I want to show that for each $n>1$ the sum $\sum_{k=1}^n (-1+\cos(k\theta))$ is negative for $0<\theta<\pi$, or, failing that, that there is some $\gamma=o(n^{-1})$ \gamma(n)=o(n^{-1})$ and some $C(n) \to -\infty$ such that the sum is negative for $\gamma <\theta<\pi$; I know indeed that \leq \theta \leq \pi$ we have
$\sum_{k=1}^n -1+\cos(k\theta) \leq C(n)$.
If we rewrite this sum can be expressed using the Dirichlet kerneland if we took $\gamma=n^{-3/4}$ say I could show , what I want , but is that: if $\gamma \leq \theta \leq \pi$ then
$-n+\frac{D_n(\theta)}{2}-\frac{1}{2} \leq C(n)$.
If I need didn't demand that $\gamma=o(n^{-1})$ for my purpose. I've been messing around with then I could do this all morningwith $\gamma=n^{-3/4}$ say, and I expect the solution won't be deep. I'd $C(n)$ could then just be happily surprised if it were though.$-n$.
I've plotted this for a few different $n$ bunch and it looks it's negative for all $\theta$, although true, and I've spent the morning trying to prove it. I suspect I may have reason to believe that shouldn't be true not just bound things using absolute values, but actually know where things are positive and that it should only be negativefor $\gamma<\theta<\pi$.
