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Let $s_n$ a sequence of positive real numbers such that $$\lim_{n\to\infty}\frac{1}{s_n}=0$$ and $$\lim_{n\to\infty}\frac{s_{[nt]}}{s_n}=t,$$ for every real $t\in[0,1]$. See here, page 4.

Question 1. Can you give a detailed proof of $$\lim_{n\to\infty}\frac{s_{[bt]}-s_{[at]}}{s_n}=b-a,$$ where $0\leq a<b\leq 1$?

If previous assumptions hold, then the sequence of ratios $$\frac{s_1}{s_n},\frac{s_2}{s_n},\ldots\frac{s_n}{s_n}$$ approach uniform distribution modulo one as $n\to\infty$, and we say that $s_n$ is pseudodistributed mod 1.

Question 2. Let $s_n=\sigma(n)$, where $\sigma(n)=\sum_{d\mid n}d$ is the sum of divisor function. Is this sequence pseudo equidistributed mod 1?

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Question $1$ is false. If $a$ is fixed, then $s_{\lfloor bt\rfloor}$ is a constant, meaning that

$$\lim_{n\to\infty}\frac{s_{\lfloor bt\rfloor}-s_{\lfloor at\rfloor}}{s_n} = \lim_{n\to\infty}\frac{s_{\lfloor bt\rfloor}}{s_n} - \lim_{n\to\infty}\frac{s_{\lfloor at\rfloor}}{s_n}=s_{\lfloor bt\rfloor}\lim_{n\to\infty}\frac1{s_n}-s_{\lfloor at\rfloor}\lim_{n\to\infty}\frac1{s_n}=0-0=0$$

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  • $\begingroup$ Then will be a typo that I write or in the paper, I don't know what say. Thanks @5xum $\endgroup$
    – user243301
    Jan 21, 2016 at 12:44

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