Let $f(n) = \sum_{i=1}^n i^{-1}$.
Would the simplest $g(n)$ be $1$?
If we let $g(n) = 1$, then for all $n > 1$, $g(n) \le f(n)$.
To construct an upper bound, observe that since $n$ is finite, $f(n) = C$ for some $C \in \mathbb R$. (Here $C$ is the $n$th harmonic number). Then by the Archimedean Property for real numbers, $\exists N \in \mathbb N$ such that $N > C$. So for all $n > N$, $f(n) \le N.g(n)$. Combining the two bounds we have that $f(n) \in \Theta(g(n))$.
Please let me know if what I have done is valid.