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Does there exist an alternative way to demonstrate and evaluate the result of the following limit without using an asymptotic expansion? $$\lim_{n\rightarrow+\infty}\dfrac{\displaystyle\sum_{k=1}^{n}\dfrac{1}{k}}{\ln n}$$ Using a spreadsheet presumably the result of the limit is $1$, and watching other posts like this and this related about Euler-Mascheroni constant $\gamma$, the result is obtained immediately. $${\displaystyle \lim_{n\rightarrow+\infty}\dfrac{\displaystyle\sum_{k=1}^{n}\dfrac{1}{k}}{\ln n}=\lim_{n\rightarrow+\infty}\dfrac{\ln n+\gamma+O\begin{pmatrix}\dfrac{1}{n}\end{pmatrix}}{\ln n}}=1$$ Is that correct? However, I would like to know if there is other way to calculate that result using calculus in a context of firsts courses of engineering careers. Maybe with convergence criteria for series?

Martin R
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Jordi Olmos
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    Compare the sum to $\int_a^b \frac{dx}{x}$, with suitable values for $a$ and $b$ (to get an upper and a lower bound). – Hans Lundmark Jun 16 '21 at 14:34
  • See also these questions, for example: https://math.stackexchange.com/questions/1564087/prove-lnn-diverges-as-quickly-as-the-harmonic-series/1564106, https://math.stackexchange.com/questions/1733438/asymptotic-equivalence-between-harmonic-numbers-and-logarithm – Hans Lundmark Jun 16 '21 at 14:37
  • Have a look at https://math.meta.stackexchange.com/q/9687/42969 about (not!) using \displaystyle and \dfrac in titles. – Martin R Jun 16 '21 at 14:37

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