August 1st, 2019, 02:20 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 592 Thanks: 87  Evaluate the limit
$\displaystyle \lim_{n\rightarrow \infty } \frac{1+1/2 +...+1/n }{\ln(n)}.$

August 1st, 2019, 06:07 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,921 Thanks: 2203 
1

August 1st, 2019, 06:35 AM  #3 
Senior Member Joined: Jun 2019 From: USA Posts: 120 Thanks: 40 
Numerator approaches 1. Denominator increases towards infinity. So 0.

August 1st, 2019, 06:48 AM  #4 
Senior Member Joined: Oct 2009 Posts: 850 Thanks: 325  
August 1st, 2019, 06:55 AM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389 
numerator is the $n$th harmonic number which is known to diverge as $n \to \infty$

August 1st, 2019, 07:21 AM  #6 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond  
August 1st, 2019, 07:35 AM  #7 
Senior Member Joined: Dec 2015 From: somewhere Posts: 592 Thanks: 87 
So many answers, but what is the method? I'm thinking about applying logarithmic integral.
Last edited by skipjack; August 2nd, 2019 at 05:48 AM. 
August 1st, 2019, 09:35 AM  #8 
Senior Member Joined: Dec 2015 From: somewhere Posts: 592 Thanks: 87 
$\displaystyle f(n\rightarrow \infty )=f(n\rightarrow n^{2} ).$ $\displaystyle \lim_{n\rightarrow \infty }=\frac{1+1/2+...+1/n^{2}}{\ln(n^{2})}=\frac{\pi^{2}}{12} \cdot \lim_{n\rightarrow \infty }\frac{1}{\ln(n)}=0$. 
August 1st, 2019, 09:42 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,921 Thanks: 2203 
That's incorrect. Read about the Eulerâ€“Mascheroni constant.

August 1st, 2019, 09:50 AM  #10 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond 
The limit is 1. If you admit that the harmonic series has an upper bound of $\log(n)+1$ the rest is trivial.


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