Does the geometric series ∑_{n=0}^∞ (1/3)^n converge, and if so, what is its sum?

Does the geometric series ∑_{n=0}^∞ (1/3)^n converge, and if so, what is its sum?

• A. Diverges
• B. Converges; sum = 2
• C. Converges; sum = 1
• D. Converges; sum = 3/2

Answer: (D)

When a geometric series has a ratio with absolute value less than 1, it converges to a finite sum. Here the first term is 1 and the common ratio is 1/3, which indeed satisfies |r| < 1, so the series converges. The sum of a geometric series is S = a / (1 − r), where a is the first term. Substituting a = 1 and r = 1/3 gives S = 1 / (1 − 1/3) = 1 / (2/3) = 3/2. So the series converges and its sum is 3/2.

When a geometric series has a ratio with absolute value less than 1, it converges to a finite sum. Here the first term is 1 and the common ratio is 1/3, which indeed satisfies |r| < 1, so the series converges. The sum of a geometric series is S = a / (1 − r), where a is the first term. Substituting a = 1 and r = 1/3 gives S = 1 / (1 − 1/3) = 1 / (2/3) = 3/2. So the series converges and its sum is 3/2.