Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics II / 1

For a nonempty set $A\subseteq \mathbb{R}$, a supremum is a real number $s$ such that $a\le s$ for every $a\in A$, and every upper bound $u$ of $A$ satisfies $s\le u$. Equivalently, $s$ is an upper bound and, for every $\epsilon >0$, some $a\in A$ satisfies $a>s-\epsilon$. By completeness of the real numbers, it exists whenever $A$ is nonempty and bounded above.

An increasing monotone sequence satisfies $a_{n+1}\ge a_n$ for every $n$; a bounded sequence satisfies $|a_n|\le M$ for some finite $M$ and every $n$. Put $s=\sup \{a_n:n\ge 1\}$. Given $\epsilon >0$, the defining property of the supremum supplies $N$ with $a_N>s-\epsilon$. For every $n\ge N$, monotonicity givesThis is exactly the definition of convergent sequence, so the bounded monotone sequence theorem gives $\boxed{a_n\to s}$.

Choose $M$ with $|b_n|\le M$. Every tail of a sequence is nonempty and bounded, so its supremum $c_n$ exists and lies in $[-M,M]$. Removing one term cannot increase a supremum, hence $c_{n+1}\le c_n$. The bounded monotone sequence theorem, applied to $-c_n$, provesThus $L$ is the limit superior of $(b_n)$.

For $\epsilon >0$, eventually $c_n<L+\epsilon$, and then $b_n\le c_n<L+\epsilon$. Only finitely many indices can satisfy $b_n\ge L+\epsilon$. If only finitely many satisfied $b_n>L-\epsilon$, some entire tail of a sequence would have $b_n\le L-\epsilon$. Its supremum would also be at most $L-\epsilon$, contradicting $c_n\ge L$.

Inductively choose $n_k>n_{k-1}$ so late that $c_{n_k}<L+1/k$ and $b_{n_k}>L-1/k$. There are infinitely many choices for the latter inequality, so both requirements can be met. Then $|b_{n_k}-L|<1/k$, proving $\boxed{b_{n_k}\to L}$. This constructs a subsequence attaining the limit superior.

At even integers $n=2k$, the terms are $1+1/(2k)\to 1$. At odd integers they are $-1-1/(2k-1)\to -1$. The largest eventual values therefore come from the subsequence at even integers, andExplicitly, every sufficiently late term is below $1+\epsilon$, while infinitely many even integers give terms greater than $1-\epsilon$.

The sine has values $1,0,-1,0$ on the four residue classes of $n$ modulo four. For $n=4k+1$, the exponential function gives $b_n=e^{1-1/n}\to e$; the other residue classes tend to $1,e^{-1},1$. All terms satisfy $b_n\le e$, and the first class approaches $e$. Hence