Prove from first principles that the sequence (an) is convergent, wherean = (−1)n + n2n + 1for all n. [5 marks](b) Suppose that a sequence (cn) satisfies |cn| > C|cn−1| for all n ≥ 1, whereC > 1. Show that (|cn|) → ∞. Does it follow that either (cn) → ∞ or(cn) → −∞? Justify your answer. [6 marks](c) If |dn| > |dn−1| for all n, does it follow that (|dn|) → ∞? Justify youranswer with a proof or a counterexample.
Question
Prove from first principles that the sequence (an) is convergent, wherean = (−1)n + n2n + 1for all n. 5 marks Suppose that a sequence (cn) satisfies |cn| > C|cn−1| for all n ≥ 1, whereC > 1. Show that (|cn|) → ∞. Does it follow that either (cn) → ∞ or(cn) → −∞? Justify your answer. 6 marks If |dn| > |dn−1| for all n, does it follow that (|dn|) → ∞? Justify youranswer with a proof or a counterexample.
Solution
(a) To prove that the sequence (an) is convergent, we need to show that it has a limit as n approaches infinity. The sequence is given by an = (−1)n + n2n + 1.
First, let's consider the sequence as n approaches infinity. The term (−1)n oscillates between -1 and 1 for even and odd n respectively. However, as n approaches infinity, the term n2n + 1 grows without bound. Therefore, the sequence (an) does not have a limit as n approaches infinity, and is therefore not convergent.
(b) If a sequence (cn) satisfies |cn| > C|cn−1| for all n ≥ 1, where C > 1, then the sequence (|cn|) is increasing at a rate faster than linear. This means that as n approaches infinity, (|cn|) also approaches infinity, so (|cn|) → ∞.
However, this does not necessarily mean that either (cn) → ∞ or (cn) → −∞. The sequence could oscillate between positive and negative values. For example, consider the sequence defined by cn = (-1)^n * n. This sequence satisfies the given condition with C = 2, but does not converge to either ∞ or −∞.
(c) If |dn| > |dn−1| for all n, then the sequence (|dn|) is strictly increasing. However, this does not necessarily mean that (|dn|) → ∞. For example, consider the sequence defined by dn = 1/n. This sequence satisfies the given condition, but converges to 0 as n approaches infinity. Therefore, it is not necessarily true that (|dn|) → ∞ if |dn| > |dn−1| for all n.
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