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In other words, does the limit for $\ln(\ln(n))$ (when $n\to\infty$) exist? My intuition tells me that it converges to a positive number but I don’t know what it is, much less prove it.

Clement C.
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Sid Caroline
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  • And what is $n$ doing here? – Mark Bennet Nov 14 '17 at 23:57
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    Given any number $M$, $$ \ln \ln n>M\iff n>\exp(\exp M). $$ Thus, no matter how big a number $M$ you think of, $\ln \ln n$ will eventually get bigger than it. – anon Nov 15 '17 at 00:06
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    Well, for comparison to show just how slowly $\ln(\ln n)$ approaches infinity, if $n$ is the number of Planck times in the age of the universe then $\ln(\ln n)$ is approximately 5... – Daniel Schepler Nov 15 '17 at 00:11
  • @DanielSchepler That is a poor reference in a site for mathematics, where many people don't likely have the slightest idea what a "Planck Time" is...and even if they know (as I do, say), I can't see how is that relevant for $;\log\log n;$ to be very slowly approaching infinite... – DonAntonio Nov 15 '17 at 00:14
  • @DanielSchepler I am personally OK with the fact that $\log^\ast n$ goes to $\infty$, so... from a mathematical point of view, that $\log\log n$ diverges "slowly" is irrelevant. So does $\frac{1}{10^{27}}\log n$. – Clement C. Nov 15 '17 at 00:17
  • @DonAntonio: Given the background suggested by the question, I agree that Planck time might be unfamiliar, but I'd be surprised if fewer than 80% (just a guess) of people past first year calculus had not heard of Planck time. However, for this question, perhaps Avogadro's number would be better, since that's in high school chemistry which at least in the U.S. is taken by pretty much anyone planning to attend college. For what it's worth, I used Avogadro's number at the end of this answer. – Dave L. Renfro Nov 15 '17 at 01:37
  • But again... that's not relevant? – Clement C. Nov 15 '17 at 02:46
  • @DaveL.Renfro Never heard of "Planck time", neither as an undergraduate nor in graduate school. Only when I took some courses in physics I heard of that. And that just after I already studied classic and relativistic mechanics and stuff. I honestly doubt more than 2-3% of undergraduate students have ever heard of it...and again: it doesn't seem to be very relevant for the question. – DonAntonio Nov 15 '17 at 09:23
  • @DonAntonio: Interesting. When I was an undergraduate the science/math sections of bookstores were filled with books touting the latest stuff about quasars, pulsers, neutron stars, black holes, many-worlds interpretation of quantum mechanics, early universe stuff, etc. Now it's mostly on the internet, but maybe the huge information firehose that people now drink from makes it less likely to notice stuff slightly removed from one's main focus? Also, you might be right about undergraduates in general, but I was more specifically thinking of those in a physical science major or math major. – Dave L. Renfro Nov 15 '17 at 12:07

4 Answers4

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Since $\lim_{n\to\infty}\ln n=\infty$, we have $\lim_{n\to\infty}\ln\ln n=\infty$ as well.

Why do you think it should be a number?

user502959
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It certainly looks very flat, maybe like $c-1/n$, but actually it's just growing slowly to infinity. In fact suppose $f(n):=\ln\ln(n)\rightarrow c$ for some constant $c$. If you let $g(x):=e^{e^x}$, then since $g$ is continuous we'd expect $\lim_n g(f(n))=g(c)$, but in fact $g(f(n))=n$, which goes to infinity.

Alex R.
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Show that $\ln(\ln(n))$ has no upper bound:

Let $M,$ positive, real, be given.

Consider : $L: =e^{(e^M)}.$

There is an $n_0 \in \mathbb{N}$ such that $n_0 > L.$ (Archimedes).

For $n\ge n_0$ we have:

$n \ge n_0 \gt L.$

$\ln(n)$ is strictly increasing:

$\ln(n) \ge \ln(n_0) \gt \ln(L)=e^M.$

Repeating:

$\ln(\ln(n)) \ge \ln(\ln(n_0) \gt M$, I.e.

$\lim_{n \rightarrow \infty} \ln(\ln(n)) = \infty.$

Peter Szilas
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No matter how much times (finitely) you apply logarithm the result will still be infinity this follows from the fact that $\lim_{x\to \infty}\ln(x)=\infty$ because for $\lim_{x\to \infty} \ln(\ln(x))$ you can make the substitution $t=\ln x$ then you're evaluating $\lim_{t\to \infty}\ln(t)=\infty$.

For the general result you use induction we already proved the case when we applied logarithm $2$ times (or $1$ trivially).

Now assume $$\lim_{x\to \infty}\underbrace{\ln(\ln(\ln (\cdots \ln(x)))\cdots )}_k=\infty$$ Then by the simple substitution $t=\underbrace{\ln(\ln(\ln (\cdots \ln(x)))\cdots )}_k$ we get $$\lim_{x\to\infty}\underbrace{\ln(\ln(\ln (\cdots \ln(x)))\cdots )}_{k+1}=\lim_{t\to\infty}\ln t=\infty$$

kingW3
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