For $p>1$, $\displaystyle \sum_{n=1}^{\infty} \dfrac1{n^p}$ converges i.e. $\displaystyle \sum_{n=1}^{N} \dfrac1{n^p} = \mathcal{O}(1)$
For $p=1$, $\displaystyle \sum_{n=1}^{\infty} \dfrac1{n}$ diverges like $\log n$ i.e. $\displaystyle \sum_{n=1}^{N} \dfrac1{n} = \log(N) + \gamma + \mathcal{O}(1/N)$
For $p<1$, $\displaystyle \sum_{n=1}^{\infty} \dfrac1{n^p}$ diverges like $N^{1-p}$ i.e. $\displaystyle \sum_{n=1}^{N} \dfrac1{n^p} = \dfrac{N^{1-p}}{1-p} + $ lower order terms.
All these can be obtained by comparing $\displaystyle \sum_{n=1}^{\infty} \dfrac1{n^p}$ with $\displaystyle \int_{x=1}^{\infty} \dfrac{dx}{x^p}$
You can make use of Euler-Maclaurin to get more accurate expansions and better error bounds.