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August 27th, 2019, 07:27 AM   #1
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A limit

Evaluate $\displaystyle \lim_{n\rightarrow \infty }n^{-1} \underbrace{\ln \ln...\ln}_{n}(n)$.

Last edited by skipjack; August 27th, 2019 at 07:54 AM.
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August 27th, 2019, 07:53 AM   #2
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Did you invent this problem?
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August 27th, 2019, 12:26 PM   #3
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Mmm... since $\ln x \to -\infty$ as $x \to 0$ at an exponential rate, I'm going to guess that this blows up. But I'm far from certain.
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August 27th, 2019, 12:35 PM   #4
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Originally Posted by v8archie View Post
Mmm... since $\ln x \to -\infty$ as $x \to 0$ at an exponential rate, I'm going to guess that this blows up. But I'm far from certain.
The limit is to infinity. $\log x$ goes to infinity as does any number of iterations of $\log$. It's an old joke that $\log \log x$ is constant for all practical purposes, the point being that it goes to infinity very slowly. But to infinity it goes nonetheless. So I don't follow your intuition. The intuition would have to be that the iterated log goes to infinity so slowly that the n in the denominator pulls it to zero.

It looks like you could prove it by using L'Hospital perhaps.
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August 27th, 2019, 01:25 PM   #5
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*sigh* Mathematicians...

It sure looks like it's going towards zero to me.
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August 27th, 2019, 02:37 PM   #6
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In fact, the magnitude is shrinking as 1/n.
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August 27th, 2019, 03:32 PM   #7
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The limit is 0. You can easily see this by noticing that $\log^{(k)}(n) \leq \log n$ for all $k \geq 1$ where the notation on the left means iteration. Combine this with the fact that $n^{-1} \log n \to 0$.
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August 27th, 2019, 03:49 PM   #8
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$\log^{(k)}(n) \leq \log n$ for all $k \geq 1$
I thought of this, but the iterated logs quickly become negative (and thus complex). If $| \log^{(k)}(n) | \leq |\log n|$, then it still works, but I wasn't sure how to determine that. Plotting was easier.
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