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.

To prove that (|cn|) → ∞, we can use the definition of a sequence tending to infinity. This states that for every real number M, there exists a natural number N such that for all n > N, |cn| > M.

Given that |cn| > C|cn−1| for all n ≥ 1, where C > 1, we can say that |cn| > C^n|c0| for all n ≥ 1. This is because we can multiply the inequalities |cn| > C|cn−1|, |cn−1| > C|cn−2|, ..., |c1| > C|c0| to get |cn| > C^n|c0|.

Now, since C > 1, C^n tends to infinity as n tends to infinity. Therefore, |cn| > C^n|c0| tends to infinity as n tends to infinity. This means that for every real number M, we can find a natural number N such that for all n > N, |cn| > M. Hence, (|cn|) → ∞.

However, this does not imply that either (cn) → ∞ or (cn) → −∞. This is because the sequence (cn) could oscillate between positive and negative values. For example, consider the sequence (-2)^n. This sequence satisfies the condition |cn| > C|cn−1| for all n ≥ 1, where C = 2. However, it does not tend to ∞ or −∞, but instead oscillates between positive and negative values. Therefore, even though (|cn|) → ∞, it does not follow that either (cn) → ∞ or (cn) → −∞.