Final answer:
To show that A = 0, we consider the sequences aₙ → A and {(−1)ⁿ aₙ}[infinity] 1. By the definition of convergence, for any ε > 0, there exists a positive integer N such that for all n ≥ N, |aₙ - A| < ε. Using the triangle inequality and the fact that the sequence {(−1)ⁿ aₙ} is convergent, we can show that A = 0.
Step-by-step explanation:
Proof:
Let aₙ → A and {(−1)ⁿ aₙ}[infinity] 1 be convergent. By definition of convergence, for any given ε > 0, there exists a positive integer N such that for all n ≥ N, |aₙ - A| < ε.
Now, consider the sequence {(−1)ⁿ aₙ}. Since this sequence is convergent, there exists a positive integer M such that for all n ≥ M, |(−1)ⁿ aₙ - L| < ε, where L is the limit of this sequence.
Since aₙ → A and {(−1)ⁿ aₙ}[infinity] 1 are both convergent, we can choose N and M such that N > M. Then, for all n ≥ N, we have:
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- |aₙ - A| < ε
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- |(−1)ⁿ aₙ - L| < ε
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- |aₙ - A + A - L| < ε (by the triangle inequality)
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- |(−1)ⁿ aₙ - L + A - L| < ε (by the triangle inequality)
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- |((−1)ⁿ aₙ + A) - 2L| < ε
Let B = ((−1)ⁿ aₙ + A). Since B is a convergent sequence, we can rewrite the above inequality as |B - 2L| < ε.
Because ε can be arbitrarily small, we must have B - 2L = 0, which means B = 2L. But B = ((−1)ⁿ aₙ + A), so we can conclude that 2L = ((−1)ⁿ aₙ + A), or L = ((−1)ⁿ aₙ + A)/2. Since the sequence {(−1)ⁿ aₙ} converges, we know that L is a finite number. Therefore, to have L = ((−1)ⁿ aₙ + A)/2, we must have A = 0.