[proofplan] We prove Weyl's criterion for this special sequence. Every non-zero exponential average is a finite geometric sum with denominator separated from zero because $\alpha$ is irrational. Hence all non-zero Fourier averages vanish, and Weyl's criterion gives equidistribution. [/proofplan] [step:Compute the exponential sums] Let $k\in\mathbb{Z}\setminus\{0\}$. Since $\alpha$ is irrational, \begin{align*} e^{2\pi i k\alpha}\neq1. \end{align*} Therefore, for $N\geq1$, \begin{align*} \frac1N\sum_{n=0}^{N-1}e^{2\pi i k n\alpha} &= \frac1N\cdot \frac{1-e^{2\pi i kN\alpha}}{1-e^{2\pi i k\alpha}}. \end{align*} The numerator is bounded in absolute value by $2$, while the denominator is a fixed non-zero complex number. Hence \begin{align*} \frac1N\sum_{n=0}^{N-1}e^{2\pi i k n\alpha}\to0. \end{align*} [/step] [step:Apply Weyl's criterion] Weyl's equidistribution criterion says that a sequence $(x_n)$ in $[0,1)$ is equidistributed modulo $1$ if and only if, for every $k\in\mathbb{Z}\setminus\{0\}$, \begin{align*} \frac1N\sum_{n=0}^{N-1}e^{2\pi i kx_n}\to0. \end{align*} Applying this criterion to $x_n=\{n\alpha\}$ and using the exponential-sum computation above gives the desired equidistribution. Thus, for every interval $[a,b)\subset[0,1)$, \begin{align*} \lim_{N\to\infty}\frac1N\#\{0\leq n<N:\{n\alpha\}\in[a,b)\}=b-a. \end{align*} [/step]