What is the sum of the geometric series sum_{n=1}^∞ (1/3)^n?
A3/2
B1
C1/2
D3
Starting at n=1 means the first term is (1/3)^1 = 1/3, not 1. So a = 1/3 and r = 1/3, giving sum = (1/3)/(1 - 1/3) = (1/3)/(2/3) = 1/2. A very common error is to use a = 1 (the n=0 term), which gives 3/2 — this is the off-by-one index mistake.
Question 2 True / False
A geometric series with ratio r = -0.5 converges because |-0.5| < 1.
TTrue
FFalse
Answer: True
The convergence condition is |r| < 1, which applies to negative ratios as well. Here |-0.5| = 0.5 < 1, so the series converges to a/(1-r) = a/1.5. Negative ratios produce alternating series that still converge as long as the terms shrink to zero.
Question 3 Short Answer
Use the geometric series formula to express the repeating decimal 0.777... as a fraction.
Think about your answer, then reveal below.
Model answer: 7/9. Write 0.777... = 7/10 + 7/100 + 7/1000 + ... This is a geometric series with a = 7/10 and r = 1/10, so the sum is (7/10)/(1 - 1/10) = (7/10)/(9/10) = 7/9.
Every repeating decimal is a geometric series in disguise. Identifying a and r and checking |r| < 1 converts the infinite decimal to a ratio of integers. This is one of the clearest real-world applications of the convergence formula.