I have to show that $ \forall N \in \mathbb{N} $ and $ \forall x \in \mathbb{R} \setminus\mathbb{Z} $ :
$$ \frac{1}{N+1}\sum_{n=0}^{N}{\sum_{k=-n}^{n}{e^{i 2 \pi k x}}} = \frac{1}{N+1} \left( \frac{\sin{((N+1)\pi x)}}{\sin \pi x} \right)^2$$
I found with geometric series that : $$ \sum_{k=-n}^{n}{e^{i 2 \pi k x}} = e^{-i \pi x n}\frac{\sin{((n+1)\pi x)}}{\sin \pi x} $$ But after calculating the next sum : $$ \sum_{n=0}^{N}{e^{-i \pi x n}\frac{\sin{((n+1)\pi x)}}{\sin \pi x}} $$ seem not feasible. Can someone give a clue if it is the right way to prove this equality.
Each geometric sum can be written as $$ \sum_{k=-n}^{n}{e^{i 2 \pi k x}} = \frac{e^{i(n+1)2\pi x}-e^{-i 2n\pi x}}{e^{i2\pi x}-1}.$$ Now, if I consider the summation from $n=0$ to $N$, this gives me: $$ \frac{\sum \limits_{i=0}^{N} e^{i(n+1)2\pi x}-\sum \limits_{i=0}^{N} e^{-i 2n\pi x}}{e^{i2\pi x}-1}.$$ Now, each of these summations are again geometric series and can be solved.