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Is the sequence defined by $a_n=\{\ln(n)\}$ dense in $[0,1]$?

Note: $\{x\}$ denotes the fractional part of $x$

math_lover
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  • It converges to infinity, so no. – Pedro Dec 18 '16 at 18:01
  • No it does not remove your downvote. – math_lover Dec 18 '16 at 18:02
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    @PedroTamaroff ${x}$ denotes the fractional part of $x$. – Noah Schweber Dec 18 '16 at 18:03
  • @NoahSchweber: false it denotes the fractional part – math_lover Dec 18 '16 at 18:03
  • @JoshuaBenabou I just edited. – Noah Schweber Dec 18 '16 at 18:03
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    This could be better if non-standard or uncommon notations were defined in this question. – MoebiusCorzer Dec 18 '16 at 18:04
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    @MoebiusCorzer I believe that notation is standard; that said, it probably should be defined in the question. – Noah Schweber Dec 18 '16 at 18:04
  • @ThomasAndrews: I have solved the problem I'm just posting it here to see what other solutions there might be – math_lover Dec 18 '16 at 18:05
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    Do you know that if $\alpha$ is irrational then ${n\alpha}$ is dense in $0,1$? If so, then the above follows if you know that $\ln 2$ is irrational. – Thomas Andrews Dec 18 '16 at 18:06
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    Then post your solution as part of the question, and ask for other solutions. @JoshuaBenabou – Thomas Andrews Dec 18 '16 at 18:06
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    @NoahSchweber Yes, as you fastly guessed what it meant, I suppose this is standard but, IMO, it is not that common. I've studied mathematics for four years and never encountered this notation in any lecture or book. Thanks for your comment anyway – MoebiusCorzer Dec 18 '16 at 18:07
  • @ThomasAndrews: since ln(n+1)-ln(n) goes to zero. given a k, its eventually less than 1/k, so well have {ln(n)} intersecting each interval of the form [i/k,(i+1)/k) for n sufficiently large, so its dense in [0,1]. – math_lover Dec 18 '16 at 18:13
  • I think additionally, the problem is that individual numbers aren't dense in a set, a set of numbers is dense in a set, so the title ${\ln n}$ seems to refer to the set of values $\ln n$, rather than the set of values. – Thomas Andrews Dec 18 '16 at 18:13
  • It's common and standard but it's not ... universal. Without context in an analysis question it just isn't immediately apparent that that is what was meant. No harm done though. – fleablood Dec 18 '16 at 19:00

3 Answers3

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Given an irrational number, $\alpha$, we know the set of $\{n\alpha\}$ is dense in $[0,1]$.

If you know that $\ln 2$ is irrational, then $\{n\ln 2\}$ is dense in $[0,1]$ and since $n\ln 2 = \ln(2^n)$, this is a subsequence of your sequence, and hence your sequence is dense in $[0,1]$.

$\ln 2$ being irrational follows from $e$ being transcendental, since if $e^{p/q}=2$ then $e$ is a root of the rational polynomial $x^p-2^q$.

This is all a "big gun" approach to this question.

You actually don't need that $e$ is transcendental to use this technique - you can show that at least one of $\ln 2$ and $\ln 3$ is irrational, because otherwise $2^p=3^q$ for some integers $p,q$. Then either $\{n\ln 2\}$ or $\{n\ln 3\}$ is dense in $[0,1]$.

This works with any base - $\{\log_b n\}$ is dense in $[0,1]$ - since at least one of $\log_b 2$ and $\log_b 3$ is irrational.

Thomas Andrews
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Hint: $\ln (n+1) - \ln n \to 0.$

zhw.
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[Just to elaborate on zhw.'s hint.]

Suppose $(\{\ln n\})_n$ is not dense in $[0,1]$. Then there is some interval $(a,b)\subset [0,1]$ such that $\{\ln n\}\not\in(a,b)$ for every $n$. As $\ln n-\ln (n-1)\to0$ when $n\to\infty$ we can choose $N$ big enough so that $$\ln n-\ln (n-1)<b-a\,\,\,\text{ if }\,\,\,n> N.\hspace{1cm}(\star)$$ Since $\ln n\to+\infty$, there is $M>N$ so that $\ln N-\ln M\geq1$, which [along with $(\star)$] tells us that $P=\{\{\ln n\}\}_{n=N}^M$ is a partition of $[0,1]$ with diameter/norm $d_P<b-a$, implying the contradiction $P\cap(a,b)\neq\phi.$