Why is $\sum_{n \leq x }1/n=\log(x)+\mathcal{O(1)}$?

Asked Mar 14 '20 at 11:47

Active Jul 05 '23 at 09:49

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I know I can compare $\sum_{n \leq x}1/n$ with $\log(x).$ I also think $\log(x) \leq\sum_{n \leq x}1/n.$

edited Jul 05 '23 at 09:49

User1865345

asked Mar 14 '20 at 11:47

AnabolicHorse

1

Or this one: https://math.stackexchange.com/q/3051851/42969 – Martin R Mar 14 '20 at 11:51
@Martin R , thanks for the help – AnabolicHorse Mar 14 '20 at 11:54

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