Question:
Does $ \lim_{n \to \infty} \frac{\operatorname{exp}(H_n)}{n+1} $ exist? If so, what is its value?
I know that the answer to the second part is $e^\gamma$, where $\gamma$ is the Euler-Mascheroni constant, but I don't know how to get there.
The solution probably involves the definition of $\gamma$: $$\gamma = \lim_{n\to\infty}\left( \sum_{k=1}^n \frac{1}{k} -\ln n\right)$$
\begin{align} \lim_{n\to\infty}\frac{\exp(H_n)}{n+1} &= \lim_{n\to\infty}\frac{\exp(H_n)}{\exp\ln(n+1)} \\ &= \lim_{n\to\infty}\left(\exp(H_n - \ln(n+1)\right)\\ &=\exp\left(\lim_{n\to\infty}\left(H_n - \ln(n+1)\right)\right)\tag{$\star$}\\ &=\exp(\gamma) \end{align} The move to line $(\star)$ is justified because $\exp(\cdot)$ is a continuous function.