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

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: