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August 1st, 2019, 02:20 AM   #1
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Evaluate the limit

$\displaystyle \lim_{n\rightarrow \infty } \frac{1+1/2 +...+1/n }{\ln(n)}.$
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August 1st, 2019, 06:07 AM   #2
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August 1st, 2019, 06:35 AM   #3
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Numerator approaches 1. Denominator increases towards infinity. So 0.
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August 1st, 2019, 06:48 AM   #4
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Quote:
Originally Posted by DarnItJimImAnEngineer View Post
Numerator approaches 1. Denominator increases towards infinity. So 0.
Numerator definitely does not approach 1.
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August 1st, 2019, 06:55 AM   #5
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numerator is the $n$th harmonic number which is known to diverge as $n \to \infty$
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August 1st, 2019, 07:21 AM   #6
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https://en.wikipedia.org/wiki/Harmon..._of_divergence
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August 1st, 2019, 07:35 AM   #7
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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.
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August 1st, 2019, 09:35 AM   #8
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$\displaystyle f(n\rightarrow \infty )=f(n\rightarrow n^{2} ).$
$\displaystyle \lim_{n\rightarrow \infty }=\frac{1+1/2+...+1/n^{2}}{\ln(n^{2})}=\frac{\pi^{2}}{12} \cdot \lim_{n\rightarrow \infty }\frac{1}{\ln(n)}=0$.
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August 1st, 2019, 09:42 AM   #9
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That's incorrect. Read about the Euler–Mascheroni constant.
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August 1st, 2019, 09:50 AM   #10
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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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