Introduces summation notation and the distinction between a sequence and a series, then develops finite arithmetic and geometric sums before showing that an infinite geometric series has a finite total exactly when its ratio is smaller than one in size.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 11 — Sequences, Probability and Counting Theory
§11.4 Series and Their Notations, pp. 1331-1345
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §11.4 Series and Their Notations §11.4, pp. 1331-1345 — the pages these objectives are drawn from
Warm-up
The question sounds like it should have an obvious answer.
Discussion prompt
Add a half, then a quarter, then an eighth, and keep going. Does the total grow without bound?
Hint: Draw a bar of length one and fill it.
Answer:
Each piece fills half the remaining gap. So the total gets closer to one but never passes it.
After ten terms the total is about 0.999, and after twenty about 0.999999. The gap shrinks towards nothing.
So infinitely many positive numbers can have a finite total — here, exactly one. The condition is that the terms shrink fast enough, and this section makes that precise.
Concept
A sequence is a list and a series is its total. When the terms shrink fast enough, adding infinitely many of them still gives a finite number.
series — the sum of the terms of a sequence, which may be finite or continue indefinitely
\[ \sum_{k=1}^{n}a_k \quad\text{or}\quad \sum_{k=1}^{\infty}a_k \]
The infinite case is the striking one, and it is genuinely surprising the first time. For geometric series the condition for a finite total is a single inequality, which makes this the cleanest possible introduction to convergence.
Figure (svg): A bar showing a length repeatedly halved, with the accumulated pieces approaching but never exceeding the whole
OpenStax, Precalculus, §11.4 Series and Their Notations §11.4, pp. 1331-1335
Section
Section 1
Concept
A summation records where to start, where to stop, which letter varies, and what expression to add at each step.
The index being a dummy variable is worth noticing. Two summations differing only in the letter used are identical, which matters when combining or comparing them.
Figure (svg): A summation expression with its index, lower limit, upper limit and summand each labelled
OpenStax, Precalculus, §11.4 Series and Their Notations §11.4, pp. 1331-1336
Picture it
Four labelled pieces.
Figure (svg): A summation expression with its index, lower limit, upper limit and summand each labelled
Reading all four before computing is what prevents the standard errors: starting at the wrong index, stopping one term early, or misreading which letter varies.
Worked example
Substitute each index in turn.
\[ \text{Expand and evaluate } \sum_{k=1}^{5}(2k+1). \]
Read the limits
Why: One to five.
Substitute each index
Why: In turn.
\[ 3, 5, 7, 9, 11 \]
Add them
Why: The total.
\[ 35 \]
Note the count
Why: Five terms, not four.
Figure (svg): A summation expression with its index, lower limit, upper limit and summand each labelled
\[ 35 \]
Verify: check the term count
Why: The limits are inclusive, so running from one to five gives five terms rather than four. Counting them explicitly before adding is what catches an off-by-one in the limits.
OpenStax, Precalculus, §11.4 Series and Their Notations §11.4, pp. 1332-1334
Prediction
A summation runs from index 3 to index 9.
Predict first
How many terms are there?
Correct: Seven.
Why: Both limits are inclusive, so the count is the difference plus one. Nine minus three is six, and adding one gives seven — the same off-by-one as counting terms against steps.
Worked example
Find the pattern, then set the limits.
\[ \text{Write } 4+7+10+13+16 \text{ in summation notation.} \]
Find the pattern
Why: Differences of three.
Write a formula
Why: Fitting the first term.
\[ 3 k + 1 \]
Check it at the first index
Why: One gives four.
Set the limits
Why: Five terms from one.
\[ 1\text{ to } 5 \]
Figure (svg): The solution to Worked example write a summation shown as a ladder of expressions, one row per legal move
\[ \sum_{k=1}^{5}(3k+1) \]
Verify: check the last term
Why: At index five the summand is 16, which is the last term given — so the upper limit is right. Checking both ends confirms the limits, where checking only one leaves the other unverified.
OpenStax, Precalculus, §11.4 Series and Their Notations §11.4, pp. 1334-1336
Trap
\[ \sum_{k=1}^{5}: \text{ four terms} \]
Treat the upper limit as a stopping point not reached
Why: The limits are read as a range that excludes its end.
One term is omitted and the total is wrong.
Both limits are inclusive. Running from one to five gives five terms.
The count is the upper limit minus the lower, plus one — the same off-by-one that appeared in §11.2's term formula.
Counting the terms explicitly before adding is the check, and it takes one line.
Faded example
Substituting the first two indices.
Fill in the blanks
k=1: \;2(1)+1=3, \quad k=2: \;2(2)+1=5
Why: Each index is substituted into the summand to produce one term. Writing the first two out confirms the summand was read correctly before computing the rest.
Sorting
Four pieces, four roles.
Sort into buckets
Sort each part.
Step zero
You are given a summation to evaluate.
Discussion prompt
What do you read before computing?
Hint: Four things.
Answer:
All four pieces: the index letter, the lower limit, the upper limit and the summand.
Then count the terms — upper minus lower plus one — before adding anything, since that count is what an off-by-one error corrupts.
Writing out the first two terms confirms the summand was read correctly. Those two checks take one line each and catch the errors that are otherwise invisible in a long sum.
Section
Section 2
Concept
A sequence is an ordered list of numbers; a series is what you get by adding them. The distinction matters because questions ask for one or the other.
The partial sums forming their own sequence is the idea that makes infinite series precise. Whether the infinite sum exists is a question about whether that sequence of partial sums approaches a limit.
Figure (svg): Three cards giving the finite and infinite geometric sum formulas and the condition separating them
OpenStax, Precalculus, §11.4 Series and Their Notations §11.4, pp. 1336-1339
Picture it
The third card connects them.
Figure (svg): Three cards giving the finite and infinite geometric sum formulas and the condition separating them
The infinite formula is the finite one after the power term has vanished, which happens only under the stated condition. Seeing them as one formula and a limit rather than two facts is what makes the condition memorable.
Worked example
Read the question carefully.
\[ \text{For } 2,\;4,\;6,\;8: \text{ what is the fourth term, and what is the sum of four terms?} \]
Identify the fourth term
Why: The value at that position.
\[ 8 \]
Identify the sum
Why: Add all four.
\[ 2 + 4 + 6 + 8 \]
Compute the sum
Why: The total.
\[ 20 \]
Compare
Why: Very different answers.
\[ 8\text{ against } 20 \]
Figure (svg): Three cards giving the finite and infinite geometric sum formulas and the condition separating them
\[ a_4=8, \; S_4=20 \]
Verify: note how different they are
Why: The two answers differ by more than a factor of two, so answering the wrong question is not a near miss. Reading for the words 'term' or 'sum' before computing is what settles which is wanted.
OpenStax, Precalculus, §11.4 Series and Their Notations §11.4, pp. 1337-1338
Sorting
A list or its total.
Sort into buckets
Sort each item.
Worked example
Each one adds the next term.
\[ \text{Find the first four partial sums of } 2,\;4,\;6,\;8. \]
First partial sum
Why: Just the first term.
\[ 2 \]
Second
Why: Add the second term.
\[ 6 \]
Third
Why: Add the third.
\[ 12 \]
Fourth
Why: Add the fourth.
\[ 20 \]
Figure (svg): The solution to Worked example build the partial sums shown as a ladder of expressions, one row per legal move
\[ 2,\;6,\;12,\;20 \]
Verify: note what the partial sums form
Why: The four values are themselves a sequence — the sequence of partial sums. Whether an infinite series has a total is exactly the question of whether that sequence settles towards a limit, which is why the idea matters.
OpenStax, Precalculus, §11.4 Series and Their Notations §11.4, pp. 1338-1339
Error analysis
A question asks for the total of the first ten terms.
Annotate
On: \( a_{10}=21 \text{, so the answer is }21 \)
The two answers are usually very different in size, so this is not a near miss. Reading the question for 'term' against 'total' or 'sum' before starting is the reliable habit.
Prediction
You compute the running totals of a sequence.
Predict first
What are those totals?
Correct: Another sequence.
Why: Each partial sum sits at a position — the first, second, third — so the partial sums form a sequence of their own. Whether an infinite series has a total is the question of whether that sequence settles.
Faded example
The third partial sum of 2, 4, 6, 8.
Fill in the blanks
S_3=2+4+6=12
Why: A partial sum adds the terms up to a stated position. The sequence of partial sums grows as long as the terms are positive, which is why convergence requires them to shrink.
Explain it to yourself
They are the tool for making infinite sums precise.
Discussion prompt
Explain what an infinite sum means in terms of them.
Hint: What can a sequence do?
Answer:
Adding infinitely many numbers cannot be done one at a time, so the infinite sum has to be defined rather than computed directly.
The definition uses the partial sums: if that sequence settles towards a particular number, the infinite sum is that number.
If the partial sums run away instead, there is no total. A good explanation notes that this turns a question about infinity into a question about a sequence's behaviour, which is something already familiar.
Section
Section 3
Concept
Arithmetic and geometric series each have a finite sum formula, developed in the previous two sections and applied here through summation notation.
Not every series belongs to one of these two families, and there is no general shortcut for the rest. Recognising which family a series belongs to — or that it belongs to neither — is the first decision.
Figure (svg): Three cards giving the finite and infinite geometric sum formulas and the condition separating them
OpenStax, Precalculus, §11.4 Series and Their Notations §11.4, pp. 1339-1341
Picture it
The first card is the finite one.
Figure (svg): Three cards giving the finite and infinite geometric sum formulas and the condition separating them
The finite formula works for any ratio except one, including ratios larger than one where no infinite sum exists. Only the infinite formula carries the size restriction.
Worked example
Identify the family, then substitute.
\[ \text{Evaluate } \sum_{k=1}^{6}3(2)^{k-1}. \]
Identify the family
Why: A constant ratio.
Read the pieces
Why: First term, ratio, count.
\[ 3, 2, 6 \]
Substitute
Why: Into the sum formula.
\[ 3(1 - 64) / (1 - 2) \]
Compute
Why: Simplify.
\[ 189 \]
Figure (svg): Three cards giving the finite and infinite geometric sum formulas and the condition separating them
\[ 189 \]
Verify: check by adding
Why: The terms are 3, 6, 12, 24, 48, 96, and adding them gives 189 — matching. Six terms is few enough to check directly, which is worth doing once to confirm the formula is being used correctly.
OpenStax, Precalculus, §11.4 Series and Their Notations §11.4, pp. 1340-1341
Prediction
A summation runs from index 1 to index 20.
Predict first
How many terms?
Correct: Twenty.
Why: Both limits are inclusive, so the count is twenty minus one plus one. The count feeds directly into both sum formulas, so getting it wrong corrupts everything after.
Worked example
The other family's formula.
\[ \text{Evaluate } \sum_{k=1}^{20}(3k+2). \]
Identify the family
Why: A constant difference.
Find the first and last terms
Why: At the two limits.
\[ 5\text{ and } 62 \]
Average them
Why: Their mean.
\[ 33.5 \]
Multiply by the count
Why: Twenty terms.
\[ 670 \]
Figure (svg): The solution to Worked example a finite arithmetic sum shown as a ladder of expressions, one row per legal move
\[ 670 \]
Verify: check the term count
Why: The limits run from one to twenty inclusive, giving twenty terms. Using nineteen would have both changed the count and used the wrong last term, so checking the count first protects two later steps.
OpenStax, Precalculus, §11.4 Series and Their Notations §11.4, pp. 1341-1341
Trap
\[ \sum_{k=1}^{5}k^2: \text{ use the arithmetic sum formula} \]
Reach for whichever formula is at hand
Why: The series' family is not checked.
The squares are neither arithmetic nor geometric, so neither formula applies.
Check the family first. The differences of the squares are 3, 5, 7 — not constant — and the ratios are not constant either.
So this series belongs to neither family and must be computed term by term, or with a formula specific to squares.
Not every series has a shortcut, and applying one that does not fit gives a confidently wrong answer.
Sorting
Identify the family first.
Sort into buckets
Sort each summand.
Faded example
First term 3, ratio 2, six terms.
Fill in the blanks
S=\frac6}})}189=\frac______=___
Why: The exponent is the term count, not one less — that distinction belongs to the term formula. Both numerator and denominator are negative, so the quotient is positive.
Explain it
Two formulas, and neither always applies.
Discussion prompt
Explain to a classmate what to check before using one.
Hint: Two tests.
Answer:
Check whether the differences are constant, which means arithmetic, or whether the ratios are, which means geometric.
If neither test passes, neither formula applies and the series must be handled another way — often just by adding the terms.
Applying a formula to a series outside its family gives a confidently wrong answer with no warning. A good explanation notes that the two tests take seconds and that not every series has a shortcut.
Section
Section 4
Concept
An infinite geometric series has a finite total precisely when its ratio is smaller than one in size, because only then do the terms shrink towards zero.
The condition includes negative ratios of small size, where the terms alternate while shrinking. What matters is the size, not the sign, which is why the condition is stated with absolute value.
Figure (svg): A contrast between geometric series that converge and those that do not, distinguished by the size of the ratio
OpenStax, Precalculus, §11.4 Series and Their Notations §11.4, pp. 1341-1344
Picture it
One inequality separates the columns.
Figure (svg): A contrast between geometric series that converge and those that do not, distinguished by the size of the ratio
The right column is not a case needing a different formula but a case with no answer at all. The series simply has no finite total.
Worked example
Check the ratio, then apply the formula.
\[ \text{Evaluate } \tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}+\cdots \]
Identify the pieces
Why: First term and ratio.
\[ \frac{1}{2}\text{ each} \]
Check the condition
Why: The ratio is below one in size.
Apply the formula
Why: First term over one minus the ratio.
\[ \frac{\frac{1}{2}}{\frac{1}{2}} \]
Compute
Why: The total.
\[ 1 \]
Figure (svg): A bar showing a length repeatedly halved, with the accumulated pieces approaching but never exceeding the whole
\[ S=1 \]
Verify: check against the picture
Why: The bar picture showed the pieces filling a length of one exactly, with each new piece halving the remaining gap. The formula and the picture agree, and the answer being a whole number is what makes this the standard first example.
OpenStax, Precalculus, §11.4 Series and Their Notations §11.4, pp. 1342-1343
Prediction
An infinite geometric series has ratio negative one third.
Predict first
Does it have a finite total?
Correct: Yes, since its size is below one.
Why: The condition is on the size of the ratio, not its sign. A negative ratio of small size makes the terms alternate while shrinking, and the partial sums still settle.
Worked example
The condition fails.
\[ \text{Does } 3+6+12+24+\cdots \text{ have a total?} \]
Find the ratio
Why: Each term doubles.
\[ r = 2 \]
Check the condition
Why: Two is not below one.
Consider the terms
Why: They grow without bound.
Conclude
Why: No finite total.
Figure (svg): The solution to Worked example a series with no total shown as a ladder of expressions, one row per legal move
\[ \text{no total} \]
Verify: check what the formula would give
Why: Applying the infinite formula anyway gives 3 over negative one, which is negative three — an absurd answer for a sum of positive terms. That absurdity is the signal that the condition was violated.
OpenStax, Precalculus, §11.4 Series and Their Notations §11.4, pp. 1343-1344
Error analysis
A student sums a growing series.
Annotate
On: \( 3+6+12+\cdots=\frac{3}{1-2}=-3 \)
The formula produces a number for any ratio except one, so it gives no error when misapplied — only a meaningless answer. Checking the condition before substituting is the only defence.
Sorting
Check the ratio's size.
Sort into buckets
Sort each ratio.
Faded example
First term one half, ratio one half.
Fill in the blanks
S=\frac21}}=\frac______=___
Why: The total is the first term divided by one minus the ratio, which applies only when the ratio's size is below one. Here both come out as one half and the total is exactly one.
Explain it
Not every infinite series has a total.
Discussion prompt
Explain to a classmate what the condition requires and why.
Hint: What must the terms do?
Answer:
The terms must shrink towards zero, which for a geometric series means the ratio's size is below one.
If the terms stay the same size or grow, each addition contributes at least as much as the last, so the partial sums run away and never settle.
The condition is on size, not sign — a negative ratio of small size alternates while shrinking, which is fine. A good explanation notes that the formula gives a number regardless, so checking the condition first is the only way to know whether that number means anything.
Section
Section 5
Concept
The infinite formula is the finite one with the ratio's power removed. That power vanishes as the count grows exactly when the ratio is smaller than one in size.
Seeing the infinite formula derived this way makes the condition inevitable rather than an added restriction. It is exactly the condition under which the removed term actually vanishes.
Figure (svg): Three cards giving the finite and infinite geometric sum formulas and the condition separating them
OpenStax, Precalculus, §11.4 Series and Their Notations §11.4, pp. 1344-1345
Picture it
The third card states the derivation.
Figure (svg): Three cards giving the finite and infinite geometric sum formulas and the condition separating them
One formula becomes the other by a term vanishing, and the condition is precisely when that happens. Nothing has to be remembered separately.
Worked example
Let the count grow.
\[ \text{Derive the infinite geometric sum formula from the finite one.} \]
Write the finite formula
Why: First term, ratio, count.
\[ a(1 - r ^{n}) / (1 - r) \]
Consider the power
Why: As the count grows.
\[ r ^{n} \]
Note when it vanishes
Why: Only for small ratios.
\[ | r | < 1 \]
Remove it
Why: The numerator becomes just the first term.
\[ \frac{a}{1 - r} \]
Figure (svg): Three cards giving the finite and infinite geometric sum formulas and the condition separating them
\[ S=\frac{a_1}{1-r} \]
Verify: check the condition's origin
Why: The power vanishes precisely when the ratio's size is below one — repeated multiplication by a small factor shrinks towards zero. So the condition is not an extra restriction but the exact circumstance under which the derivation is valid.
OpenStax, Precalculus, §11.4 Series and Their Notations §11.4, pp. 1344-1345
Prediction
The ratio is one half and the count grows.
Predict first
What does the ratio's power do?
Correct: Shrinks towards zero.
Why: Repeated multiplication by a factor below one in size drives the result towards zero. That vanishing is exactly what turns the finite formula into the infinite one.
Worked example
An infinite series in disguise.
\[ \text{Write } 0.333\ldots \text{ as a fraction.} \]
Write it as a sum
Why: Three tenths, three hundredths, and so on.
\[ 0.3 + 0.03 +... \]
Identify the ratio
Why: Each term is a tenth of the last.
\[ r = 0.1 \]
Check the condition
Why: Below one.
Apply the formula
Why: First term over one minus the ratio.
\[ \frac{0.3}{0.9} \]
Figure (svg): The solution to Worked example a repeating decimal shown as a ladder of expressions, one row per legal move
\[ \tfrac{1}{3} \]
Verify: check by division
Why: Dividing one by three gives 0.333 repeating, confirming the conversion. Every repeating decimal is an infinite geometric series, which is why every one of them equals a fraction.
OpenStax, Precalculus, §11.4 Series and Their Notations §11.4, pp. 1345-1345
Trap
\[ \text{the formula has a restriction, so remember it separately} \]
Memorise the condition alongside the formula
Why: The connection to the derivation is not made.
The condition is then easy to forget, and the formula gets applied where it means nothing.
The condition is where the derivation is valid. The removed power term vanishes only for ratios below one in size.
So the restriction is inevitable rather than added — the formula is a limit, and the limit exists only under that condition.
Understanding the derivation makes the condition unforgettable, because it is the same fact stated twice.
Faded example
As an infinite geometric series.
Fill in the blanks
0.333\ldots=\frac13}}=\frac______=\frac______}
Why: Each digit contributes a tenth of the previous term's value, so the ratio is one tenth. Every repeating decimal is such a series, which is why every one equals a fraction.
Sorting
The finite and infinite formulas differ.
Sort into buckets
Sort each feature.
Explain it to yourself
Every repeating decimal equals a fraction.
Discussion prompt
Explain why, using infinite series.
Hint: What kind of series is a repeating decimal?
Answer:
A repeating decimal is an infinite geometric series: each repetition contributes a fixed fraction of the previous one's value.
The ratio is a power of a tenth, which is always below one in size, so the series always converges.
And the infinite sum formula produces a quotient of two numbers, which is a fraction. A good explanation notes that this proves every repeating decimal is rational, which is otherwise a surprising claim to justify.
Comparison
Fill the blanks from memory. One formula becomes the other by a limit.
Comparison matrix
| finite geometric sum | infinite geometric sum | |
|---|---|---|
| the numerator | first term times one minus the power | just the first term |
| condition on the ratio | anything except one | size below one |
| why the condition | the denominator would vanish | the power term must vanish |
| exists for growing ratios | yes | no |
The third row is what makes the second memorable. The infinite formula's condition is exactly where its derivation is valid, not an extra rule.
Pattern
Five steps, and the third is where infinite series differ.
Step 3 has no counterpart for finite sums, where any ratio works. It is the only place a series can turn out to have no answer at all.
OpenStax Algebra and Trigonometry 2e, §13.4 Series and Their Notations §13.4
Check
Summation notation.
Check your understanding
A summation runs from index 1 to index 6. How many terms are there?
Answer: A
Why: Both limits are inclusive, so the count is the upper minus the lower plus one. The summand affects the values but never the count.
Check
Convergence.
Check your understanding
When does an infinite geometric series have a finite total?
Answer: A
Why: Only then do the terms shrink towards zero and the partial sums settle. The condition is on size rather than sign, so a small negative ratio also works.
Check
Sequence against series.
Check your understanding
What is a series?
Answer: A
Why: A sequence is the list and a series is its total. Confusing them means answering a question about a total with a single term, and the two are usually very different in size.
Real world
A drug reaching a steady level in the body is a geometric series.
Discussion prompt
A patient takes the same dose daily and the body clears a fixed fraction each day. Why does the level stabilise?
Hint: What does each past dose contribute now?
Answer:
Each past dose still contributes something, but reduced by the clearance factor once per day since it was taken — so the contributions form a geometric series.
Because the clearance factor is below one, the series converges and the total approaches a finite steady level rather than growing without bound.
That steady level is the first term over one minus the retained fraction, which is exactly the infinite sum formula. Dosing schedules are designed by choosing a dose that puts that steady level in the therapeutic range, so the convergence condition is what makes repeated dosing safe at all.
Commit first
State your confidence along with your answer.
Predict first
Why must a geometric series' ratio be below one in size to have an infinite total?
Correct: Only then does the ratio's power vanish as the count grows.
Why: The infinite formula is the finite one with that power removed, and removing it is only valid when it actually goes to zero. That happens precisely for ratios below one in size, which is why the condition is inevitable rather than added.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate does not believe infinitely many positive numbers can add to a finite total.
Hint: Draw something.
Answer:
Draw a bar of length one and fill half of it, then half of what remains, then half of that. Each piece is positive and there are infinitely many.
But the filled part never passes the whole bar — each piece fills half the remaining gap, so the gap shrinks towards nothing without ever being exceeded.
So the total is exactly one. A good explanation stresses that the condition is the terms shrinking fast enough: adding a half each time instead would pass any bound, and the picture would fail.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The fourth is the section's genuinely new result and the one worth understanding rather than memorising, since its condition follows from its derivation.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Draw a summation with all four parts labelled. Beside it, write the finite geometric sum formula and derive the infinite one by letting the count grow, marking where the condition comes from. Underneath, draw the halving bar and write what its total is, then convert one repeating decimal to a fraction.
If your derivation shows the condition arising from the power term vanishing, rather than being stated alongside, the section's central idea is understood rather than recorded.
Recap
Five things, and the last is the one that generalises furthest.
| if you remember one thing | it should be this |
|---|---|
| about notation | both limits are inclusive, so count with a plus one |
| about the distinction | a sequence is a list, a series is its total |
| about convergence | the ratio's size must be below one |
| about the condition | it is where the derivation is valid, not an added rule |
Section 11.5 changes subject to counting: how many ways something can happen, which is the foundation the probability of §11.7 rests on.
OpenStax, Precalculus, §11.4 Series and Their Notations §11.4, pp. 1331-1345 — everything on these slides traces back here
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