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On page 25, in the book Quantum Field Theory for the Gifted Amateur by Tom Lancaster and Stephen. J Blundell, it states the following:

We impose periodic boundary conditions forcing $e^{ikja}=e^{ik(j+N)a}$. The wave vector therefore takes the values $2\pi m/Na$, where $m$ is an integer in the range $-N/2<m\le N/2$. Note that \begin{equation} \sum_je^{ikja}=N\delta_{k,0}. \end{equation}

Clearly this is true for $k=0$, but I can't see how this holds true for $k\ne 0$.

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    $\begingroup$ Note that this identity is purely mathematical once one has put in the "right" momenta. It is widely used in all of physics, including plain quantum theory or classical electrodynamics. $\endgroup$
    – Jonas Greitemann
    Commented May 24, 2015 at 8:47
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    $\begingroup$ This question (v2) seems like an archetype of a math problem encountered in many areas of physics, e.g. crystals, and which the community consistently wants to not migrate to Math.SE, cf. this meta post. $\endgroup$
    – Qmechanic ♦
    Commented May 25, 2015 at 13:41

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I guess you can use the formula for the sum of a geometric progression for $k$ non equal to 0.

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    $\begingroup$ So: For $k\ne 0$ \begin{equation} \sum_{j=1}^N(e^{ika})^j=\frac{(1-(e^{ika})^N)e^{ika}}{1-e^{ika}}=\frac{e^{ika}-e^{ik(1+N)a}}{1-e^{ika}}=0 \end{equation} where the final equality follows from the imposed boundary conditions forcing $e^{ikja}=e^{ik(j+N)a}$ and thus $e^{ika}-e^{ik(1+N)a}=0$ where $j=1$. $\endgroup$
    – Joseph Dewdney
    Commented May 25, 2015 at 8:23

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