# Asymptotic analysis of $e^{H_n}$

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.

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

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 $$n\to +\infty$$
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 $$e^{H(n)}\sim ne^\gamma$$