Here is how I interpret the question: Let $(b_n)$ be an increasing sequence of reals, $s_n=b_n-\lfloor b_n\rfloor$ be the fractional part of $b_n$ and $a_n=b_{n+1}-b_n.$

On the assumption that $(a_n)$ is decreasing (or at least non-increasing), what conditions on the growth rate of $(b_n)$ (or decay rate of $(a_n)$) allow us to conclude that $(s_n)$ is uniformly distributed in $[0,1]?$

Here I only consider the case that $\lim a_n=0$ aside from a quick mention of the classic case $b_n=n\alpha$.

Define $N_{b}$ to be the last $n$ with $b_n \lt b.$

**Conjecture**: Given $\lim a_n=0$, the condition $\lim_{y \rightarrow \infty}\frac{N_{y}}{N_{y+1}}=1$ is necessary and sufficient for $(s_n)$ to be uniformly distributed.

In case $b_n=\log{n}$ it turns out that $b_n$ grows too slowly (equivalently, $a_n$ decays too rapidly). Here $\frac{N_{y}}{N_{y+1}}=\frac{e^{y}}{e^{y+1}}=\frac1e.$ I note below that the distribution fluctuates with $\frac{\liminf r(\cdot)}{\limsup r(\cdot)}=\frac1e.$ For this sequence $a_n=\log{\frac{n+1}{n}}=\frac{1}{n}+O(\frac{1}{n^2})$ The previous question concerned $a_n=\frac{1}{n}.$ Then the same proof works there but some minor details are slightly more involved.

I claim that $b_n=n^p$ for $0 \lt p \lt 1$ does lead to a uniform distribution.

For a given sequence of real numbers $(s_n)_{n=1}^\infty \subset [0,1]$, let $A(x,N)=\left|\lbrace n \le N: s_n \le x \rbrace \right|$ and let $r(x,N)= \frac{A(x,N)}{N}.$ The sequence is said to have a limiting distribution $r(x)$ if $\displaystyle \lim_{N \rightarrow \infty}r_N(x)=r(x)$ for every $0 \leq x \leq 1$.

Let $0 \lt \epsilon$ be small . Then for $K \in \mathbb{N},$ we have that $A(\epsilon,N_{K+\epsilon})=A(\epsilon,N_{K+1}).$ So $r(\epsilon,N)$ fluctuates with a local maximum at each $r(\epsilon,N_{K+\epsilon})$ and a local minimum at $r(\epsilon,N_{K+1}).$ Furthermore $$\frac{r(\epsilon,N_{K+1})}{r(\epsilon,N_{K+\epsilon})}=\frac{N_{K+\epsilon}}{N_{K+1}}.$$

If there is any limiting distribution at all then the second ratio must converge to 1 in which case for any fixed $z$, $\lim_{y \rightarrow \infty}\frac{N_{y}}{N_{y+z}}=1.$ This establishes that the condition is necessary. It helps to think of $\epsilon$ as small, but all that is really used is that $0 \lt \epsilon \lt 1.$

Assume for simplicity that $\epsilon$ is not one of the countable number of values $s_n$, Then we can also express the common value above as $A(\epsilon,N_{K+\epsilon})=A(\epsilon,N_{K+1})=\sum_{J=0}^K(N_{J+\epsilon}-N_J)$

With a little work this establishes the claim made above for $b_n=n^p$ for $0 \lt p \lt 1$. It also establishes that in the case $b_n=\log n$ we have $\frac{N_y}{N_{y+1}}=\frac1e.$
$$\liminf r(x,N)=\frac{e^x-1}{e-1} \text{ and }\limsup r(x,N) =\frac{e^{1-x}(e^x-1)}{e-1}.$$ Recall that $\frac{N_{y}}{N_{y+1}}=\frac{e^{y}}{e^{y+1}}=\frac1e$ in this case.

There has been some mention of Weyl's criterion. It is very useful for the extreme case $b_n=n \alpha$ in which $a_n$ is identically $\alpha.$ Then $(s_n)$ is uniformly distributed if and only if $\alpha \gt 0$ is irrational. The criterion is that $$\lim_{N \rightarrow \infty}\frac1N \sum_{j=1}^N e^{2\pi i \ell b_j}=0\text{ for all integers }\ell \gt 0.$$ I do not see that it helps for $b_n=\sqrt{n}$ or $b_n=\sqrt{2n}$ although it does apply to $b_n=1,\ 3/2,4/2,\ 7/3,8/3,9/3,\ 13/4,14/4,15/14,16/4,\ 21/5 \cdots$ which has $b_n \approx \sqrt{2n}$.