Working with Geometric Series: Study Guide (Calculus BC 10.2)

BC-only material. Geometric series are part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

This topic applies the definition from Topic 10.1 to the most important family of series. If any row is shaky, revisit it from the Calculus BC roadmap first.

Prerequisite	Topic	What you use it for here
Convergence means lim Sₙ exists	10.1	Every decision on this page
Limits at infinity	1.15	lim rⁿ as n → ∞
Laws of exponents	—	Rewriting terms into the form a·rⁿ
Logarithms	—	Solving rⁿ ≤ c for n

Notation on this page: Σ from n = 0 to ∞ of a·rⁿ means a + ar + ar² + …. Here a is the first term and r is the common ratio.

What makes a series geometric

A series is geometric when each term is the previous term multiplied by the same number r, the common ratio. To test a series, divide each term by the one before it:

Series	Ratios of successive terms	Geometric?
2 + 6 + 18 + 54 + …	6/2 = 3, 18/6 = 3, 54/18 = 3	Yes, r = 3
6 − 3 + 3/2 − 3/4 + …	−1/2 each time	Yes, r = −1/2
1 + 1/2 + 1/3 + 1/4 + …	1/2, then 2/3, …	No: the ratio changes

A negative ratio makes the signs alternate. The pizza series from Topic 10.1, ½ + ¼ + ⅛ + …, is geometric with a = ½ and r = ½.

In standard form, a geometric series is written Σ from n = 0 to ∞ of a·rⁿ. Starting at n = 0 makes the first term a·r⁰ = a.

The partial sum formula

Let Sₙ be the sum of the first n terms:

Sₙ = a + ar + ar² + … + arⁿ⁻¹

Multiply both sides by r:

r·Sₙ = ar + ar² + … + arⁿ⁻¹ + arⁿ

Subtract the second line from the first. Everything cancels except the first term of the top line and the last term of the bottom line:

Sₙ − r·Sₙ = a − arⁿ, so Sₙ(1 − r) = a(1 − rⁿ)

Sₙ = a(1 − rⁿ)/(1 − r), for r ≠ 1

Check with the pizza series: a = ½, r = ½, n = 10 gives S₁₀ = ½(1 − 1/1024)/(½) = 1023/1024, as found in Topic 10.1. ✓

When does a geometric series converge?

By the definition from Topic 10.1, the series converges exactly when lim Sₙ exists. In the formula, only rⁿ depends on n, so everything depends on what rⁿ does. Assume a ≠ 0.

Ratio	What rⁿ does	Partial sums	Verdict
|r| < 1	rⁿ → 0	Sₙ → a(1 − 0)/(1 − r)	Converges to a/(1 − r)
r = 1	— (formula not valid)	Sₙ = na, grows without bound	Diverges
r = −1	Alternates 1, −1	Sₙ = a, 0, a, 0, …	Diverges (no limit)
|r| > 1	|rⁿ| → ∞	Sₙ is unbounded	Diverges

If |r| < 1, then Σ from n = 0 to ∞ of a·rⁿ = a/(1 − r). If |r| ≥ 1 and a ≠ 0, the series diverges.

A memory aid that also handles other starting indices:

Sum = (first term) / (1 − common ratio), valid only when |r| < 1.

Worked example 1: rewrite, then decide (no calculator)

Question. Decide whether each series converges. Give the sum of each convergent series.

(a) Σ from n = 1 to ∞ of 3ⁿ⁺¹/5ⁿ (b) Σ from n = 0 to ∞ of 4ⁿ/3ⁿ⁺¹ (c) Σ from n = 2 to ∞ of 8(−3/4)ⁿ

(a)

1. Write out terms. n = 1: 9/5. n = 2: 27/25. n = 3: 81/125.
2. Ratio. (27/25) ÷ (9/5) = 3/5. Or rewrite: 3ⁿ⁺¹/5ⁿ = 3 · (3/5)ⁿ, so r = 3/5.
3. Decide. |3/5| < 1, so the series converges.
4. Sum. First term ÷ (1 − r) = (9/5)/(1 − 3/5) = (9/5)/(2/5) = 9/2.

(b)

1. Rewrite. 4ⁿ/3ⁿ⁺¹ = (1/3)(4/3)ⁿ. First term 1/3, ratio 4/3.
2. Decide. |4/3| > 1, so the series diverges. The partial sums grow without bound (S₁₀ is already about 16.8).
3. Warning. Blindly using a/(1 − r) gives (1/3)/(1 − 4/3) = −1, a negative “sum” of positive terms. The formula is meaningless when |r| ≥ 1.

(c)

1. First term. The series starts at n = 2, so the first term is 8(−3/4)² = 8 · 9/16 = 9/2, not 8.
2. Decide. r = −3/4 and |r| < 1, so it converges.
3. Sum. (9/2)/(1 − (−3/4)) = (9/2)/(7/4) = 18/7.

Check for (c). Using 8 as the first term would give 8/(7/4) = 32/7. That is the sum from n = 0, which includes the two extra terms 8 and −6. Indeed 32/7 − 8 + 6 = 18/7. ✓

Worked example 2: a slowing flywheel (calculator allowed)

Context. A fictional exercise bike has a heavy flywheel. After a single push, it turns 40 times in the first minute. Model: in each later minute it turns 0.9 times as many rotations as in the minute before, forever.

(a) According to the model, how many rotations does the flywheel make in total? (b) How many rotations does it make in the first 10 minutes? (c) After how many whole minutes has it completed at least 350 rotations?

1. Identify the series. Rotations per minute: 40, 40(0.9), 40(0.9)², …. Geometric with a = 40 and r = 0.9.
2. (a) Total. |0.9| < 1, so the series converges: total = 40/(1 − 0.9) = 40/0.1 = 400 rotations.
3. (b) First 10 minutes. This is a partial sum, not the full sum: S₁₀ = 40(1 − 0.9¹⁰)/(1 − 0.9) = 400(1 − 0.9¹⁰) = 400(1 − 0.34868…) ≈ 260.529 rotations.
4. (c) Set up an inequality. Sₙ = 400(1 − 0.9ⁿ) ≥ 350 gives 0.9ⁿ ≤ 0.125. Take ln of both sides: n·ln 0.9 ≤ ln 0.125. Because ln 0.9 < 0, dividing reverses the inequality: n ≥ ln 0.125 / ln 0.9 ≈ 19.74. So the smallest whole number is n = 20 minutes.
5. Check (c). S₁₉ ≈ 345.966 (not yet 350) and S₂₀ ≈ 351.369. ✓

Interpretation. The model says the flywheel never quite reaches 400 rotations, but it gets as close as you like. After 20 minutes, about 48.6 rotations are still “to come”, spread over the rest of time. A real flywheel would stop after a finite time, so the infinite total is a property of the model, not a measurement.

Worked example 3: a ratio that contains x

Question. Consider Σ from n = 0 to ∞ of ((x − 1)/3)ⁿ. (a) For which values of x does the series converge? (b) Find the sum in terms of x. (c) Find the value of x for which the sum is 2.

1. Identify a and r. First term (n = 0) is 1. Ratio r = (x − 1)/3.
2. (a) Convergence condition. Need |r| < 1: |(x − 1)/3| < 1, so |x − 1| < 3, so −2 < x < 4. At the ends: x = 4 gives r = 1 and x = −2 gives r = −1. Both diverge, so the endpoints are excluded.
3. (b) Sum. 1/(1 − (x − 1)/3) = 3/(3 − (x − 1)) = 3/(4 − x), valid only for −2 < x < 4.
4. (c) Solve. 3/(4 − x) = 2 gives 4 − x = 3/2, so x = 5/2. Check it is allowed: at x = 5/2, r = (3/2)/3 = ½, and |½| < 1. ✓ Sum = 1/(1 − ½) = 2. ✓

Why check? Solving 3/(4 − x) = ½ gives x = −2, but then r = −1 and the series diverges. The formula 3/(4 − x) only describes the series inside the interval of convergence. This idea returns in the power series topics later in Unit 10.

Repeating decimals are geometric series

A repeating decimal is a geometric series in disguise. For example,

0.454545… = 0.45 + 0.0045 + 0.000045 + … , with a = 0.45 and r = 0.01.

Since |r| < 1, the sum is 0.45/(1 − 0.01) = 0.45/0.99 = 45/99 = 5/11. Check: 5 ÷ 11 = 0.4545…. ✓

Common misconceptions

• Using the coefficient as the first term. In Σ from n = 2 of 8(−3/4)ⁿ, the first term is 9/2, not 8. Always substitute the starting index.
• Applying a/(1 − r) when |r| ≥ 1. The formula then gives nonsense, such as a negative sum of positive terms. Check |r| < 1 first and say so in your answer.
• Including the endpoints |r| = 1. r = 1 and r = −1 both diverge. The condition is strict: |r| < 1.
• Dropping the minus sign of a negative ratio. For r = −3/4, 1 − r = 1 + 3/4 = 7/4, not 1/4.
• Finding the ratio upside down. r = (later term)/(earlier term), not the other way round.
• Confusing Sₙ with the sum. “How many in the first 10 minutes?” needs S₁₀; “how many in total?” needs a/(1 − r).
• Forgetting to reverse the inequality when dividing by ln r, which is negative for 0 < r < 1.
• Thinking alternating signs mean divergence. Σ 6(−½)ⁿ converges to 4 (Figure 1).

Where this leads

Geometric series give you a reliable benchmark. Next, Topic 10.3, The nth Term Test for Divergence, gives a quick test for divergence that works for any series. Later, geometric series are the comparison series in Topic 10.6, the model behind the ratio test in Topic 10.8, and the starting point for power series such as 1/(1 − x) = 1 + x + x² + … in Topics 10.13 to 10.15. Use the Calculus BC roadmap to see the order.

Try the practice questions now, then use the revision notes and the checklist to consolidate.