\omega}{2}\right)^N}}{1-e^{-\frac{j\omega}{2}}}\\
&=A\frac{1-e^{-\frac{j\omega N}{2}}}{1-e^{-\frac{j\omega}{2}}}\\
&=-\frac{A}{e^{-\frac{j\omega}{2}}}\cdot\frac{1-e^{-\frac{j\omega N}{2}}}{1-e^\frac{j\omega}{2}}\\
&=\color{blue}{-e^{\frac{j\omega}{2}}A\frac{1-e^{-\frac{j\omega N}{2}}}{1-e^{\frac{j\omega}{2}}}}\tag{1}
\end{align*}
  we observe that (1) coincides with OP's expression  $A\frac{1-e^{-\frac{j\omega N}{2}}}{1-e^{\frac{j\omega}{2}}}$ in the first line iff $-e^{\frac{j\omega}{2}}=1$.

A note to OP's line
\begin{align*}
X(\omega) = A  \displaystyle\sum\limits_{n=0}^{-(N-1)} e^{j \frac{1}{2} \omega n} 
\end{align*}
The usual meaning of $\sum_{n=a}^b f(n)=\sum_{a\leq  n \leq b}f(n)=0$ if $b<a$. So, you probably want to write $\sum_{n=-(N-1)}^0e^{j \frac{1}{2} \omega n}$. But this does not match the standard form of the finite geometric sum formula.
A: 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
$$
\eqalign{
  & {{1 - r^{\, - 3} } \over {1 - r}} = \quad \sum\nolimits_0^{ - 3} {r^{\,k} }  =  - {{r^{\, - 3} } \over {1 - r}} + {{r^{\,0} } \over {1 - r}} =   \cr 
  &  =  - \sum\nolimits_{ - 3}^0 {r^{\,k} }  = \sum\limits_{ - 3\, \le \,k\; < \,0} {r^{\,k} }  = \sum\limits_{ - 3\, \le \,k\; \le \, - 1} {r^{\,k} }  =   \cr 
  &  = \sum\limits_{1\, \le \,k\; \le \,3} {\left( {1/r} \right)^{\,k} }  = \sum\limits_{0\, \le \,k\; \le \,3} {\left( {1/r} \right)^{\,k} }  - 1 = {{1 - \left( {1/r} \right)^{\,4} } \over {1 - \left( {1/r} \right)}} - 1 =   \cr 
  &  = {{r\left( {r^{\,4}  - 1} \right)} \over {r^{\,4} \left( {r - 1} \right)}} - 1 = {{r^{\,3}  - 1} \over {r^{\,3} \left( {r - 1} \right)}} \cr} 
$$
