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I want to show that $\sum_{k=-N}^{N}e^{ikx}=\frac{\sin((N+\frac{1}{2})x)}{\sin(\frac{x}{2})}$ for $N\in \mathbb{N}$

Any tips on how to proceed?

I tried doing it in two ways:

First using the sum of polynomials formula:

($1+r+r^2+...+r^n=\frac{1-r^{n+1}}{1-r}$) while letting $r=e^{ix}$

but the calculation got too tedious.

Second by replacing $e^{ix}$ with Euler's formula directly, I get:

$\sum_{k=-N}^{N}e^{ikx}=1+2\sum_{k=1}^{N}\cos(kx)$

which isn't anywhere close the required result.

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  • $\begingroup$ Possible Duplicate of This. $\endgroup$
    – Mark Viola
    Commented Mar 16, 2017 at 2:31

2 Answers 2

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You gave up too soon on the geometric series. With $r = e^{ix}$,

$$\eqalign{ r^{-N} + r^{-N+1} + \ldots + r^N &= \frac{r^{N+1} - r^{-N}}{r-1}\cr &= \frac{r^{N+1/2} - r^{-N-1/2}}{r^{1/2} - r^{-1/2}}\cr &= \frac{e^{i(N+1/2)x} - e^{-i(N+1/2)x}}{e^{ix/2} - e^{-ix/2}}\cr &= \frac{\sin((N+1/2)x}{\sin(x/2)}}$$

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  • $\begingroup$ Thanks, had to multiply by $r^{-1/2}$, didn't see that one. $\endgroup$
    – Omrane
    Commented Mar 16, 2017 at 11:29
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Creative telescoping is (almost) always the answer. You may notice that

$$\left(\sin\frac{x}{2}\right)\cdot\left(1+2\sum_{k=1}^{N}\cos(kx)\right) = \sin\frac{x}{2}+\sum_{k=1}^{N}\left[\sin\left(\left(k+\frac{1}{2}\right)x\right)-\sin\left(\left(k-\frac{1}{2}\right)x\right)\right]$$ and easily draw your conclusions.

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