I want to show that there is some $\gamma(n)=o(n^{-1})$ and some $C(n) \to -\infty$ such that for $\gamma \leq \theta \leq \pi$ we have


$\sum\_{k=1}^n -1+\cos(k\theta) \leq C(n)$.


If we rewrite this using the Dirichlet kernel, what I want is that: if $\gamma \leq \theta \leq \pi$ then


$-n+\frac{D\_n(\theta)}{2}-\frac{1}{2} \leq C(n)$.


If I didn't demand that $\gamma=o(n^{-1})$ then I could do this with $\gamma=n^{-3/4}$ say, and $C(n)$ could then just be $-n$.


I've plotted this a bunch and it looks true, and I've spent the morning trying to prove it. I suspect I may have to not just bound things using absolute values, but actually know where things are positive and negative



Bound on trigonometric sum
To elaborate on my comment: First, $\cos k \theta - 1 = -2 \sin^2 \frac 12 k \theta,$
so the inequality can be rephrased as saying that 


$\lim\_{n\rightarrow \infty} \sum\_{k=1}^n \sin^2 k \theta/2 = \infty.$


For $\theta=O(1)$ you already know how to do this. For $\theta \ll 1/n,$ as per @Matt's suggestion one can use the Taylor series approximation: each term is of order $k^2 \theta^2,$ so the sum is of order $\theta^2 n^3,$ which indicates that it goes to infinity for $\theta = O(n^{-3/2}).$ If $\theta = o(1),$ but $\theta=O(1/n),$ use the Riemann sum approximation, to show that the sum is asymptotic to $\frac{1}{\theta} \int\_0^{n \theta} \sin^2 x dx.$ The integrand is positive, so is the integral. The upper limit is bigger than some constant, $1/\theta$ grows at least linearly....


