August 27th, 2019, 07:27 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 642 Thanks: 91  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. 
August 27th, 2019, 07:53 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,972 Thanks: 2222 
Did you invent this problem?

August 27th, 2019, 12:26 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra 
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.

August 27th, 2019, 12:35 PM  #4  
Senior Member Joined: Aug 2012 Posts: 2,393 Thanks: 749  Quote:
It looks like you could prove it by using L'Hospital perhaps.  
August 27th, 2019, 01:25 PM  #5 
Senior Member Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 
*sigh* Mathematicians... It sure looks like it's going towards zero to me. 
August 27th, 2019, 02:37 PM  #6 
Senior Member Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 
In fact, the magnitude is shrinking as 1/n.

August 27th, 2019, 03:32 PM  #7 
Senior Member Joined: Sep 2016 From: USA Posts: 645 Thanks: 408 Math Focus: Dynamical systems, analytic function theory, numerics 
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 $n^{1} \log n \to 0$.

August 27th, 2019, 03:49 PM  #8 
Senior Member Joined: Jun 2019 From: USA Posts: 213 Thanks: 90  

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