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To be precise, what is the asymptotic behavior of $e^{H_n}$, as $n$ tends to infinity, where $e$ is the Euler's number, a mathematical constant approximately equal to $2.71828$ and $H_n$ is the $n$-th Harmonic number?

Moreover, what is its limit and how fast does it grow like exponentially, sub-exponentially or super-exponentially?

Hint - Using the L'Hôpital's rule, big O notation, small o notation and the fact that $H_n \sim \ln n + \gamma + \frac1{2n}$, where $\gamma$ is the Euler-Mascheroni constant, might help in answering this question.

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    $\begingroup$ Please use MathJax syntax, so $e^{H_n}$ for $e^{H_n}$ for example $\endgroup$ Oct 22, 2023 at 13:09
  • $\begingroup$ @BenjaminWang Ok thanks a lot, am doing that. $\endgroup$ Oct 22, 2023 at 13:11

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We have $H_n=\ln(n)+\gamma+a_n$ and $a_n\to 0$ as $n\to +\infty$. So

$$e^{H_n}=e^{\ln(n)+\gamma+a_n}=ne^\gamma e^{a_n}\sim ne^\gamma$$ Here I used the fact: if $a_n\to 0$ then $e^{a_n}\to 1$ as $n\to +\infty$

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Another way of writing @GaloisGroup's answer:

Start with $$\lim_{n\rightarrow\infty}(H(n)-\ln n)=\gamma$$exponentiate:$$\lim_{n\rightarrow\infty}\frac{e^{H(n)}}n=e^\gamma\quad\text{ or }\quad\frac{e^{H(n)}}n\sim e^\gamma$$Multiplying by $n$, we get $e^{H(n)}\sim ne^\gamma$

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