Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 1 / 9C (Analysis I)

Use the least-upper-bound property as the least upper bound axiom: every nonempty set of real numbers bounded above has a least upper bound in $\mathbb{R}$. It implies convergence of a bounded increasing sequence: if $L$ is the supremum of its terms, then for every $\epsilon >0$ some term exceeds $L-\epsilon$, and all subsequent terms lie between that term and $L$. A bounded decreasing sequence converges by applying this argument to its negative.

The alternating series test states that if $b_n\ge 0$, $b_{n+1}\le b_n$, and $b_n\to 0$, then ${\sum }_{n\ge 1}(-1{)}^{n-1}{b}_n$ converges. To prove it, let $S_N$ denote its partial sums. The even partial sums are increasing because $S_{2m+2}-S_{2m}=b_{2m+1}-b_{2m+2}\ge 0$. The odd partial sums are decreasing because $S_{2m+3}-S_{2m+1}=-b_{2m+2}+b_{2m+3}\le 0$. Moreover,Completeness of the real numbers therefore gives a limit of a sequence for each subsequence, denoted $L_e,L_o$ respectively for the even and odd subsequences. Their difference is $S_{2m+1}-S_{2m}=b_{2m+1}\to 0$, so $L_e=L_o$. Every sufficiently late partial sum belongs to one of these two subsequences, proving convergence of the full sequence of partial sums. Multiplying the terms by $-1$ leaves convergence unchanged.

The positive terms $b_n=1/\log (n+1)$ decrease to zero. The alternating series test just proved therefore gives convergence of the signed series. It is conditional convergence: $\log (n+1)\le n$, so $b_n\ge 1/n$, and the absolute-value series dominates the divergent harmonic series.

The comparison test for series used here says that if $0\le c_n\le d_n$, convergence of $\sum d_n$ implies convergence of $\sum c_n$, while divergence of $\sum c_n$ implies divergence of $\sum d_n$. This follows by comparing increasing partial sums. For completeness, the harmonic series diverges because every block from $2^k$ to $2^{k+1}-1$ has sum at least $1/2$.

The absolute-value series can be grouped in pairs of nonnegative terms:Indeed, $1/n^2\le {\int }_{n-1}^n{x}^{-2}dx$ for $n\ge 2$, so these positive partial sums are bounded by two. Completeness of the real numbers proves their convergence. The triangle inequality bounds every late signed sum by the corresponding late absolute-value sum, so absolute convergence implies convergence. Thus the series converges absolutely.

Group three consecutive terms. Their block sum isSince $2n-1\ge n$ and $4n-1\ge 3n$ for $n\ge 1$, we have $|b_n|\le 1/(2n^2)$. The comparison test for series and the preceding proof of convergence of $\sum n^{-2}$ show that $\sum b_n$ converges absolutely.

This establishes convergence of the original partial sums at indices $3N$. Each intermediate partial sum differs from one of these by at most two of the original terms; those terms tend to zero. Thus all partial sums have the same limit of a sequence, which is the mechanism of convergence from fixed-length series blocks. The original series is not absolutely convergent, since its absolute-value series contains ${\sum }_n(2n-1{)}^{-1}\ge \frac{1}{2}{\sum }_n{n}^{-1}$. Therefore the original series converges conditionally.

Let $S_N={\sum }_{k=1}^N{a}_k$, with $S_0=0$. The indices in the $n$th dyadic block have one sign, soA convergent sequence of partial sums must be a Cauchy sequence: for every $\epsilon >0$, all sufficiently late pairs of partial sums differ by less than $\epsilon$. The displayed differences violate this necessary condition with $\epsilon =1/4$, at arbitrarily late endpoints. Thus the series diverges, even though its individual terms tend to zero. This dyadic-block alternation of reciprocal terms has growing block lengths; the fixed-length block argument of the previous part does not apply.