The sum of Dirichlet's kernel

Question

Assume $f \in L^1(\mathbb{T})$
I define: $$S_N = \sum \limits_{k = -N}^{k=N} c_ke^{2 \pi ikx},$$ where $c_k = \int \limits_{0}^{1} f(x)e^{2 \pi ikx} \mbox{d}x$.
I managed to show that: $$S_N = \int \limits_{0}^{1} f(x-t)D_N(t) \mbox{d}t.$$ Where $D_N(t) = \sum \limits_{k=N}^Ne^{2 \pi ikx}$ (Dirichlet's kernel).
Now I would like to show that: $$D_N(t) = \frac{\sin \big(\pi(2N + 1)t \big)}{\sin(\pi t)}$$ I did show that: $$D_n(t) = 1 + 2\sum \limits_{k = 1}^N \cos(2 \pi kt)$$ I don't know where to go from here however. I would appreciate any hints or tips.

Answer 1

$$D_N(t) = \sum_{k=-N}^Ne^{2\pi i k t} =e^{-2\pi i N t}\sum_{k=0}^{2N}e^{2\pi i k t}=e^{-2\pi i N t}\frac{e^{2\pi i (2N+1) t}-1}{e^{2\pi i t}-1}=\frac{\sin \big(\pi(2N + 1)t \big)}{\sin(\pi t)}.$$

Answer 2

Hint: For $x\in\mathbb{Z}$ it is obvious that $D_N(x)=2N+1$. For $x\not\in\mathbb{Z}$: Let $z=e^{2\pi ix}$ Factorize with $z^{-N}$. It is $D_N(x)=z^{-N}\cdot\displaystyle{\sum_{k=0}^{2N}z^k=z^{-N}\cdot\frac{z^{2N+1}-1}{z-1}}$. Can you pick it up from this point?