Proving central limit theorem in a specific case( fourier analysis course)

Question

I am stuck with the following problem: Show that for any given $R > 0$ $$\lim_{N \to \infty}\sup_{\left|\xi \right| \le R} \left |\left(\frac{1}{6}\sum_{k=1}^{6}e^{2\pi i\xi \frac{k-\frac{7}{2} }{\sqrt{N}} }\right)^N -e^{\frac{-35\pi^2\xi^2}{6}}\right|=0.$$I have tried to taylor expand $ \eta \mapsto e^{i\eta} $ and then noticed that all of the terms that had the imaginary part raised to an odd power cancelled(as the terms $k-\frac{7}{2} $ when summing from 1 to 6 has a negative counterpart) but i did not end up with anything good in the end. Any ideas? Thanks in advance.

Answer 1

I will show what a possible start could be and let you fill in the technical details. Using $z + z^* = 2 \mathrm{Re}(z)$ we have $$ \left(\frac{1}{6} \sum_{k = 1}^{6}{e^{2 \pi i \xi \frac{k - \frac{7}{2}}{\sqrt{N}}}} \right)^N = \left(\frac{1}{3} \sum_{k \in \{1, 3, 5\}} \frac{e^{\pi i \xi \frac{k}{\sqrt{N}}} + e^{-\pi i \xi \frac{k}{\sqrt{N}}}}{2} \right)^N = \left(\frac{1}{3} \sum_{k \in \{1,3,5\} } \cos\left(\pi \xi \frac{k}{\sqrt{N}}\right) \right)^N. $$ Remembering the (first two summands of the) power series representation of cosine, we get $$ \cos\left(\pi \xi \frac{k}{\sqrt{N}}\right) \approx 1 - \frac{k^2}{2}\frac{\pi^2 \xi^2 }{N}. $$ Therefore $$ \left(\frac{1}{6} \sum_{k = 1}^{6}{e^{2 \pi i \xi \frac{k - \frac{7}{2}}{\sqrt{N}}}} \right)^N \approx \left(1 - \frac{1}{3}\left(\frac{1^2}{2} + \frac{3^2}{2} + \frac{5^2}{2} \right)\frac{\pi^2 \xi^2 }{N} \right)^N = \left(1 - \frac{35 \pi^2 \xi^2 }{6 N}\right)^N \longrightarrow e^{-\frac{35 \pi^2 \xi^2}{6}}. $$