Let $n\in\mathbb{Z}^{+}$ and $\displaystyle S_n = \sum_{k=1}^{n}\frac{n-k+1}{k}$. Finding $\Theta(S_n)$
PS: I found $\mathcal{O}(S_n) = n^2$. Thus, having $(n-k+1)/k = (n+1)/k -1 \leq n$.
$\rightarrow S_n = \sum_{k = 1} ^ {n}n = n^2$. But I cant find $\mathcal{\Omega}(S_n)$, so I cant also find $\mathcal{\Theta}(S_n)$.