Discrete Fourier Transform of generalised Hamming Window The generalised Hamming Window is defined as:
$$  w(n) =
\begin{cases}
\alpha - (1 - \alpha)\cos(2 \pi n /N),  & \text{if $ 0 \leq n \leq N$} \\
0, & \text{otherwise}
\end{cases} $$
with $ 0 \leq \alpha \leq 1$.
By using the formula for DFT: $X[k] = \sum_{k = 0}^{N-1} x[n] e^{-2 \pi i n k / N}$ I tried to obtain the DFT of $w(n)$. The problem is that if I plug $w(n)$ into the DFT, rearrange and sum the geometric series, I get the result:
$$  X[k] =
\begin{cases}
\alpha N,  & \text{if $ k = 0 $} \\
0, & \text{otherwise}
\end{cases} + \frac{\alpha-1}{2} \left[ \frac{1 - e^{-2 \pi i (k-1)}}{1 - e^{-2 \pi i (k-1) / N}} + \frac{1 - e^{-2 \pi i (k+1)}}{1 - e^{-2 \pi i (k+1) / N}} \right] $$
However, I keep getting the result
$$  X[k] =
\begin{cases}
\alpha N ,  & \text{if $ k = 0 $} \\
0, & \text{otherwise}
\end{cases} $$
as the bracket on the right has both numerators equal to zero for all $k$s.
Question: This is definitely not the right result. What is the correct result? Optional: Where did I make a mistake? Thanks!
 A: Taking the DFT of the generalised Hamming window, leads to
$$\begin{align}X[k]&=\sum_{n=0}^{N-1}\left(\alpha-(1-\alpha)\cos\left(\frac{2\pi n }{N}\right)\right)e^{-2\pi ink/N}
\\&=\alpha\sum_{n=0}^{N-1} e^{-2\pi ink/N}+\frac{\alpha-1}{2}\sum_{n=0}^{N-1}\left( e^{-2\pi in(k-1)/N}+e^{-2\pi in(k+1)/N}\right)\end{align}$$
Now, similar to what you have done using the sum of $N$ terms of a geometric progression,  for non-zero $m$ we have
$$\sum_{n=0}^{N-1} e^{-2\pi in m/N}=0$$
However if $m=0$, the sum becomes
$$\sum_{n=0}^{N-1} e^{-2\pi in (0)/N}=N$$
so that for integer $m\geq 0$ we can use the Dirac Delta function $\delta(m)$, where $\delta(m)=1$ for $m=0$, but $\delta(m)=0$ for $m>0$.
$$\sum_{n=0}^{N-1} e^{-2\pi inm/N}=N\delta(m)$$
Thus the DFT of the Hamming window is
$$X[k]=\alpha N\delta(k)+\frac{(\alpha-1)N}{2}(\delta(k-1)+\delta(k+1))$$
where there will be only three values of $k$ for which $X[k]$ will be non-zero:-
$$\begin{align}X[0]&=\alpha N \\ X[1]&=\frac{(\alpha-1)N}{2}\\ X[-1]&=X(N-1)=\frac{(\alpha-1)N}{2}\end{align}$$
Note that the modulo $2\pi$ nature of the DFT leads to the value of $k=-1$ being equivalent to $k = N-1$, as we restrict the range of integer $k$ to be from $0$ to $N-1$.
In terms of the mistake, your first sum of the geometric series is not defined for when $(k-1)=0$, as the denominator will be $0$ in addition to the numerator being $0$. In a similar fashion, for the second sum of the geometric series, the denominator will be $0$ when $(k+1)=0$, as well as the numerator. For all other values of $k$, only the numerator will be $0$.    
A: These geometric series are known as Dirichlet sincs or digital sincs:
$$
\begin{align}
\sum_{n=0}^{N-1} e^{-j2\pi n(k-1)/N} = & \,\,\frac {1 - e^{-j2\pi (k-1)}} {1 - e^{-j2\pi(k-1)/N}} \\
= & \,\,\frac { e^{-j\pi(k-1)} \left( e^{j\pi(k-1)} - e^{-j\pi(k-1)} \right) } {e^{-j\pi(k-1)/N} \left( e^{j\pi(k-1)/N} - e^{-j\pi(k-1)/N} \right)} \\
= & \,\,e^{-j\pi (k-1)(N-1)/N} \frac { \sin( \pi(k-1) ) } { \sin (\pi(k-1)/N ) }
\end{align}
$$
