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

For a strictly decreasing function, $f(n+1)<f(x)<f(n)$ in the interior of $[n,n+1]$. Integrating over the unit interval and summing from $n=r$ to $s-1$ gives the integral comparison for a nonincreasing nonnegative function:Geometrically, right-endpoint rectangles lie below the curve and left-endpoint rectangles above it. If “decreasing” permits flat pieces, use non-strict inequalities unless strictness is ensured on some interval. When $r=1$, the endpoint value $f(1)$ is also needed; the power function below is defined there.

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.

For fixed $p>0$, the partial sums are increasing and bounded above by $1+1/p$. The monotone convergence theorem for sequences therefore proves convergence to a finite sum $S(p)$. Passing to the limit $N\to \infty$ in the two bounds givesThe squeeze theorem now yieldsSince $S(p)=\zeta (1+p)$, this is the real-axis singular limit of the Riemann zeta function at one. The limiting sum is taken at fixed $p$ before sending $p$ to zero; setting $p=0$ inside the convergent-series argument would instead give the divergent harmonic series.