Let (an) be the sequence given byan = 1√2n + 1 − 1√2nProve from first principles that (an) is Cauchy. [5 marks](b) Suppose that (bn) is a sequence of integers which is Cauchy. Prove that(bn) is eventually constant, i.e. bn = bm for all sufficiently large m and

(a) A sequence (an) is Cauchy if for every positive number ε, there exists a positive integer N such that for all natural numbers m, n > N, the absolute difference |am - an| < ε.

Given the sequence an = 1/√2n + 1 - 1/√2n, let's prove it's Cauchy.

First, we need to find the absolute difference |am - an|:

|am - an| = |1/√2m + 1 - 1/√2m - (1/√2n + 1 - 1/√2n)|

Simplify this to:

|am - an| = |(1/√2m + 1 - 1/√2n + 1)|

This can be further simplified to:

|am - an| = |(√2n - √2m) / (√2m*√2n)|

As m, n > N, we can make the difference as small as we want by choosing a sufficiently large N. Therefore, the sequence (an) is Cauchy.

(b) If a sequence (bn) of integers is Cauchy, then for every positive number ε, there exists a positive integer N such that for all natural numbers m, n > N, the absolute difference |bm - bn| < ε.

Since (bn) is a sequence of integers, the absolute difference |bm - bn| is always an integer. If we choose ε to be less than 1, then for all m, n > N, |bm - bn| must be 0, because it's the only integer less than 1. This means bm = bn for all m, n > N, so the sequence is eventually constant.