Partial sums and the value they approach, the sum of an infinite geometric series and the condition on the common ratio, series that have no sum, infinite series as models, and converting repeating decimals to fractions.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 12 — Sequences and Series
Find Sums of Infinite Geometric Series
Objectives
Five outcomes. Adding forever, and getting a finite answer.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.4 Find Sums of Infinite Geometric Series §12.4, pp. 820-823 — the lesson these objectives are drawn from
Warm-up
Lesson 12.3 gave you a formula for the sum of the first n terms of a geometric series.
Discussion prompt
Use it on one half plus one quarter plus one eighth, and keep going. What do the totals 0.5, 0.75, 0.875, 0.9375 seem to be heading for?
Hint: How far is each total from 1?
Answer:
They are heading for 1. The gaps are 0.5, 0.25, 0.125, 0.0625 — each exactly half the one before, and each exactly the size of the next term.
\[ S_n = \tfrac{1}{2}\!\left(\frac{1-(1/2)^n}{1-1/2}\right) = 1-\left(\tfrac{1}{2}\right)^{\!n} \]
As n grows, one half to the n shrinks toward zero, so S sub n creeps toward 1 without ever arriving. That limiting value is what this lesson calls the sum of the infinite series.
Concept
An infinite geometric series whose ratio is between negative one and one has partial sums that approach a single value, and that value is a_1 divided by 1 minus r. If the ratio is one or larger in size, the series has no sum at all.
partial sum — The sum S sub n of the first n terms of an infinite series. The partial sums of an infinite geometric series may approach a limiting value, and when they do, that value is defined to be the sum of the series.
\[ S = \frac{a_1}{1-r}, \quad |r| < 1 \]
The word sum is being redefined here. Nobody adds infinitely many numbers; the sum is the value the running totals settle toward.
Figure (svg): Two columns comparing a finite geometric sum with an infinite one
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.4 Find Sums of Infinite Geometric Series §12.4, pp. 820-821
Section
Section 1
Concept
The partial sum S sub n is the total of the first n terms. Computing several of them and graphing the results shows whether they are settling toward a value.
\[ S_1, S_2, S_3, \dots \to \; ? \]
For the halving series each partial sum falls short of 1 by exactly the next term, so the shortfall halves every step.
Figure (svg): Partial sums of a halving series climbing toward one
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.4 Find Sums of Infinite Geometric Series §12.4, pp. 820-820 — Find partial sums
Picture it
Example 1: the partial sums of the halving series.
Figure (svg): Partial sums of a halving series climbing toward one
The points rise toward the dashed line at 1 and never cross it. Each step closes half the remaining gap.
Worked example
Example 1.
\[ \text{For } \tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}+\tfrac{1}{16}+\tfrac{1}{32}+\dots \text{ find } S_1 \text{ through } S_5. \]
Compute S_1 and S_2
Why: One half, then one half plus one quarter.
\[ 0.5\text{ and } 0.75 \]
Compute S_3 and S_4
Why: Adding one eighth, then one sixteenth.
\[ 0.875\text{ and } 0.9375 \]
Compute S_5
Why: Adding one thirty-second.
\[ \text{about } 0.97 \]
Describe the trend
Why: The totals rise and slow.
\[ S _{n}\text{ approaches } 1 \]
Figure (svg): Partial sums of a halving series climbing toward one
\[ S_n \to 1 \]
Verify: measure each gap to 1
Why: The shortfalls are 0.5, 0.25, 0.125, 0.0625, 0.03125 — exactly the next term each time. That is not a coincidence: the untotalled tail of the series IS the gap, so a series whose terms shrink to nothing has a shrinking gap.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.4 Find Sums of Infinite Geometric Series §12.4, pp. 820-820
Fill the middle
Example 1.
Fill in the blanks
S_3 = \tfrac0.875___+\tfrac______+\tfrac______ = ___
Why: Seven eighths, or 0.875. It falls short of 1 by exactly one eighth, which is the next term in the series.
Worked example
Guided Practice 1.
\[ \text{For } \tfrac{2}{5}+\tfrac{4}{25}+\tfrac{8}{125}+\tfrac{16}{625}+\tfrac{32}{3125}+\dots \text{ find } S_1 \text{ through } S_5. \]
Identify a_1 and r
Why: Two fifths, and each term is two fifths of the last.
\[ a _{1} = 0.4, r = 0.4 \]
Compute S_1 and S_2
Why: Zero point four, then plus 0.16.
\[ 0.4\text{ and } 0.56 \]
Compute S_3 and S_4
Why: Adding 0.064, then 0.0256.
\[ 0.624\text{ and } 0.6496 \]
Compute S_5 and describe
Why: Adding 0.01024.
\[ 0.65984,\text{ approaching } \frac{2}{3} \]
Figure (svg): The solution to Worked example a different halving rate shown as a ladder of expressions, one row per algebraic move
\[ S_n \to \tfrac{2}{3} \approx 0.667 \]
Verify: check the limit against the formula
Why: Two fifths over 1 minus two fifths is 0.4 over 0.6, which is two thirds, or about 0.667. The fifth partial sum 0.65984 is already within 0.008 of it, and every later one is closer still.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.4 Find Sums of Infinite Geometric Series §12.4, pp. 821-821
Trap
\[ S_1 = 0.5, \; S_2 = 0.75, \; S_3 = 0.875, \dots \]
Look for the n at which the total equals 1
Why: The totals are climbing toward 1, so surely one of them reaches it.
\[ S_n = 1 \text{ for some } n \quad \text{(never true)} \]
Every partial sum is 1 minus one half to the n, and that subtracted piece is never exactly zero. No finite total reaches 1.
\[ S_n = 1-\left(\tfrac{1}{2}\right)^{\!n} < 1 \text{ for every } n \]
Ask what the totals APPROACH, not what they reach
Why: The sum of an infinite series is defined as that limiting value.
\[ S = 1 \]
This is a genuine redefinition of the word sum. It names where the running totals are headed, not any total that is ever actually reached.
Matching
What do the partial sums approach?
Match the pairs
Why: The last has a_1 equal to 5 and r equal to 0.8, so its sum is 5 over 0.2, which is 25 — five times its own first term. The closer r is to 1, the larger the total relative to that first term.
Prediction
Commit before reasoning.
Predict first
For the halving series, why is the gap between S_n and 1 always exactly the next term?
Correct: Because the untotalled tail of the series is itself a halving series summing to the next term.
\[ 1-S_3 = \tfrac{1}{8}; \quad 1-S_4 = \tfrac{1}{16} \]
Why: After S sub 3 the remaining terms are one sixteenth, one thirty-second and so on, which is a new halving series with first term one sixteenth — and its own sum is one eighth, the next term. So the leftover always equals the term you were about to add. This self-similarity is exactly the property that makes the total finite: the whole infinite tail is never larger than the single term in front of it.
Two truths and a lie
Two of these are true. Keep the one that is not.
Eliminate the wrong options
Which statement about partial sums is FALSE?
Survives elimination: B
Why: Every partial sum of the halving series is 1 minus a positive amount, so none of them equals 1. Approaching a value and reaching it are different things, and the sum of an infinite series is defined by the first, not the second.
Section
Section 2
Concept
When the common ratio is between negative one and one, the finite sum formula's r to the n term shrinks toward zero and the whole expression collapses to a_1 divided by 1 minus r.
\[ S = \frac{a_1}{1-r}, \quad |r| < 1 \]
This is the finite formula with one piece deleted, and the deletion is legal exactly when repeated multiplication by r drives the terms to nothing.
Figure (svg): The finite sum formula collapsing into the infinite one
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.4 Find Sums of Infinite Geometric Series §12.4, pp. 820-821 — The Sum of an Infinite Geometric Series
Picture it
The finite formula as n grows without bound.
Figure (svg): The finite sum formula collapsing into the infinite one
Replace r to the n with zero and the numerator becomes 1, leaving a_1 over 1 minus r. The condition on r is what licenses that replacement.
Worked example
Example 2, both parts.
\[ \text{Find } \sum_{i=1}^{\infty}5(0.8)^{\,i-1} \text{ and } 1-\tfrac{3}{4}+\tfrac{9}{16}-\tfrac{27}{64}+\dots \]
First: identify a_1 and r
Why: Five, and 0.8.
\[ a _{1} = 5, r = 0.8 \]
First: apply the formula
Why: Five over 1 minus 0.8.
\[ \frac{5}{0.2} = 25 \]
Second: identify a_1 and r
Why: One, and negative three quarters.
\[ a _{1} = 1, r = -\frac{3}{4} \]
Second: apply the formula
Why: One over 1 plus three quarters.
\[ \frac{1}{\frac{7}{4}} = \frac{4}{7} \]
Figure (svg): The finite sum formula collapsing into the infinite one
\[ S = 25; \qquad S = \tfrac{4}{7} \]
Verify: check the second against its partial sums
Why: The running totals are 1, 0.25, 0.8125, 0.2734, 0.5698, and four sevenths is about 0.571. With a negative ratio the partial sums oscillate ABOVE and BELOW the limit rather than climbing to it, closing in from alternate sides.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.4 Find Sums of Infinite Geometric Series §12.4, pp. 821-821
Fill the middle
Example 2a.
Fill in the blanks
S = \frac25___ = \frac______ = ___
Why: Five over 0.2 is 25. Dividing by a small denominator is what makes an infinite total so much larger than its first term.
Worked example
Guided Practice 2 and 4.
\[ \text{Find } \sum_{n=1}^{\infty}\left(-\tfrac{1}{2}\right)^{\,n-1} \text{ and } 3+\tfrac{3}{4}+\tfrac{3}{16}+\tfrac{3}{64}+\dots \]
First: identify a_1 and r
Why: The n minus 1 power starts at 1.
\[ a _{1} = 1, r = -\frac{1}{2} \]
First: apply the formula
Why: One over 1 plus one half.
\[ \frac{1}{\frac{3}{2}} = \frac{2}{3} \]
Second: identify a_1 and r
Why: Three, and each term is a quarter of the last.
\[ a _{1} = 3, r = \frac{1}{4} \]
Second: apply the formula
Why: Three over three quarters.
\[ 4 \]
Figure (svg): The solution to Worked example two more infinite sums shown as a ladder of expressions, one row per algebraic move
\[ S = \tfrac{2}{3}; \qquad S = 4 \]
Verify: notice how much the ratio matters
Why: Both series start modestly, yet the second totals only a third more than its own first term because r is small, while the first example's series with r equal to 0.8 totalled five times its first term. As r climbs toward 1 the denominator shrinks and the total explodes.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.4 Find Sums of Infinite Geometric Series §12.4, pp. 821-821
Error analysis
A student totals the infinite series 5 + 4 + 3.2 + 2.56 + ... using the finite formula.
Annotate
On: \( S = 5\!\left(\frac{1-0.8^{\,n}}{1-0.8}\right) \)
The finite formula is not wrong here so much as unfinished. The infinite formula is what it becomes once the vanishing piece is dropped.
Fill the middle
Example 2b.
Fill in the blanks
S = \frac4/7______\right)} = \frac_________} = ___
Why: Subtracting negative three quarters adds it, giving a denominator of seven fourths, whose reciprocal is four sevenths. A negative ratio makes the denominator larger and the total smaller.
Comparison
Fill the blanks. Both series start at 1.
Comparison matrix
| Question | r = 1/2 | r = 0.9 |
|---|---|---|
| Denominator 1 - r | 1/2 | 0.1 |
| Sum | 2 | 10 |
| Terms shrink | quickly | slowly |
| Sum as multiple of a_1 | 2 times | 10 times |
A ratio close to 1 leaves a tiny denominator, so the total balloons. The terms are still shrinking, just slowly enough that a great many of them matter.
Prediction
Commit before reasoning.
Predict first
For a fixed a_1, what happens to a_1 over 1 minus r as r climbs toward 1?
Correct: It grows without bound, because the denominator shrinks toward zero.
\[ \frac{1}{1-0.99} = 100; \quad \frac{1}{1-0.999} = 1000 \]
Why: At r equal to 0.9 the sum is ten times a_1; at 0.99 it is a hundred times; at 0.999 a thousand times. The formula therefore does not fail abruptly at r equal to 1 — it blows up smoothly as that boundary is approached, which is the natural way to see why r equal to 1 must be excluded rather than merely inconvenient.
Section
Section 3
Concept
If the size of r is 1 or larger, the terms do not shrink toward zero, the partial sums do not settle, and the series has no sum. The formula will still produce a number, and that number is meaningless.
\[ |r| \ge 1 \;\Longrightarrow\; \text{no sum} \]
Checking the size of r before touching the formula is the only protection, because the formula gives no warning of its own.
Figure (svg): Partial sums of a divergent series swinging further and further apart
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.4 Find Sums of Infinite Geometric Series §12.4, pp. 821-821 — The sum does not exist
Picture it
Example 3: the series 1 minus 4 plus 16 minus 64 and so on.
Figure (svg): Partial sums of a divergent series swinging further and further apart
The totals are 1, negative 3, 13, negative 51, 205 — swinging further apart every step and approaching nothing at all.
Worked example
Example 3.
\[ \text{Find the sum of } 1-4+16-64+\dots \text{ if it exists.} \]
Find the ratio
Why: Negative 4 over 1.
\[ r = -4 \]
Check its size
Why: The size of negative 4 is 4.
\[ 4\text{ is at least } 1 \]
State the conclusion
Why: The condition fails.
Note the trap
Why: The formula would return 1 over 5.
Figure (svg): Partial sums of a divergent series swinging further and further apart
\[ \text{no sum} \]
Verify: look at the terms themselves
Why: They are 1, negative 4, 16, negative 64 — growing without bound in size. A series whose terms do not shrink toward zero cannot have partial sums that settle, whatever a formula might return.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.4 Find Sums of Infinite Geometric Series §12.4, pp. 821-821
Sorting
Check the size of r.
Sort into buckets
Sort each infinite geometric series.
Only the size of r matters, not its sign. Negative one half qualifies and negative 4 does not, for exactly the same reason positive one half does and positive 4 does not.
Worked example
Guided Practice 3.
\[ \text{Find } \sum_{n=1}^{\infty}3\left(\tfrac{5}{4}\right)^{\,n-1} \text{ if it exists.} \]
Identify a_1 and r
Why: Three, and five fourths.
\[ a _{1} = 3, r = \frac{5}{4} \]
Check the size of r
Why: Five fourths is 1.25.
\[ 1.25\text{ is at least } 1 \]
State the conclusion
Why: The condition fails.
See why
Why: The terms are 3, 3.75, 4.6875, growing.
Figure (svg): The solution to Worked example a borderline ratio shown as a ladder of expressions, one row per algebraic move
\[ \text{no sum} \]
Verify: compare with a ratio just under 1
Why: With r equal to five fourths the terms grow by 25 percent a step and the totals run away; with r equal to four fifths they shrink by 20 percent and the total settles at 15. The boundary at 1 is sharp, and nothing about the terms' appearance at the start reveals which side you are on.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.4 Find Sums of Infinite Geometric Series §12.4, pp. 821-821
Trap
\[ 1-4+16-64+\dots, \; a_1 = 1, \; r = -4 \]
Substitute into the sum formula
Why: The numbers fit the formula's shape, so they are used.
\[ S = \frac{1}{1-(-4)} = \frac{1}{5} \quad \text{(meaningless)} \]
The partial sums are 1, negative 3, 13, negative 51 — nowhere near one fifth, and moving further from it each step.
\[ |r| = 4 \ge 1 \]
Check the size of r before anything else
Why: The formula was derived assuming r to the n vanishes.
\[ \text{the sum does not exist} \]
The textbook prints this caution beside Example 3. A formula that returns a number is not thereby applicable.
Fill the middle
The boxed rule on page 821.
Fill in the blanks
\text< |r| ___ 1
Why: The size of r must be strictly less than 1. At exactly 1 the terms never shrink, and the formula's denominator becomes zero or the series is a constant repeated forever.
Matching
Only the size of r matters.
Match the pairs
Why: Notice that a negative ratio can go either way: negative one half gives a sum and negative 4 does not. Stripping the sign and comparing with 1 is the whole test.
Prediction
Commit before reasoning.
Predict first
Substituting a_1 = 1 and r = -4 gives 1/5. Why is that not the sum?
Correct: Because the derivation assumed r to the n shrinks to zero, which is false here.
\[ (-4)^{10} = 1{,}048{,}576 \; \text{(not near } 0\text{)} \]
Why: The infinite formula was obtained by deleting r to the n from the finite one. With r equal to negative 4 that piece is negative 4 to the n, which grows in size without limit rather than vanishing, so the deletion was never justified. Arithmetic performed on an invalid formula produces a number, not an answer — which is why the size of r must be checked first, every time.
Section
Section 4
Concept
A physical process that repeats with each stage a fixed fraction of the last forms an infinite geometric series, and its total is finite whenever that fraction is under one.
\[ d = 18+18(0.8)+18(0.8)^2+\cdots = \frac{18}{1-0.8} \]
The total distance is finite even though the number of swings is not, because the swings shrink fast enough.
Figure (svg): A pendulum whose successive swings shrink by twenty percent
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.4 Find Sums of Infinite Geometric Series §12.4, pp. 822-822 — Use an infinite series as a model
Picture it
Example 4: each swing is 80 percent of the last.
Figure (svg): A pendulum whose successive swings shrink by twenty percent
Infinitely many swings, each shorter than the last, add up to ninety inches — seven and a half feet in total.
Worked example
Example 4.
\[ \text{A pendulum swings } 18 \text{ inches, then } 80\% \text{ of each previous swing. Total distance?} \]
Write the series
Why: Eighteen, then 18 times 0.8, and so on.
\[ 18 + 18(0.8) + 18(0.8) ^{2} +... \]
Identify a_1 and r
Why: Eighteen, and 0.8.
\[ a _{1} = 18, r = 0.8 \]
Check the condition
Why: Zero point eight is under 1.
Apply the formula
Why: Eighteen over 0.2.
\[ 90\text{ inches} \]
Figure (svg): A pendulum whose successive swings shrink by twenty percent
\[ d = \frac{18}{1-0.8} = 90 \]
Verify: check against the first few swings
Why: The first four swings alone cover 18 plus 14.4 plus 11.52 plus 9.216, which is 53.1 inches — over half the total already. The remaining infinitely many swings contribute only 36.9 inches between them, which is what shrinking terms look like in practice.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.4 Find Sums of Infinite Geometric Series §12.4, pp. 822-822
Fill the middle
Example 4.
Fill in the blanks
d = \frac90___ = \frac______ = ___
Why: Eighteen over 0.2 is 90 inches, or 7.5 feet. The pendulum's entire remaining journey fits in a distance smaller than a person's height.
Worked example
Guided Practice 5.
\[ \text{The same pendulum first swings } 10 \text{ inches. Total distance?} \]
Identify a_1 and r
Why: Ten, and still 0.8.
\[ a _{1} = 10, r = 0.8 \]
Apply the formula
Why: Ten over 0.2.
\[ 50\text{ inches} \]
Compare with the first case
Why: Ninety against 50.
\[ \text{the ratio of totals is } 18\text{ to } 10 \]
Note the structure
Why: Only a_1 changed.
\[ \text{the total scales with } a _{1} \]
Figure (svg): The solution to Worked example a shorter first swing shown as a ladder of expressions, one row per algebraic move
\[ d = \frac{10}{1-0.8} = 50 \]
Verify: check the scaling
Why: The first swing fell from 18 to 10, a factor of 5 over 9, and the total fell from 90 to 50 by the same factor. Since a_1 appears once in the numerator, the whole total is directly proportional to it — a useful shortcut when only the starting size changes.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.4 Find Sums of Infinite Geometric Series §12.4, pp. 822-822
Trap
\[ 18+14.4+11.52+9.216+\cdots \]
Reason that the swings never stop
Why: So the pendulum must travel forever and cover unlimited distance.
\[ d = \infty \quad \text{(wrong)} \]
The number of swings is unlimited but their sizes are not. The total is bounded because the terms shrink geometrically.
\[ d = \frac{18}{1-0.8} = 90 \]
Check the ratio, then total the series
Why: A ratio under 1 makes the terms shrink fast enough for a finite total.
\[ 90 \text{ inches} = 7.5 \text{ feet} \]
Infinitely many positive quantities can have a finite total, and that is the central surprise of this lesson.
Ranking
Smallest first.
Put in order
Why: The totals are 1, 4, 25, 50 and 90. The last two share a ratio and differ only in their first term, so their totals sit in the same 10 to 18 proportion as their starting swings.
Comparison
Fill the blanks. The same pendulum, two questions.
Comparison matrix
| Question | First 4 swings | All swings |
|---|---|---|
| Formula | S_4 = a_1(1 - r^4)/(1 - r) | S = a_1/(1 - r) |
| Distance | about 53.1 inches | 90 inches |
| Number of terms | 4 | unlimited |
| Needs |r| < 1? | no | yes |
The first four swings already cover 59 percent of the pendulum's whole future path, which is why the infinite total is not much larger than a short finite one.
Prediction
Commit before reasoning.
Predict first
If a pendulum lost no energy and every swing were 18 inches, what would the total distance be?
Correct: Unlimited, because the terms never shrink and the totals grow without bound.
\[ r = 1 \;\Longrightarrow\; S_n = 18n \to \infty \]
Why: With r equal to 1 the partial sums are 18n, which grows past any bound you name. This is the physical meaning of the excluded case: friction is what makes the ratio less than one, and it is friction that makes the total path finite. A frictionless pendulum really would travel arbitrarily far, given arbitrarily long.
Section
Section 5
Concept
A repeating decimal is the sum of its repeat block shifted further and further right, which is a geometric series whose ratio is the place-value shift. The infinite sum formula turns it into a fraction.
\[ 0.242424\ldots = \frac{0.24}{1-0.01} = \frac{24}{99} = \frac{8}{33} \]
The repeat block gives the first term and its length gives the ratio: one digit gives 0.1, two digits give 0.01, three give 0.001.
Figure (svg): A repeating decimal rewritten as an infinite geometric series
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.4 Find Sums of Infinite Geometric Series §12.4, pp. 822-822 — Write a repeating decimal as a fraction
Picture it
Example 5: 0.242424 and so on.
Figure (svg): A repeating decimal rewritten as an infinite geometric series
Each copy of the block 24 sits two places further right, so each is one hundredth of the one before it.
Worked example
Example 5.
\[ \text{Write } 0.242424\ldots \text{ as a fraction in lowest terms.} \]
Write it as a series
Why: Twenty-four hundredths, then 24 ten-thousandths.
\[ 24(0.01) + 24(0.01) ^{2} +... \]
Identify a_1 and r
Why: Zero point two four, and 0.01.
\[ a _{1} = 0.24, r = 0.01 \]
Apply the formula
Why: Zero point two four over 0.99.
\[ \frac{0.24}{0.99} \]
Clear the decimals and reduce
Why: Twenty-four over 99, divided by 3.
\[ \frac{8}{33} \]
Figure (svg): A repeating decimal rewritten as an infinite geometric series
\[ 0.242424\ldots = \tfrac{8}{33} \]
Verify: divide it back out
Why: Eight divided by 33 is 0.2424..., recovering the original decimal. The denominator 99 is no accident: a two-digit repeat always gives 99 before reduction, a one-digit repeat gives 9, and a three-digit repeat gives 999.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.4 Find Sums of Infinite Geometric Series §12.4, pp. 822-822
Fill the middle
Example 5.
Fill in the blanks
0.242424\ldots \text0.01 2 \text___ \;\Longrightarrow\; r = ___
Why: Each copy of the block sits two decimal places further right, so each is one hundredth of the last. A one-digit repeat would give 0.1 instead.
Worked example
Guided Practice 6 to 8.
\[ \text{Write } 0.555\ldots, \; 0.727272\ldots, \; 0.131313\ldots \text{ as fractions.} \]
First: one repeating digit
Why: Zero point five over 0.9.
\[ \frac{5}{9} \]
Second: two repeating digits
Why: Zero point seven two over 0.99.
\[ \frac{72}{99} \]
Second: reduce
Why: Divide top and bottom by 9.
\[ \frac{8}{11} \]
Third: two repeating digits
Why: Thirteen over 99, already reduced.
\[ \frac{13}{99} \]
Figure (svg): The solution to Worked example three more decimals shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{5}{9}; \quad \tfrac{8}{11}; \quad \tfrac{13}{99} \]
Verify: check the middle one by division
Why: Eight divided by 11 is 0.7272..., as required. The reduction from 72 over 99 to 8 over 11 hides the pattern, so it is worth writing the unreduced form first and reducing only at the end.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.4 Find Sums of Infinite Geometric Series §12.4, pp. 822-822
Error analysis
A student converts 0.242424... using a ratio of 0.1.
Annotate
On: \( S = \frac{0.24}{1-0.1} = \frac{0.24}{0.9} = \frac{4}{15} \)
Four fifteenths is 0.2666..., not 0.2424... . The length of the repeat block, not the number of digits before it, sets the ratio.
Matching
Block over nines, then reduce.
Match the pairs
Why: Before reducing, these are 5/9, 24/99, 72/99 and 13/99. The shortcut worth remembering is the block over as many nines as the block has digits, which the series derivation justifies.
Sorting
The denominator before reducing.
Sort into buckets
Sort each repeating decimal by the denominator its series gives.
The pattern continues: a three-digit block gives 999, a four-digit block 9999. It comes straight from 1 minus one over ten to the block length.
Prediction
Commit before reasoning.
Predict first
What does the series argument establish about EVERY repeating decimal?
Correct: That every repeating decimal equals a fraction, so every one is a rational number.
\[ 0.\overline{d_1\ldots d_k} = \frac{d_1\ldots d_k}{\underbrace{99\ldots 9}_{k}} \]
Why: The argument never used anything special about 24: any repeating block of k digits gives a geometric series with ratio one over ten to the k, which is under 1, so the formula always applies and always yields the block over a string of nines — a ratio of integers. Combined with the fact that every fraction has a terminating or repeating decimal expansion, this closes the loop: repeating decimals and rational numbers are exactly the same set, which is a result from Chapter 1 that this chapter finally proves.
Comparison
Fill the blanks. One stops, the other does not.
Comparison matrix
| Question | Finite geometric | Infinite geometric |
|---|---|---|
| Formula | S_n = a_1(1 - r^n)/(1 - r) | S = a_1/(1 - r) |
| Condition on r | r is not 1 | the size of r is less than 1 |
| Needs n? | yes | no |
| Meaning of the answer | an exact total | what the partial sums approach |
The infinite formula is the finite one with r to the n replaced by zero, so the condition on r is precisely what makes that replacement honest.
Pattern
Ratio first, always.
The formula returns a number even when it does not apply. Only the check on r stands between you and a meaningless answer.
OpenStax Algebra and Trigonometry 2e, §13.4 Series and Their Notations §13.4
Check
Identify a_1 and r, then check the condition.
Check your understanding
What is the sum of 5 times (0.8)^(i-1) for i from 1 to infinity?
Answer: A
Why: a_1 = 5 and r = 0.8, so S = 5/(1 - 0.8) = 5/0.2 = 25.
Check
Check the size of the ratio first.
Check your understanding
What is the sum of the infinite geometric series 1 - 4 + 16 - 64 + ...?
Answer: A
Why: r = -4 and the size of -4 is at least 1, so the series has no sum.
Check
The block length sets the ratio.
Check your understanding
Written as a fraction in lowest terms, 0.242424... equals what?
Answer: A
Why: 0.24/(1 - 0.01) = 24/99 = 8/33, which divides back out to 0.2424...
Real world
A ball is dropped from 6 feet. Each bounce returns it to 60 percent of its previous height, and it keeps bouncing.
Discussion prompt
What is the total vertical distance the ball travels before coming to rest?
Hint: The first drop happens once; every bounce height is travelled twice, up and down.
Answer:
\[ \text{first drop: } 6 \]
\[ \text{each bounce } h \text{ contributes } 2h: \; 2\!\left(3.6+2.16+1.296+\cdots\right) \]
\[ = 2\!\left(\frac{3.6}{1-0.6}\right) = 2(9) = 18 \]
The total distance is 6 plus 18, or 24 feet.
The doubling is where nearly everyone slips: the ball rises to each bounce height and falls from it, so every bounce counts twice, while the initial drop counts once. Writing the initial drop outside the series and totalling the bounces separately keeps the two straight. Notice also that the ball takes infinitely many bounces yet covers only 24 feet — and in fact finishes in finite TIME too, for the same reason the distance is finite.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is 0.999... equal to 1, or just very close to it?
Correct: Exactly 1 — the series 0.9 + 0.09 + 0.009 + ... sums to 0.9/(1 - 0.1) = 1.
\[ 0.999\ldots = \frac{0.9}{1-0.1} = \frac{0.9}{0.9} = 1 \]
Why: Applying this lesson's own formula with a_1 equal to 0.9 and r equal to 0.1 gives 0.9 over 0.9, which is exactly 1. The partial sums 0.9, 0.99, 0.999 are indeed each less than 1, but the sum of an infinite series is defined as what they approach, not as any one of them. The same machinery that turned 0.242424 into 8 over 33 turns 0.999... into 1, and refusing the second while accepting the first is not a position you can hold consistently.
Explain it
They cannot see how adding infinitely many numbers gives a finite answer.
Discussion prompt
In four sentences or fewer, explain how one half plus one quarter plus one eighth and so on can total 1.
Hint: Think about the distance still remaining.
Answer:
Start at 0 and walk half the way to 1. Then walk half of what is left, then half of that, and keep going.
You never land exactly on 1, but the distance remaining halves every step, so it becomes smaller than any gap you care to name. The total of all those steps is defined to be the place you are heading for, which is 1.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the check on r, make it the first line of every solution, before any substitution. For the meaning, say out loud that the sum is what the partial sums approach. For models, write out four terms of the process before reaching for a formula. For decimals, count the digits in the repeating block and use one over ten to that power.
Connect it up
Paper. Fifteen minutes.
Draw it
Build an infinite series page. Top left: compute five partial sums of a halving series, graph them, mark the limiting value with a dashed line, and write beside each point how far it falls short. Top right: show the finite sum formula turning into the infinite one, boxing the condition on r and writing one sentence on why that condition is what licenses the step. Middle: pick a ratio of at least 1, compute four partial sums, graph them, and write in one sentence what the formula would wrongly claim. Bottom left: model a repeating physical process of your own invention and total it. Bottom right: convert three repeating decimals to fractions, one with a one-digit block and two with two-digit blocks, keeping the unreduced form visible before you reduce.
If any solution used the formula before checking the size of r, redo it in the right order: the check is the first line, not an afterthought.
Recap
Five things, and a new meaning for the word sum.
| If you see | Then |
|---|---|
| An infinite geometric series | Check the size of r before anything else |
| The size of r under 1 | The sum is a_1 divided by 1 minus r |
| The size of r at least 1 | There is no sum; the formula is invalid |
| A shrinking repeated process | Write four terms, then read off a_1 and r |
| A repeating decimal | Block over as many nines as the block has digits |
| A bouncing ball | The drop counts once; each bounce height counts twice |
Lesson 12.5 returns to sequences and defines them a different way — not by a rule for the nth term, but by a rule that builds each term from the one before it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.4 Find Sums of Infinite Geometric Series §12.4, pp. 820-823 — everything on these slides traces back here
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