Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 1 / 4F (Analysis I)

Let $s_N={\sum }_{n=1}^N(-1{)}^{n-1}{b}_n$. Pairing consecutive terms shows $s_{2m}\ge 0$ andThe even partial sums increase and the odd partial sums decrease. They satisfy $s_{2m}\le s_{2m+1}\le b_1$, whileBoth are monotone bounded sequences and hence have limits, and the displayed difference makes their limits equal. Every partial sum belongs to one subsequence, so the alternating series converges. This is a proof of the alternating series test; the monotonicity and zero-limit hypotheses play separate roles.

The summand is positive, so it suffices to show that its partial sums are unbounded. On the block $2^k\le n<2^{k+1}$, for $k\ge 1$,There are $2^k$ terms in that block, so its sum is at least $1/[2(k+1)\log 2]$. Adding blocks gives a constant multiple of the divergent harmonic series. HenceEquivalently the integral test for convergence gives ${\int }_2^{R}dx/(x\log x)=\log \log R-\log \log 2\to \infty$. The block proof supplies the divergence without assuming the integral test.