Vinogradov's Theorem/Minor Arcs/Lemma 1

Lemma
For $\beta \in \R$, define $\|\beta\| = \min\{|n - \beta| : n \in \Z\}$.

Then for any $\alpha \in \R$,


 * $\displaystyle \left| \sum_{k = N_1}^{N_2} e(\alpha k) \right| \leq \min\left\{ N_2 - N_1, \frac1{2\|\alpha\|} \right\}$

Proof
The bound $N_2 - N_1$ is trivial; since $|e(\alpha k)| = 1$ for all $k$, by the Triangle Inequality we have:


 * $\displaystyle \left| \sum_{k = N_1}^{N_2} e(\alpha k) \right| \leq \sum_{k = N_1}^{N_2}1 = N_2 - N_1$

To show the second bound we evaluate the sum as a geometric series.

We have $e(\alpha k) = e(\alpha)^k$, so by Sum of Geometric Progression,

Now by the polar form of a complex number,


 * $e(\alpha) - 1 = e(\alpha/2) [e(\alpha/2) - e(-\alpha / 2)]$

and


 * $e(\alpha/2) - e(-\alpha / 2) = \exp(\pi i \alpha) - \exp(-\pi i \alpha) = 2i\sin(\pi \alpha)$

Therefore we have


 * $\displaystyle \left\vert \sum_{k = N_1}^{N_2} e(\alpha k) \right\vert \leq \frac{1}{|\sin(\pi \alpha)|}$

We know that the sine function is concave on $[0,\pi/2]$, so $\sin(\pi \alpha) \geq 2\alpha$ for $\alpha \in [0,1/2]$.

By definition there is $n \in \Z$ such that $\alpha = n + \|\alpha\|$, and $\|\alpha\| \in [0,1/2]$, so

This completes the proof.