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I would like to compute the Fourier series given as

$f(2\pi k/n)= \sum\limits_{\substack{k=-N}}^{N}{t_{k} e^{(i2 \pi k^2/n)}} $

I have in my work that $t_{k}=t_{-k}$, then we are in a symmetric case, then the series can be expressed as :

$f(2\pi k/n)= 1/2+ 2 \sum\limits_{\substack{k=1}}^{N}{t_{k} \cos(2 \pi k^2/n)}$.

Now the question is what could be a possible approach to find the sum of this series $\sum\limits_{\substack{k=1}}^{N}{\cos(2 \pi k^2/n)} $?

I don't take into account $t_{k}$ because they have all the same value in my work.

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    [This result](http://www.wolframalpha.com/input/?i=%5Csum_%7Bk%3D1%7D%5E%7Bn%7Dcos(2+*+pi+*+k%5E2+%2Fn)) shows no sum from the series. – 2017-02-20
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    @Adola Yes there is no closed form for this Sum, but perhaps we can find a good estimate of the sum. – 2017-02-20
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    Talking about an estimate, is there any relation between $n$ and $N$ e.g. one much larger than the other... ? – 2017-02-20
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    @xaxa Yes n is larger than N, there is a relation between them, when n change, the big N change also. – 2017-02-20
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    If not only $n \gg N$ but $n^2 \gg 2\pi^2 N^4/5$ you can expand $\cos \phi \approx 1-\phi^2/2$ and use the formula for the sum of $k^4$. If also $N \gg 1$ I got $\sum_{k=1}^N \cos(2\pi k^2/n) \approx N (1 - 2\pi^2 N^4/5n^2)$ – 2017-02-21

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