# Dirichlet Kernel : Trigonometric Identity Question

I'm reading the proof for the trigonometric identity of the Dirichlet Kernel. There is just one step that I don't understand fully and I would appreciate it if someone could explain it for me.

We know that the sum of a limited geometric series is

$$\displaystyle \sum _{n=0} ^N ar^n = a \frac{1 - r^{n+1}}{1-r}$$

In particular we know that

$$\displaystyle \sum _{n = -N} ^N r^n = r^{-N} \frac{1-r^{2N+1}}{1-r}$$

The rest of the proof is pretty straightforward but it has been a while since I have dealt with series and I don't really understand how the second equation is obtained.

Thanks!

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## 1 Answer

\begin{align} \sum_{n=-N}^N r^n & = r^{-N} + r^{-N+1} + r^{-N+2} + \cdots + r^{-1} + 1 + r + \cdots + r^{N-1} + r^N\\ & = r^{-N}\left(1 + r + r^{2} + \cdots + r^{N-1} + r^N + r^{N+1} + \cdots + r^{2N-1} + r^{2N} \right)\\ & = r^{-N} \left(\sum_{n=0}^{2N} r^n \right)\\ & = r^{-N} \left(\dfrac{1-r^{2N+1}}{1-r} \right) \end{align}

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Haha. Thank you! – docjay Nov 19 '12 at 21:14