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I am currently taking first course in real analysis following Ross's Elementary Analysis textbook. When I was introduced to lim sup and lim inf, I found it hard to manage to play around or make meaningful conclusions from them because the terms are not in explicit forms but are in form of suprema and infima of a set.

For example, how to start tackling this problem: if $\lim \sup |a_n| > 0$, prove that $\lim \sup |a_n|^\frac{1}{n} \geq 1$. What might be the first ideas that I should try thinking about

Generally, how ones can approach lim sup and lim inf, starting intuitively and working technically?

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    $\begingroup$ If you let $\alpha = \limsup |a_n|$, then that implies that for any $\varepsilon$, there are infinitely many terms of the sequence that are $> \alpha - \epsilon$. So the fact that $\alpha > 0$ just means that there is some $\beta > 0$ such that there are infinitely many terms of the sequence $|a_n|$ that are $> \beta$. Your desired conclusion is that for any $\varepsilon$, there are infinitely many terms of the sequence $|a_n|^{1/n}$ that are $> 1 - \varepsilon$. $\endgroup$ – user49640 Jun 14 '17 at 21:19
  • $\begingroup$ The plural of supremem is suprema. Analogous, that of infimum is infima. $\endgroup$ – Displayname Jun 15 '17 at 8:00
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    $\begingroup$ Here is another answer you may have a look at: math.stackexchange.com/a/1893725/72031 $\endgroup$ – Paramanand Singh Jun 15 '17 at 13:05
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My analysis teacher taught us about these concepts using the ideas of "eventually" and "frequently". A sequence is "eventually" in a set if there exists some $N$ such that for every $n>N$, we have $a_n$ in the set. A sequence is "frequently" in a set if, for every $N$, there exists an $n>N$ with $a_n$ in the set.

Step one is to meditate on that paragraph until it makes sense with your understanding of the words "eventually" and "frequently".

Once you have that, $\lim\sup$ works this way. To say $\lim\sup\{a_n\}=c$ means that the sequence $\{a_n\}$ is eventually less than $c+\epsilon$, and frequently greater than $c-\epsilon$, for any small $\epsilon>0$.

The first claim, that the sequence is eventually less than $c+\epsilon$, means that there is no accumulation point of the sequence strictly larger than $c$. Adding in the other fact, that it is frequently greater than $c-\epsilon$, we know that $c$ is an accumulation point itself.

For $\lim\inf$, just turn all that on its head. If $\lim\inf\{a_n\}=c$, then $\{a_n\}$ is eventually greater than $c-\epsilon$, and it is frequently less than $c+\epsilon$, again for any small $\epsilon>0$.

Once you have these definitions in your head, you can gain intuition by playing around with practice sequences that have various kinds of limit points. You should be able to find such exercises in any decent analysis textbook. If you google "lim sup lim inf practice problems", you can find many more.


A nice thing about these concepts, "eventually" and "frequently", is that they can be easily adapted to talk about other kinds of limits in mathematics, such as limits of sequences of sets, and possibly even direct limits of algebraic and topological structures, although I have less experience with those, so I'm not too sure.

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  • $\begingroup$ That's how I learned it too. $\endgroup$ – Ryan Reich Jun 15 '17 at 5:30
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    $\begingroup$ I can add that "frequently" and "eventually" are kind of opposites. Any sequence fulfills one and only one of the two properties "Eventually in $N$" and "Frequently outside $N$" (also known as "Frequently in the complement of $N$"). Whether that helps making sense of the terms or just causes further confusion, I don't know. $\endgroup$ – Arthur Jun 15 '17 at 10:38
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    $\begingroup$ The terms frequently and eventually are a good choice for the usual mathematical terms "infinitely many $n$" and "sufficiently large $n$" especially when when has a problem understanding the latter terms. +1 $\endgroup$ – Paramanand Singh Jun 15 '17 at 13:03
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    $\begingroup$ That explanation should be included in a textbook somewhere. $\endgroup$ – Robert Wolfe May 11 '18 at 15:29
  • $\begingroup$ @RobertWolfe, yes, it's beautiful. Thanks G Tony. I had never understood the "eventually less than $c+\epsilon$" part, and for years I thought I understood the concepts--until just now, when a step in a proof confused me. What I needed was exactly that point. $\endgroup$ – Mars Jul 25 at 17:53
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The number $\limsup|a_n|$ is the supremum of the accumulation points of the sequence $\{|a_n|\}$. So if you are told that $\limsup|a_n|=c>0$, you know that there is a subsequence $\{|a_{n_k}|\}$ with limit $c$, and no subsequence has a greater limit. The first fact is already enough to tackle your problem.

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  • $\begingroup$ That's true for sequences that are bounded above. For sequences unbounded above, you'd have to regard $+\infty$ as an "accumulation point", which is not usual. $\endgroup$ – Robert Israel Jun 14 '17 at 22:25
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    $\begingroup$ It is usual to me, and standard when you consider the extended real numbers in measure theory. $\endgroup$ – Martin Argerami Jun 14 '17 at 22:40
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I would write here $\limsup|a_n|=k$. Then $k>0$. What this means is that for any $\delta>0$, $|a_n|>k-\delta$ infinitely often, but $|a_n|<k+\delta$ for all sufficiently large $n$. Thus $|a_n|^{1/n}>(k-\delta)^{1/n}$ infinitely often but $|a_n|<(k+\delta)^{1/n}$ for all sufficiently large $n$. Can we choose a useful $\delta$ which will tell us something about the sequence $(|a_n|^{1/n})$?

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    $\begingroup$ To the O.P. : Try $\delta=k/2.$ $\endgroup$ – DanielWainfleet Jun 14 '17 at 22:27
  • $\begingroup$ I think the point that I was missing is that "for sufficiently large n" means $\left\lvert a_n \right\rvert ^{1/n} \to 1$, and then you can say something about the inequality with $\delta$. Is that correct? $\endgroup$ – Burnsba Jun 15 '17 at 12:52
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An idea for you to work out: that $\;\limsup\limits_{n\to\infty}a_n=q>0\;$ means that $\;q\;$ is the biggest of all the partial limits of $\;\{a_n\}\;$ , and in particular that there exists a subsequence

$$\{a_{n_k}\}\subset\{a_n\}\;\;\;s.t.\;\;\;\lim_{x\to\infty}a_{n_k}=q$$

and this last means that for $\;\epsilon=\frac q2\;$ , for example, there exists $\;K\in\Bbb N\;$ such that if $\;k>K\;$ then

$$\;|a_{n_k}-q|<\epsilon\iff 0<q-\epsilon<a_{n_k}<q+\epsilon$$

and from here

$$1\xleftarrow[\infty\leftarrow n]{}\sqrt[n]{q-\epsilon}\le\sqrt[n]{a_{n_k}}\le\sqrt[n]{q+\epsilon}\xrightarrow[n\to\infty]{}1$$

Deduce now what you asked...

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