Questions: Geometric Sequences and Series

Question 1

A student uses the infinite geometric series formula S = a₁/(1−r) on the series 2 + 4 + 8 + 16 + ... and gets S = 2/(1−2) = −2. What is the error?

• A. The formula requires a₁ = 1; the student should divide everything by 2 first
• B. The formula only applies when |r| < 1; this series has r = 2, so it diverges and has no finite sum
• C. The formula gives a negative result because r > 1; the student should use S = a₁/(r−1) instead
• D. The formula is correct; −2 is the Ramanujan sum of this series

Answer: (B)

The infinite series formula S = a₁/(1−r) is only valid when |r| < 1. Here r = 2 and |r| = 2 > 1, so the partial sums grow without bound — the series diverges. Plugging r = 2 into the formula gives −2, which is arithmetically valid but meaningless as a sum. The formula's derivation relies on r^n → 0 as n → ∞, which only occurs when |r| < 1. Students who forget this condition generate nonsensical 'answers' by blindly applying the formula.

Question 2

The repeating decimal 0.272727... is an infinite geometric series. What is its exact fractional value?

• A. 3/11
• B. 27/100
• C. 272/999
• D. 3/10

Answer: (A)

0.272727... = 27/100 + 27/10000 + 27/1000000 + .... This is a geometric series with a₁ = 27/100 and r = 1/100. Since |r| = 0.01 < 1, the formula applies: S = (27/100)/(1 − 1/100) = (27/100)/(99/100) = 27/99 = 3/11. Verify: 3/11 = 0.272727..., confirming the formula gives the exact fractional equivalent of a repeating decimal. This technique works for any repeating decimal.

Question 3

Nearly every infinite geometric series with a positive common ratio has a finite sum.

• T. True
• F. False

Answer: False

Only geometric series with |r| < 1 have finite sums. If r ≥ 1, the terms do not shrink to zero — each term is at least as large as the previous one — and the partial sums grow without bound. For example, 1 + 2 + 4 + 8 + ... has r = 2 and diverges. Even r = 1 produces an infinite sum (1 + 1 + 1 + ... = ∞). A positive common ratio is no guarantee of convergence; the required condition is strictly |r| < 1.

Question 4

A geometric sequence with r = −0.5 models exponential decay, because the terms decrease in magnitude with each step.

• T. True
• F. False

Answer: True

A geometric sequence a_n = a₁·r^(n−1) is the discrete analogue of an exponential function, and r is the multiplicative factor at each step. For r = −0.5, the absolute value |r| = 0.5 < 1, so the terms shrink toward zero — this is exponential decay in magnitude. The terms also alternate in sign, so it's oscillating decay. The connection to exponential functions is genuine: the geometric sequence traces an exponential curve when plotted (with alternating sign). The key property of all geometric sequences — constant ratio between consecutive terms — is exactly the discrete version of constant multiplicative growth or decay.

Question 5

Why does the infinite geometric series formula S = a₁/(1−r) only produce valid results when |r| < 1, and what happens when this condition fails?

Model answer: The formula comes from taking the limit of S_n = a₁(1−r^n)/(1−r) as n → ∞. This limit only exists if r^n → 0, which requires |r| < 1. When |r| < 1, each term is a strictly smaller fraction of the previous one, so partial sums converge to a fixed value. When |r| ≥ 1, the terms don't shrink: if |r| > 1, partial sums grow without bound; if |r| = 1, every term has the same magnitude and sums grow linearly (or oscillate). The formula produces a finite number algebraically in these cases, but that number does not represent any ordinary sum.

Plugging r = 2 into the formula gives S = a₁/(1−2) = −a₁, a finite negative number — but the actual partial sums 2, 6, 14, 30, ... grow without bound. The formula's algebraic output is disconnected from the actual behavior of the series. This is why checking |r| < 1 before applying the formula is not optional — it's what determines whether the formula means anything at all.