Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 2 / 10

The function $x^{-3}$ is strictly decreasing for positive $x$. On $a<x<a+1$, this gives $x^{-3}<a^{-3}<(x-1{)}^{-3}$. Integrate across that unit interval to obtainApplying this to every integer $n\ge a$ and summing proves the integral test tail boundsBoth remain strict: the sum of the positive differences between rectangles and integrals is already bounded below by its positive first difference. Convergence of the integrals and the p-series makes the passage to infinity valid.

The shifted integral representation of a p-series tail evaluates the defining integral asIt is strictly increasing in $k$, since $I_p^{\prime }(k)=(5-k{)}^{-p}>0$. From the bounds just proved with $a=5$, $I_3(0)<{\sum }_{n=5}^{\infty }{n}^{-3}<I_3(1)$. Since the middle expression equals $I_3(k_3)$, monotonicity proves

Subtract the first four terms from the given full p-series sums. WriteandSolving $I_p(k_p)=S_p$ yields $k_p=5-[(p-1)S_p{]}^{-1/(p-1)}$. ThereforeThe unrounded shifts are approximately $0.4817158633$ and $0.4638861447$.

Use the stipulated interpolation $k_3\approx (k_2+k_4)/2$. The unrounded values give $k_3\approx 0.4728010040$, and henceThus a sensible reported estimate isUsing the rounded shifts from part (ii) instead gives $k_3\approx 0.473$ and a sum about $1.202059745$, with the same six-significant-figure estimate. The interpolation of shifts is an assumption of this approximation; the integral test bounds alone do not prove that many digits of accuracy.