Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 4 / 7E / a

Every convergent sequence is bounded. Conversely, let $(x_n)$ be increasing and bounded above, and set $L=\sup \{x_n:n\ge 1\}$. For any $\epsilon >0$, $L-\epsilon$ is not an upper bound, so some $x_N>L-\epsilon$. Monotonicity then givesfor all $n\ge N$. Hence $|x_n-L|<\epsilon$, proving the monotone bounded sequence criterion.

LetFor $n\ge 3$, the binomial theorem givesConsequently $(1000n^2+4)/1.1^n\to 0$: given $\epsilon >0$, the displayed rational upper bound is below $\epsilon$ for all sufficiently large $n$. Thus, directly from the definition of convergence,