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 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 Thanks from romsek
 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. Thanks from idontknow
August 1st, 2019, 06:48 AM   #4
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 Originally Posted by DarnItJimImAnEngineer Numerator approaches 1. Denominator increases towards infinity. So 0.
Numerator definitely does not approach 1.

 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 $n \to \infty$ Thanks from idontknow
 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 Thanks from idontknow
 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. Thanks from idontknow
 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. Thanks from idontknow

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