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2 another typo

# NegativityofBoundon trigonometric sum

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 $\gamma<\theta<\pi$.

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# Negativity of trigonometric sum

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 $\gamma=o(n^{-1})$ such that the sum is negative for $\gamma<\theta<\pi$; I know indeed that this sum can be expressed using the Dirichlet kernel and if we took $\gamma=n^{-3/4}$ say I could show what I want, but I need $\gamma=o(n^{-1})$ for my purpose. I've been messing around with this all morning, and I expect the solution won't be deep. I'd be happily surprised if it were though.

I've plotted this for a few different $n$ and it looks it's negative for all $\theta$, although I have reason to believe that shouldn't be true and that it should only be negative for $\gamma<\theta<\pi$.