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Show from first principles that (an) is convergent, where an = cos(n) sin (n)/nfor all n ≥ 1. [3 marks](b) Let ℓ, m ∈ R and suppose that (cn) and (dn) are sequences such that(cn + dn) → ℓ + m. Show that (cn) → ℓ if and only if (dn) → m. Anyresults you use must be accurately stated. [4 marks](c) If (xn + yn) is a convergent sequence, must (xn − yn) also be convergent?Give a proof or a counterexample, with justification.

Question

Show from first principles that (an) is convergent, where an = cos(n) sin (n)/nfor all n ≥ 1. 3 marks Let ℓ, m ∈ R and suppose that (cn) and (dn) are sequences such that(cn + dn) → ℓ + m. Show that (cn) → ℓ if and only if (dn) → m. Anyresults you use must be accurately stated. 4 marks If (xn + yn) is a convergent sequence, must (xn − yn) also be convergent?Give a proof or a counterexample, with justification.

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Solution

(a) To show that the sequence (an) is convergent, we need to show that it satisfies the definition of a convergent sequence, i.e., for every positive number ε, there exists a positive integer N such that for all n ≥ N, |an - L| < ε, where L is the limit of the sequence.

The sequence an = cos(n) sin(n)/n is a product of a bounded sequence (cos(n) sin(n)) and a sequence (1/n) that converges to 0. By the limit laws, the product of a bounded sequence and a sequence that converges to 0 also converges to 0. Therefore, the sequence (an) is convergent.

(b) Suppose that (cn + dn) → ℓ + m. This means that for every positive number ε, there exists a positive integer N such that for all n ≥ N, |(cn + dn) - (ℓ + m)| < ε.

If (cn) → ℓ, then for every positive number ε, there exists a positive integer N1 such that for all n ≥ N1, |cn - ℓ| < ε. Similarly, if (dn) → m, then for every positive number ε, there exists a positive integer N2 such that for all n ≥ N2, |dn - m| < ε.

Therefore, if (cn) → ℓ and (dn) → m, then (cn + dn) → ℓ + m. Conversely, if (cn + dn) → ℓ + m, then (cn) → ℓ and (dn) → m.

(c) The sequence (xn + yn) being convergent does not necessarily imply that (xn - yn) is also convergent. For example, consider the sequences xn = n and yn = -n. Then (xn + yn) = 0 for all n, which is a convergent sequence. However, (xn - yn) = 2n, which is a divergent sequence. Therefore, (xn + yn) being convergent does not necessarily imply that (xn - yn) is also convergent.

This problem has been solved

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