Using this in the context of Fourier transforms. This should probably be an easy derivation for you guys, but I forget how to derive it.
$$ \sum_{n=0}^{L-1}e^{-j\frac{2\pi k}{L}n} = \frac{1-e^{-j2\pi k}}{1-e^{\frac{-j2\pi k}{L}}} $$
which somehow can also be reduced to $$L\delta [k] $$
When $k=0$, the summation is simply $L$.
For $k \neq 0$, write the numerator as: $e^{-j \pi k} \left( e^{j \pi k} - e^{-j \pi k} \right)$ = $e^{-j \pi k} \times 2j \sin(k \pi)$, which is $0$ for $k \neq 0$, because $\sin(k \pi)=0$.
Thus, you have a summation which is $L$ for $k=0$ and $0$ for $k \neq 0$. This can be expressed as $L \delta[k]$.
Also, the denominator should read $1- e^{-j \frac{2 \pi k}{L}}$.