If (xn) and (yn) are two Cauchy sequences, prove from first principles that(xn + yn) is Cauchy. [6 marks](b) Show that (an) is Cauchy, where an = 2n/3n. Clearly state any resultsused from lectures or the Supplementary Materials

Sure, here is the step by step solution:

(a) Proof that (xn + yn) is Cauchy:

1. By definition, a sequence (xn) is Cauchy if for every ε > 0, there exists a natural number N such that for all natural numbers m, n > N, |xm - xn| < ε.

2. Given that (xn) and (yn) are Cauchy, for every ε > 0, there exist natural numbers N1 and N2 such that for all natural numbers m, n > N1, |xm - xn| < ε/2 and for all natural numbers m, n > N2, |ym - yn| < ε/2.

3. Let N = max(N1, N2). Then for all natural numbers m, n > N, we have |xm - xn| < ε/2 and |ym - yn| < ε/2.

4. Now consider the sequence (xn + yn). For all natural numbers m, n > N, |(xm + ym) - (xn + yn)| = |(xm - xn) + (ym - yn)|.

5. By the triangle inequality, |(xm - xn) + (ym - yn)| ≤ |xm - xn| + |ym - yn| < ε/2 + ε/2 = ε.

6. Therefore, (xn + yn) is a Cauchy sequence.

(b) Show that (an) is Cauchy, where an = 2n/3n:

1. The sequence (an) can be rewritten as (2/3)^n.

2. This is a geometric sequence with common ratio r = 2/3, where |r| < 1.

3. From the properties of geometric sequences, we know that such a sequence is always Cauchy.

4. Therefore, (an) is a Cauchy sequence.