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## Evaluate the sum or explain why it diverges: Sigma^infinity_k = 3(-3/2)^k

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Answer:Step-by-step explanation:Remember that in the geometric serie if | r | < 1 the serie converges and if | r | ≥1 the serie diverges.

I suppose that the serie starts at 0, so using the geometric serie with r = | | > 1 the serie diverges.