Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 1 / 9F (Analysis I)

The comparison test for series says that an eventual upper bound by a convergent positive series proves convergence, while an eventual lower bound by a divergent positive series proves divergence. The p-series $\sum n^{-s}$ converges exactly when $s>1$.

If $p>1$, then $(\log n{)}^q\ge 1$ for all sufficiently large $n$, soand the series converges. If $0<p<1$, choose $\eta =(1-p)/2>0$. The allowed logarithmic bound, applied with exponent $\eta /q$, gives $(\log n{)}^q<n^{\eta }$ eventually. Thusand the series diverges.

For $p=1$, use the integral test for convergence: for an eventually positive decreasing function $g$, $\sum g(n)$ and $\int g(x)dx$ either both converge or both diverge. Here $g(x)=1/[x(\log x{)}^q]$ is decreasing for $x>1$, and the substitution $u=\log x$ givesThis integral converges exactly when $q>1$. Hence the polynomial-logarithmic series convergence classification isAll other allowed positive $p,q$ give divergence.

For any fixed $r>0$, $\log y=o(y^{1/r})$ as $y\to \infty$. For example, l'Hôpital's rule applied to $\log y/y^{1/r}$ gives the limit zero. With $y=\log n$, this implies $(\log \log n{)}^r<\log n$ eventually. Thereforefor all sufficiently large $n$. The comparison series diverges by the integral test for convergence, sinceThe comparison test for series givesOne may obtain the same growth bound by extending the given logarithmic estimate to real $y$ through $\lceil y\rceil$ and choosing any exponent smaller than $1/r$; no special convergence test for iterated logarithms is needed.