I want to prove the following: let $a$ be an irrational constant and $m$ an integer. Then
$$\lim_{n \to\infty} \frac{1}{n}\sum_{k=0}^{n-1}e^{2\pi m i (x+ka)} = \begin{cases} 0, & m\not=0 \\ 1, & m=0 \end{cases}$$ for any $x$. Intuitively I can understand this result: because $a$ is irrational, the points $x+ka$ mod $1$ are dense in $[0,1]$. So the sum on the left hand side is just the average of the values of $e^{2\pi m i x}$ on the interval $[0,1]$. And this average is of course $0$ (both the real and imaginary parts are just sine waves), except when $m=0$ and thus $e^{2\pi m i x} = 1$.
However, I am having difficulty formalizing the idea. In particular, here is something that troubles me: I know that the points $x+ka$ mod $1$ are dense in $[0,1]$, but how do I prove that they are "evenly distributed" across the interval, so that we get an unweighted average?
A good way to do this would be to look for cancellations. In particular, we can use the property that we can find a $n$ such that $na$ is arbitrarily close to $\frac{1}2$. By this and continuity, we could choose $n$ such that $|e^{2\pi i na}+1|<\varepsilon$ for any $\varepsilon>0$. Then we could take the sum: $$\sum_{n=0}^{2k-1}e^{2\pi na}$$ and rearrange to: $$\sum_{n=0}^{k-1}e^{2\pi i na} + e^{2\pi i(n+k)a} $$ $$\sum_{n=0}^{k-1}e^{2\pi ina} (1+e^{2\pi ka}) $$ where each term has absolute value less than $\varepsilon$, and so the sum must have absolute value less than $k\varepsilon$. If you generalize this argument a little, you can prove the desired result.