If $\sqrt{h_{n+1}}-\sqrt{h_n} \rightarrow 0$ and $\alpha$ irrational, then $\{h_n \alpha\}$ is equidistributed mod 1

Question

Here $h_n$ is a sequence of increasing integer numbers. The brackets represent the fractional part function. I am looking for a reference about this statement (which is obvious if $h_n = n$), or a proof. I kind of remember this argument being used to prove that $\{p_n \alpha\}$ is equidistributed mod $1$ if $p_n$ is the $n$-th prime. Actually it was posted on Math.StackExchange but can't find the post anymore.

Answer 1

The conclusion that $\{h_n \alpha\}$ is equidistributed mod 1 in [0,1) is not true. Let's select an irrational $\alpha$ which is slightly larger than $\frac13$ so that $\frac13\lt\alpha\lt\frac12$, in sequence {1,2,3,4,...}, removing all number n so that $\{n \alpha\} \gt \frac23$ to generate a sequence $\{h_n\}$. Now we get an increasing sequence $\{h_n\}$ so that $1\le h_{n+1}-h_n\le 2$ so that $\lim_{n\to\infty}\sqrt{h_{n+1}}-\sqrt{h_n}\le \lim_{n\to\infty}\frac2{\sqrt{h_{n+1}}+\sqrt{h_n}}\le\lim_{n\to\infty}\frac1{\sqrt{n}}=0$ but $\{h_n\alpha\}$ is not equidistributed in [0,1) since $(\frac23,1)$ is not covered