For what values of x does the series below converge? [infinity] Σ n=1 (2x - 3)n n SOLUTION Let an = (2x - 3)n / n. Then an+1 an = · n (2x - 3)n = 1 | 2x - 3 | → as n → [infinity] By the Ratio test, the given series is absolutely convergent, and therefore convergent, when | 2x - 3 | < and divergent when | 2x - 3 | > . Now | 2x - 3 | < 1 ↔ < 2x - 3 < ↔ < x < so the series converges when < x < and diverges when x < or x > . The Ratio Test gives no information when | 2x - 3 | = 1, so we must consider x = 1 and x = 2 separately. If we put x = 2 in the serie