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Is there a way to perform the finite sum $\sum_{m = 1}^n \exp(2 \pi i k (\sqrt5) ^m)$?, m even.

I am trying to show a specific sequence is not equidistributed, and so I'd like to show that Weyl's criterion fails, but I am not sure how to perform this sum, since it is not a geometric series.

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    Maybe you could just tell us what sequence is the one whose distribution you are investigating?2012-03-26

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As written, the sum is $\left\lfloor\frac{n}{2}\right\rfloor$ since $\exp(2\pi ik5^{m/2})=1$ for $m$ even.

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    Oh. Duh. ${}{}$2012-03-26
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An elaboration on what has been stated: $ \begin{align} \sum_{m=2, m \text{ even}}^{n}\exp\left(2\pi i k \left(\sqrt{5}\right)^m\right)&=\sum_{m=2, m \text{ even}}^{n}e^{2\pi ik \cdot 5^{\frac{m}{2}}}\\ &=\sum_{m=2,m \text{ even}}^{n}\left(e^{2\pi i}\right)^{k\cdot 5^{\frac{m}{2}}}\\ e^{2\pi i}&=1\\ \therefore \sum_{m=2,m \text{ even}}^{n}\left(e^{2\pi i}\right)^{k\cdot 5^{\frac{m}{2}}}&=\sum_{m=2, m \text{ even}}^{n}1^{k\cdot 5^{\frac{m}{2}}}\\ &=\underbrace{1+1+1+\dots}_{\frac{n}{2} \text{ ones}}\\ &=\left \{ \begin{array}{cl} \frac{n}{2}&n \text{ even }\\ \left\lfloor\frac{n}{2}\right\rfloor &n \text{ odd} \end{array} \right. \end{align} $