Let $N$ be a sufficiently large natural number and let $k \in \mathbb{N}$ such that $k | N$. Suppose I have a sequence $\{ \alpha_j \}_{j=1}^N \subseteq [0,1)$, which satisfies $$ \# \{ j \in \{1, ..., N\} : \alpha_j \in [l/k, (l+1)/k) \} = N/k $$ for $l = 0, ..., k-1$. What is a bound for $$ \sum_{j=1}^N cos(2 \pi \alpha_j) ? $$
I was guessing that if $k$ is large enough, then perhaps there would be enough cancelation to make the sum small. I would appreciate any help with it. Thank you!
Let $$I_l=\left\{j\in\{1,\ldots,N\}:\alpha_j\in\left[\frac{l}{k},\frac{l+1}{k}\right]\right\},\quad\hbox{for $l=0,1,\ldots k-1$.}$$ and let $b_l=\dfrac{2l+1}{2k}$; the midpoint of the interval $\left[\frac{l}{k},\frac{l+1}{k}\right]$. Now, clearly we have $$\sum_{j\in I_l}\vert\cos(2\pi\alpha_j)-\cos(2\pi b_l)\vert \leq \#(I_l)(2\pi)\max_{j\in I_l}|\alpha_j-b_l|\leq\frac{\pi}{k}\#(I_l)=\frac{\pi N}{k^2} $$ Adding these inequalities, for $l=0,1,\ldots k-1$, we get $$ \left|\sum_{j=1}^N\cos(2\pi\alpha_j)-\frac{N}{k}\sum_{l=0}^{k-1}\cos(2\pi b_l)\right| \leq \frac{\pi N}{k} $$ Finally, it is an easy exercise to prove that $$\sum_{l=0}^{k-1}\cos(2\pi b_l)=0,$$ Therefore, $$ \left|\sum_{j=1}^N\cos(2\pi\alpha_j)\right| \leq \frac{\pi N}{k}. $$ This confirms the fact that when $k$ is big, a lot of cancellation intervene to reduce the sum.$\qquad\square$