Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics II / 2

A sequence $(a_n)$ of real numbers is a convergent sequence with limit of a sequence $\ell$ whenDivergence to positive infinity meansIn both definitions the same $N$ must control every later term. The second condition is not convergence to a real number.

Suppose $a_n\to \ell$. With tolerance $1$, choose $N$ such that $|a_n-\ell |<1$ whenever $n\ge N$. The triangle inequality gives $|a_n|\le |a_n-\ell |+|\ell |<1+|\ell |$ for these terms. SetAn empty sum is zero. This finite $B$ satisfies $|a_n|\le B$ for every $n$, so every convergent sequence is a bounded sequence.

Put $u=c^{-1}-1>0$, so $c=(1+u{)}^{-1}$. By Bernoulli's inequality, $(1+u{)}^n\ge 1+nu$, and henceMore explicitly, for $\epsilon >0$ choose an integer $N>1/(u\epsilon )$. For $n\ge N$, the right-hand side is less than $\epsilon$. Thus the geometric sequence has limit of a sequence zero, either directly by the definition or by the squeeze theorem.

Given $\epsilon >0$, the two assumed limits of sequences supply thresholds $N_1,N_2$ such that $|a_n-\ell |<\epsilon$ for $n\ge N_1$ and $|c_n-\ell |<\epsilon$ for $n\ge N_2$. For $n\ge \max \{N_1,N_2\}$,Thus $|b_n-\ell |<\epsilon$ eventually, proving the squeeze theorem:

False. Take the positive sequence $\boxed{a_n=n}$. Thenbut $a_n\to +\infty$, so it has no finite limit of a sequence. Successive ratios close to one do not prevent cumulative growth.

False. Take the positive geometric sequence $\boxed{a_n=2^{-n}}$. It tends to zero, whileA positive limit of a sequence would make the claim true since $a_{n+1}/a_n$ would then tend to $\ell /\ell =1$, but positivity of each individual term does not exclude a zero limit of a sequence.

True. Since $a_n^n\to 2$, there is $N_0$ such that $1<a_n^n<3$ for $n\ge N_0$. Positivity then gives $a_n>1$ on this tail.

Fix $\epsilon >0$. If $a_n\ge 1+\epsilon$, Bernoulli's inequality givesFor $n>2/\epsilon$, this contradicts $a_n^n<3$. Therefore eventually $1<a_n<1+\epsilon$, proving

True. Choose $N$ such that $\ell /2<a_n<3\ell /2$ for every $n\ge N$. All the finitely many preceding terms are also positive. Define the positive minimum and finite maximumEvery term lies between $\eta$ and $B$, so every ratio is at most $B/\eta$. Consequently the required supremum satisfies

False. Choose the positive sequenceIt has limit of a sequence one, but Bernoulli's inequality givesThus $(a_n^n)$ is not a bounded sequence.