Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 2 / 10 / b

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