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Calculus II · Unit 4A · lesson

Geometric Series

Concept

Learning objectives

recognize, test, and sum geometric series and interpret the common ratio.

Geometric Series

Explanation

Repeated proportional change creates the simplest infinite series

A geometric series multiplies each term by the same ratio. That repeated scaling makes the partial sums algebraically manageable: multiplying the finite sum by the ratio lines up almost every term for cancellation. The resulting formula reveals both convergence and the exact sum.

The ratio controls everything. When ∣r∣<1|r|<1, powers of rr shrink to zero and the partial sums settle. When ∣r∣≥1|r|\ge1, the terms do not approach zero, so the series cannot converge. In applications, the ratio may represent retained energy after each bounce, a repeated discount, a reflection coefficient, or the fraction of material remaining after each stage.

Bridge

Repeated scaling creates an exact formula

A geometric series multiplies each term by a constant ratio rr. Multiplying a finite partial sum by rr shifts its terms, so subtraction leaves only the first term and a final remainder.

The condition ∣r∣<1|r|<1 means that remainder vanishes. When ∣r∣≥1|r|\ge1, the terms do not shrink to zero, so convergence is impossible.

The remaining tail is a scaled copy. Self-similar interval filling and a shrinking tail.
Read this graph as text

The remaining tail is a scaled copy. Lengths one-half, one-quarter, one-eighth, and so on fill a unit interval. Self-similar interval filling and a shrinking tail.

Written labels, distinct line styles, markers, and fill patterns communicate every relationship in the remaining tail is a scaled copy; color is never the only cue.

Why it matters: Self-similar interval filling and a shrinking tail.

The remaining tail is a scaled copy

Lengths one-half, one-quarter, one-eighth, and so on fill a unit interval.

The remaining tail is a scaled copy. Self-similar interval filling and a shrinking tail.

Proof idea

Finite identity first, limit second

For SN=a+ar+⋯+arNS_N=a+ar+\cdots+ar^N, subtraction gives

(1−r)SN=a−arN+1.(1-r)S_N=a-ar^{N+1}.

Thus SN=a(1−rN+1)/(1−r)S_N=a(1-r^{N+1})/(1-r). Only after this finite formula is established do we let N→∞N\to\infty.

A geometric series repeatedly takes the same fraction. Self-similar interval filling and a shrinking tail.
Read this graph as text

A geometric series repeatedly takes the same fraction. Successive pieces occupy one half of what remains, producing areas 1/2,1/4,1/8, whose total fills one whole square. Self-similar interval filling and a shrinking tail.

Written labels, distinct line styles, markers, and fill patterns communicate every relationship in a geometric series repeatedly takes the same fraction; color is never the only cue.

Why it matters: Self-similar interval filling and a shrinking tail.

A geometric series repeatedly takes the same fraction

Successive pieces occupy one half of what remains, producing areas 1/2,1/4,1/8,…1/2,1/4,1/8,\ldots whose total fills one whole square.

A geometric series repeatedly takes the same fraction. Self-similar interval filling and a shrinking tail.

How to read the visual

The uncovered strip after each step is again half as wide as before. The self-similarity is the geometric ratio in visual form.

Concept

Geometric-series formula

For ∣r∣<1|r|<1,

∑n=0∞arn=a1−r.\sum_{n=0}^{\infty}ar^n=\frac{a}{1-r}.

If ∣r∣≥1|r|\ge1, the series diverges.

Guided walkthrough

Sum a shifted geometric series

Consider

∑n=2∞5(13)n.\sum_{n=2}^{\infty}5\left(\frac13\right)^n.

The first included term is 5/95/9, and the ratio is 1/31/3. Therefore

5/91−1/3=56.\frac{5/9}{1-1/3}=\frac56.

Using the actual first term avoids an indexing error.

Worked example

A repeating decimal is a convergent series

The decimal 0.272727…0.272727\ldots equals

27100+271002+271003+⋯ .\frac{27}{100}+\frac{27}{100^2}+\frac{27}{100^3}+\cdots.

This is geometric with first term 27/10027/100 and ratio 1/1001/100, so

0.272727…=27/1001−1/100=311.0.272727\ldots=\frac{27/100}{1-1/100}=\frac3{11}.
Common mistake

Use the first included term

For a series starting at n=3n=3, the first term is obtained by substituting 33. A shifted index changes the first term even when the ratio is unchanged.

Interactive checku4a-geometric_series-01

Evaluate ∑n=0∞3(1/2)n\sum_{n=0}^{\infty}3(1/2)^n.

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Show hint

Use a/(1−r)a/(1-r).

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Exercise

Determine whether ∑4(−1.2)n\sum 4(-1.2)^n converges.

Exercise

Write 0.272727…0.272727\ldots as a geometric series and fraction.

Exercise

A ball rebounds to 70 percent of its previous height. Model total vertical distance after a drop from 10 meters.

Exercise

Explain why the first term depends on the starting index.

After the explanation

Use the section idea

Reading lens

Build every infinite sum from finite partial sums, and expose geometric or telescoping structure before taking a limit.

Mental model

A series converges exactly when its sequence of partial sums approaches a finite value.

Decision

Check the term limit first, then look for an exact partial-sum pattern before selecting a comparison test.

Common trap

Concluding convergence from terms approaching zero or canceling inside an unwritten infinite expression.

Check yourself

Can you write the relevant finite partial sum and identify which terms survive?

Practice this skill

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