Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics II / 2

If $L<1$, set $\epsilon =(1-L)/2>0$. The definition of limit of a sequence gives $a_n<L+\epsilon =(1+L)/2<1$ for all sufficiently large $n$, contradicting $a_n\ge 1$. Hence $\boxed{L\ge 1}$.

Factor the difference of cubes. Since both cube roots are at least one,Given $\epsilon >0$, the definition of limit of a sequence supplies $N$ with $|a_n-L|<3\epsilon$ for $n\ge N$. The displayed inequality then gives $|a_n^{1/3}-L^{1/3}|<\epsilon$, proving directly that $\boxed{a_n^{1/3}\to L^{1/3}}$.

Let $\epsilon >0$. By the definition of Cauchy sequence, choose an integer $K\ge 1$ such that $|x_j-x_k|<\epsilon /2$ whenever $j,k\ge K$. Apply the given inequality with $N=K+1$. It supplies $n>K+1$ for whichBoth $n$ and $K$ are in the Cauchy sequence estimate, soFor every $m\ge K$ the triangle inequality givesThis is the definition of convergence to zero: $\boxed{x_m\to 0}$. The proof uses only the Cauchy sequence estimate, positivity, and the stipulated inequality; it does not assume the general completeness of the real numbers theorem for Cauchy sequences. Taking $N=K+1\ge 2$ also avoids the empty initial set in a literal reading of the condition for very small $N$.

Put $u_n=11^n\log n/(2^{2n+1}n!)$ for $n\ge 2$. The factorial givesThe logarithm identityfollows from continuity of the logarithm at one and $\log n\to +\infty$. The product rule for limits of sequences therefore makes the displayed ratio tend to zero. In particular, some $N$ satisfies $0<u_{n+1}\le u_n/2$ for all $n\ge N$. By mathematical induction, $0<u_{N+j}\le u_N{2}^{-j}$. A geometric sequence with ratio in $(0,1)$ tends to zero, so the squeeze theorem proves

The sequence does not converge. For $n=2^k$, the largest prime factor is $p(n)=2$, while the sum of prime factors with multiplicity is $s(n)=2k$. This subsequence therefore has ratio $1/k\to 0$.

There are infinitely many prime numbers: if there were only $q_1,\dots ,q_r$, a prime factor of $q_1\cdots q_r+1$ would be absent from the list. Listing the prime numbers in increasing order gives a subsequence of indices $n=q_j$ for which $p(n)=s(n)=q_j$, so the ratio is constantly one. Every subsequence of a convergent sequence has the same limit of a sequence. The two distinct subsequential limits $\boxed{0\text{ and }1}$ rule out convergence.