I've never thought that I would have difficulties to read such a simple formula, which goes as follows1:

A well-known unsolved problem in number theory concerns the distriubtion of $(3/2)^n\pmod1$. The sequence is believed to be uniformly distributed, which is the case for almost all real numbers $\theta^n\pmod1$, but it is not even known to be dense in $[0,1]$. One of the few positive results known for (non-integer) rational $\theta=p/q$ is that of Vijayaraghavan ($1940$), who showed that the set $(p/q)^n\pmod1$ has infinitely many limit points. Vijayaraghavan later remarked that it was striking that one could not even decide whether of not $(3/2)^n\pmod1$ has infinitely limit points in $[0,1/2)$ or in $[1/2,1)$. Both these latter assertions would follow if one could show that $$\limsup_{n\to\infty}\left\{\left(\frac32\right)^n\right\}-\liminf_{n\to\infty}\left\{\left(\frac32\right)^n\right\}>\frac12.$$

I assume surely correctly, that the curly braces mean the fractional part. And $\lim_{n \to \infty} \sup()$ the highest occuring fractional value (or better its limit) and the other the lowest occuring fractional value (or better its limit) . But of course already for small $n$ the left expression approaches $1$ and the right expression approaches $0$, which makes its difference larger than $1/2$.

So obviously I must misread something elementary. Just trying to remove the tomato from my eyes...

1 Flatto, Lagarias, 1995 "On the range of fractional parts { ξ(p/q) n }"

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    $\begingroup$ The whole point of that paragraph is to point out that while it is widely believed that $\{(3/2)^n\}$ approaches both $1$ and $0$ as $n$ grows, we simply do not know. Maybe all terms are larger than $\frac12$ from some point on. We have no idea. $\endgroup$ – Arthur Apr 30 '19 at 12:19
  • $\begingroup$ @MartinR : I always understood the $\lim \inf()$ means the "global" minimum. But it seems I've never got that correctly...? $\endgroup$ – Gottfried Helms Apr 30 '19 at 12:23
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    $\begingroup$ No, $\liminf_{n\to \infty}$ and $\limsup_{n\to \infty}$ means the eventual min / max (or, rather, inf and sup instead of min and max). They are what the (global) $\inf$ and $\sup$ converge to as you successively delete entries from the beginning of the sequence. $\endgroup$ – Arthur Apr 30 '19 at 12:26
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    $\begingroup$ I begin my last message with please, and I've included links to related Mathematics Meta discussions so that you can know more about that. I don't see any harshness in my previous comments. What I'm politely asking for is to be nice to others, including the visually impaired, those with disorders in reading and writing, searching engines, etc. They can't see images, but they can hear from screen readers. The main point is that images aren't SEO friendly, that is, they can't be searched. They might hinder future users from finding this post $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 30 '19 at 18:36
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    $\begingroup$ As T Bongers points out, I've misused the word "forbidden", and I apologize for that. I should have said "discouraged since there's no explicit mechanism that stops the upload of text images. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 30 '19 at 20:45

Given a sequence $a_n$, $\limsup_{n\to\infty}a_n$ is an eventual supremum, not a global supremum of $a_n$. What I mean by that is that it doesn't care about what the early terms do, only what happens as $n\to \infty$.

Slightly more rigorously, if you successively delete entries from the beginning of the sequence $a_n$ (i.e. start at $n = 2$ instead of $n = 1$, then start at $n = 3$ instead of $n = 2$, and so on), then the (global) $\sup a_n$ might stay the same, or it might decrease. Whatever it converges to (be it a real number or $-\infty$), that is $\limsup_{n\to \infty}a_n$

Example: $$ a_n = \frac 1n + (-1)^n\\ \begin{array}{|c|cccccc} \hline n & 1&2&3&4&5&\cdots\\ \hline a_n&0&\frac32&-\frac23&\frac54&-\frac45&\cdots\\\hline \end{array} $$ The largest term in this sequence is $\frac32$, for $n = 2$. So $\sup a_n = \frac32$. However, if we delete the first two terms, then the new largest term is $\frac 54$ for $n = 4$. And so on. The largest term of whatever is left keeps shrinking as we keep deleting terms, and it converges to $1$.

$\liminf_{n\to\infty}a_n$ works the same way, only with $\inf$. Using the example above again, we see that there is no smallest term. But there is a (global) $\inf a_n = -1$. This will stay exactly where it is as we delete terms, as it is the tail end of the sequence which gives it this $\inf$. So $\liminf_{n\to\infty}a_n = -1$.

And be careful labeling $\inf$ and $\sup$ (and their $\lim$ variations) as "min" and "max". It is not the same (although they fill much of the same role). The sequence $a_n$ doesn't have a min, as it never becomes $-1$, but comes ever closer to it, so the $\inf$ is $-1$.

Going back to your original sequence, the whole point of that paragraph is to point out that while it is widely suspected that $\{(3/2)^n\}$ approaches both $1$ and $0$ as $n$ grows, we simply do not know. Maybe all terms are larger than $\frac12$ from some point on. An in that case, the $\liminf$ would be $\frac12$, or even larger.

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  • $\begingroup$ Many thanks for that thorough explanations! Funny, I have certain difficulties to grasp this firmly. Seems I've to do some exercises with more good examples to get familiar with it. $\endgroup$ – Gottfried Helms Apr 30 '19 at 17:06
  • $\begingroup$ Hmm,what about the fractional part of $ \{N \cdot \beta \}$ (where $\beta= \log_2(3)$) ? This is oscillating between $0$ and $1$ as well, and we know, that at such $N$ indicated by the convergents of the continued fraction of $ \beta$ we get ever more extreme examples approaching $0$ or $1$ with increase of $N$ (but of course no actual convergence to some limit). Is this cont-frac-effect enough for having a $\lim \sup()$ / $\lim \inf()$ ? $\endgroup$ – Gottfried Helms May 1 '19 at 2:12
  • $\begingroup$ I've now looked at the wikipedia (german) from where I thought I remember some short explanation a couple of years ago. Now it has excellent explanation and illustration such that I think the question in my previous comment is to be answered in the positive. (Or not?) $\endgroup$ – Gottfried Helms May 1 '19 at 2:30

" But of course already for small n the left expression approaches 1 and the right expression aproaches 0, which makes its difference larger than 1/2."

But we care aobut BIG n!


There is not always a relationship between $\lim \sup$/$\lim \inf$ and global maximums/minimums. You may want to think about it in the following way:

For very big $n$ (meaning, as big of $n$ as we arbitrarely decide to choose), our sequence/function will be oscilating in some range, with its lower boundary being $\lim \inf$ and its upper boundary being $\lim \sup$ Higher values than the $\lim \sup$ may be achieved for smaller $n$

As an example for that, let $f(x):=\frac{1}{x}$ for every $x>=1$ We have a global maximum point at $x=1$ with $\lim \sup$ and $\lim \inf$ (when $n \to \infty$) being both 0

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  • $\begingroup$ Possibly the problem of my understanding is that I assume $\lim \sup()$ and $\lim \inf()$ as somehow global maximum/minimum , but which cannot be less extreme as some local maximum/minimum... Can you take on this likely problem of mine? $\endgroup$ – Gottfried Helms Apr 30 '19 at 12:26
  • $\begingroup$ I editted the question to provide some info about $\lim \sup$ and $\lim \inf$ $\endgroup$ – David Apr 30 '19 at 20:25
  • $\begingroup$ Hmm, thanks also for your extended answer. I must confess there is some subtlelety that I've missed completetely when I've met that concept first time and have read just over it (surely asked wikipedia once...). I'll need some more time for chewing this with more examples - thanks again! $\endgroup$ – Gottfried Helms May 1 '19 at 1:30
  • $\begingroup$ Just looked again into wikipedia, that article (in german) has much very nice explanation this days... I think that shall suffice for all my needs! $\endgroup$ – Gottfried Helms May 1 '19 at 2:27

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