Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics II / 3

A series ${\sum }_{n=1}^{\infty }{a}_n$ converges if its partial sums $s_N={\sum }_{n=1}^N{a}_n$ have a finite limit of a sequence $S$ as $N\to \infty$; its sum is then $S$. It has absolute convergence if the series ${\sum }_{n=1}^{\infty }|a_n|$ converges, where $|a_n|$ is the absolute value of the term.

Assume first that the original nonnegative series converges to $S$. Its partial sums $s_m$ increase to $S$. Write $t_N={\sum }_{n=1}^N{a}_{g(n)}$ for the partial sums of the rearrangement of a series. Since $g$ is an injection, the indices $g(1),\dots ,g(N)$ are distinct, soThus $(t_N)$ is an increasing bounded sequence, and the bounded monotone sequence theorem gives $t_N\to T\le S$.

For any fixed $m$, surjectivity of $g$ givesFor $N\ge N_m$, nonnegativity implies $t_N\ge s_m$. Taking limits of sequences first in $N$ and then in $m$ gives $T\ge S$. Hence $T=S$.

Conversely, $g^{-1}$ is a bijection, and the original series is the rearrangement of a series of $\sum b_n$ defined by $g^{-1}$. The same argument proves its convergence whenever $\sum b_n$ converges. Both series converge together, and their sums are equal.

In level $n$, each group isand there are $2^n$ such groups. The partial sums at completed groups are therefore zero. Inside a group of level $n$, the only other partial sums are $2^{-(n+1)}$ and $2^{-n}$. Every level contains finitely many terms, so the level containing the $N$th term tends to infinity with $N$. Thus all partial sums lie between zero and a bound tending to zero. By the squeeze theorem, the series has sum $\boxed{0}$.

For the next two parts, label the positive occurrences of size $2^{-(n+1)}$ by $P_n$ and the negative occurrences of size $-2^{-n}$ by $N_n$. There are exactly $2^{n+1}$ occurrences in $P_n$ and $2^n$ in $N_n$. Occurrences with equal value retain separate labels, so checking their counts verifies a bijection of terms.

Place the single term in $N_0$ first. For $n=0,1,2,\dots$ in order, append $2^{n+1}$ copies of the pairThis uses all of $P_n$ and all of $N_{n+1}$ exactly once, so it is an explicit rearrangement of a series. Its beginning isEvery completed pair leaves the partial sum at $-1$. Within a level-$n$ pair the partial sum differs from $-1$ by at most $2^{-(n+1)}$. As $n\to \infty$ this tends to zero, proving that the rearranged series has sum $\boxed{-1}$.

Start with all of $N_0$ and $N_1$, namely $-1,-1/2,-1/2$, whose sum is $-2$. For $n=0,1,2,\dots$ in order, append $2^{n+1}$ copies of the tripleThis uses all $2^{n+1}$ occurrences of $P_n$ and all $2^{n+2}$ occurrences of $N_{n+2}$, exactly once. Together with the initial terms, every original occurrence is used, establishing a rearrangement of a series. Its beginning isEach triple has sum zero. At its intermediate terms the partial sum is at most $2^{-(n+1)}$ above $-2$. This bound tends to zero, so the whole series, rather than just the subsequence of sums at completed triples, has sum $\boxed{-2}$.