I just want to know the proof that $\left\vert\sum_{x\le Q} e(\alpha x)\right\vert\ll \lVert \alpha\rVert^{-1}$ for $\alpha\not \in\mathbb Z$, for any positive integer $Q$. Here $\lVert x\rVert$ means the least distance between $x$ and an integer, and $e(x)=\exp(2\pi i x)$.
Is it also possible to show that $$\left\vert\sum_{x\le Q} e(2\alpha x)\right\vert\ll \dfrac{1}{\lVert \alpha\rVert}?$$
I doubt perhaps it cannot? Because if we let $m=\exp(2\pi i \alpha)$, $$\left| \sum_{x\le Q} e(2\alpha x) \right|=\left|\dfrac{m^{2Q}+1}{m+1}\right|\cdot\left|\sum_{x\le Q} e(\alpha x)\right|,$$ but the first term on the right is not bounded by a constant? Is this term bounded by $Q^\epsilon$ (as I've seen from the book involving this that I want to eliminate $\epsilon$ if possible).
Edited: Is it okay as follows? With geometric series calculation we have $$\left|\sum_{x\le Q}e(\alpha x)\right|=\sqrt{\dfrac{1-\cos(2\pi\alpha Q)}{1-\cos(2\pi\alpha)}}.$$
Then we just need to check where $\alpha\rightarrow 0$ for $$\lim_{\alpha\rightarrow 0}\dfrac{\alpha^2}{1-\cos(2\pi\alpha)}=\lim_{\alpha\rightarrow 0}\dfrac{1}{2\pi^2\cos(2\pi\alpha)}=\dfrac{1}{2\pi^2},$$ by applying L'hospital law? And that thing should be bounded by $\frac{1}{\lVert \alpha \rVert}$? And I mean we just need to show $$\left|\dfrac{\alpha^2}{1-\cos(2\pi\alpha)}\right|=O(1),$$ for $\alpha\in [-1/2,1/2]-\left\{0\right\}$, which seems to me pretty possible?
--> The graph with the filled hole Graph plot