Question

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

Answer 1

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 = | $\frac{-3}{2}$ | > 1 the serie diverges.