If $(x_n)$ is ud mod $1$ and $(y_n)$ satisfies
$$\lim_{N\to \infty} \frac 1N \sum_{n=1}^N |y_n|=0$$
then $(x_n+y_n)$ is ud mod $1$.
Since this comes as an exercise after the first two sections of the first chapter of Uniform Distribution of Sequences, I guess we need to use the Weyl criterion. So, we want to show that
$$\lim_{N\to \infty} \frac{1}{N} \sum_{n=1}^N e^{2\pi ih(x_n+y_n)}=0$$
for all non-zero integers $h$.
We can note that $\sum_{n=1}^N |y_n|$ is a monotone sequence, and hence it either converges or diverges to $\infty$. If the series converges, then $\lim y_n=0$ which completes the proof (by Lemma $1.1$). But, I don't know what to do if the sum diverges to $\infty$. I need some help with it.
Best Answer
It's indeed a reasonable approach to use Weyl's criterion. Simply note that $$\big| e^{2i\pi h(x_n+y_n)} - e^{2i\pi hx_n}\big| = \big|e^{2i\pi h y_n} - 1\big| = \big|e^{i\pi h y_n} - e^{-i\pi h y_n}\big| = 2|\sin(\pi h y_n)| \le 2\pi h |y_n|.$$
Hence $$\Big|\frac{1}{N}\sum\limits_{n=1}^N e^{2i\pi h(x_n+y_n)} - \frac{1}{N}\sum\limits_{n=1}^N e^{2i\pi h x_n} \Big| \le 2\pi h \frac{1}{N} \sum\limits_{n=1}^N |y_n|$$
Since $\frac{1}{N}\sum\limits_{n=1}^N e^{2i\pi h x_n} \underset{N \to \infty}{\longrightarrow} 0$ and $\frac{1}{N} \sum\limits_{n=1}^N |y_n| \underset{N \to \infty}{\longrightarrow} 0$, we conclude that $\frac{1}{N}\sum\limits_{n=1}^N e^{2i\pi h(x_n+y_n)} \underset{N \to \infty}{\longrightarrow} 0$.