I came across this chain of inequalities in notes I am reading.

$|\sum_{1 \leq n \leq N}e^{2 \pi i \alpha n}| \leq \frac{|1 - e^{2 \pi i \alpha N}|}{|1 - e^{2 \pi i \alpha}|} \leq \frac{\sin(\pi \alpha N)}{\sin(\pi \alpha)} \leq \frac{1}{||\alpha||}$.

Here $\alpha$ is a nonzero real number and $||\alpha ||$ is the distance from $\alpha$ to the nearest integer.

The first inequality I was able to verify since it follows from the formula for the sum of a finite geometric series.

I am not sure how to verify the 2nd and 3rd inequalities. I tried exploiting that the numerator and denominator of the 2nd term is a difference of two squares and using the relation $e^{i\theta} = \cos \theta + i\sin \theta$ but I didn't have success. No idea why the 3rd one is true either.

A similar question which has been answered considers the final inequality, but I did not understand a few steps in the solution. I post it here for reference:

$\left|{\sin(\pi \alpha N)}/{\sin(\pi \alpha)}\right| \leq {1}/{2 \| \alpha \|}$

  • $\begingroup$ I think there is a typo here $\frac{|1 - e^{2 \pi i \alpha N}|}{|1 - e^{2 \pi i \alpha}|} \leq \frac{\sin(\pi \alpha n)}{\sin(\pi \alpha)}$ which should be $\frac{|1 - e^{2 \pi i \alpha N}|}{|1 - e^{2 \pi i \alpha}|}= \frac{|\sin(\pi \alpha n)|}{|\sin(\pi \alpha)|}$. $\endgroup$
    – user
    Aug 29, 2021 at 8:56

1 Answer 1


We have that by $\sin x=\frac{e^{ix}-e^{-{ix}}}{2i}$

$$ \frac{|1 - e^{2 \pi i \alpha N}|}{|1 - e^{2 \pi i \alpha}|} = \frac{|e^{ \pi i \alpha N}|}{|e^{ \pi i \alpha }|}\frac{|\sin( \pi \alpha N)|}{|\sin( \pi \alpha)|}=\frac{|\sin( \pi \alpha N)|}{|\sin( \pi \alpha)|} $$

and since $|\sin x|\le 1$

$$\frac{|\sin( \pi \alpha N)|}{|\sin( \pi \alpha)|} \le \frac{1}{|\sin( \pi \alpha)|} $$

we also have

$$|\sin( \pi \alpha)|=|\sin\left( \pi ||\alpha)||\right)|$$

indeed $|\sin(\pi + \theta)|=|\sin (\theta)|$ and

  • for $\alpha\in[0,\frac12]$: $||\alpha)||=\alpha$
  • for $\alpha\in[\frac12,1]$: $||\alpha)||=1-\alpha$
  • for $\alpha\in[1,\frac32]$: $||\alpha)||=\alpha-1$
  • $\dots$

finally we use that $|\sin( x)|\geq \frac{2}{\pi} x$ as $-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}$ therefore

$$\frac{1}{|\sin( \pi \alpha)|} =\frac{1}{|\sin( \pi ||\alpha||)|} \le \frac1{\frac2 \pi \pi ||\alpha||}=\frac1{2||\alpha||}$$

Refer also to


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