A geometric series has ratio r with |r|<1. Which statement is true?

• A. It converges to a/(1 - r) for all a with |r|<1
• B. It converges to a/(1 - r) only if a ≠ 0
• C. It diverges
• D. It converges only if r > 0

Answer: (A)

When you have a geometric series with first term a and common ratio r, the sum exists and equals a/(1 − r) whenever |r| < 1. This comes from the partial sums S_n = a(1 − r^n)/(1 − r); as n grows, r^n → 0 if |r| < 1, so S_n → a/(1 − r). Since |r| < 1 is given, the series converges to a/(1 − r) for all a, including a = 0 (which gives 0, matching the formula). It also converges when r is negative, as long as |r| < 1, so requiring r > 0 is not necessary. The statement about divergence is incorrect under the given condition.