Solution
= Solution
For $f(x)=x^{-1-p}$ with $p>0$, the <integral comparison> on $[1,N]$ gives $\sum_{n=2}^Nn^{-1-p}<I_N<\sum_{n=1}^{N-1}n^{-1-p}$, where $I_N=(1-N^{-p})/p$. Add the missing endpoint term on each side to obtain
$$
\boxed{\frac{1-N^{-p}}p+N^{-1-p}<\sum_{n=1}^Nn^{-1-p}<\frac{1-N^{-p}}p+1.}
$$
The lower bound uses the right inequality for $I_N$ and adds $N^{-1-p}$; the upper uses the left inequality and adds one.