The foundation half of BC Unit 10, built around one distinction: the terms of a series are one sequence, the partial sums are another, and only the partial sums decide. It covers sequences and their limits, the monotone bounded rule, series as limits of partial sums, telescoping and geometric series with their exact totals, the n-th term test and the one-way arrow it lives on, the grouping proof that the harmonic series diverges, and the p-series family used as the reference for every comparison that follows. Every limit, sum and partial sum in the deck was re-derived and checked numerically.
Subject: Calculus BC · 67 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Calculus BC - Unit 10
What it means to add up forever, and how to tell when the total is actually a number
Objectives
Adding infinitely many numbers is not an operation anyone can perform. This deck replaces it with something you can perform, then builds the first tools for deciding when the replacement produces a number.
College Board, AP Calculus AB and BC Course and Exam Description, Unit 10 Unit 10 — The AP Calculus BC unit this deck opens.
Section
The ingredients, before any adding happens
Warm-up
Everything in this deck rests on a limit you already know how to take. Take it before reading on.
\[ \lim_{x \to \infty} \frac{3x+1}{x+2} \]
Discussion prompt
What is this limit, and what is the single move that gets you there?
Hint: Divide the top and the bottom by the highest power in the denominator.
Answer:
\[ \lim_{x \to \infty} \frac{3x+1}{x+2} = 3 \]
Divide the top and the bottom by the highest power in the denominator. Every leftover piece with a variable underneath it fades to nothing, and only the ratio of the leading coefficients survives. That one move reappears on almost every slide in this part.
Concept
A sequence is an infinite list of numbers in a fixed order. The order is the whole point: there is a first entry, a second, a third, and so on forever, and every entry has an address.
Sequence — A function whose inputs are the positive whole numbers. The output at input n is written with n as a subscript rather than in brackets.
\[ a_1, \; a_2, \; a_3, \; \dots, \; a_n, \; \dots \]
The rule producing the list is either a formula in n, or a recipe that builds each entry out of the one before it. Both kinds turn up in this course, and they need different techniques.
Notation
Four separate pieces of information are packed into one short line. Pull them apart once and the notation stops being mysterious.
Annotate
On: \( a_n = \frac{3n+1}{n+2} \)
Read every subscript as an address, every time. Treating it as a factor is the first mistake this unit invites.
Intuition
A sequence is the same kind of object as a function, with one restriction: you may only feed it whole numbers. That restriction is why its graph is a scatter of separate dots rather than an unbroken curve.
\[ f(x) = \frac{3x+1}{x+2} \qquad \text{versus} \qquad a_n = \frac{3n+1}{n+2} \]
Everything you learned about end behaviour still applies. If the continuous function settles toward a height as its input runs off to the right, then the dots sitting on that curve settle toward the same height. That is the bridge from Calculus AB into this unit, and it is worth crossing deliberately.
Picture it
The dots below are the entries of the sequence from the last slide. Watch what they do rather than what they are.
Figure (svg): Scatter plot of sequence terms climbing steadily toward a dashed horizontal line at height three without ever reaching it.
Converging is a claim about the gap closing forever, not about the gap ever being closed.
Prediction
Look at the first few entries and the tenth before you commit to an answer.
\[ a_1 = \tfrac{4}{3}, \quad a_2 = \tfrac{7}{4}, \quad a_3 = 2, \quad a_4 = \tfrac{13}{6}, \quad a_{10} = \tfrac{31}{12} \]
Predict first
If the list continues forever, where do the entries end up?
Correct: They settle toward 3
Why: Divide the top and the bottom by n. The formula becomes three plus one over n, all divided by one plus two over n. Both small pieces fade to zero, leaving three divided by one. The entries climb toward 3 from below and never pass it.
Worked example
The sequence from the last three slides, done properly rather than by eye.
\[ a_n = \frac{3n+1}{n+2} \]
Divide every term on the top and the bottom by n, the highest power of n in the denominator.
Why: Dividing top and bottom by the same nonzero quantity leaves the value of the fraction alone, and it converts the runaway pieces into shrinking ones.
\[ a_n = \frac{\,3 + \frac{1}{n}\,}{\,1 + \frac{2}{n}\,} \]
Send n off to infinity and read what each piece does separately.
Why: One over n and two over n both shrink to zero, because a fixed number divided by an ever larger number becomes arbitrarily small.
\[ \lim_{n \to \infty} a_n = \frac{3 + 0}{1 + 0} = 3 \]
Verify: evaluate the hundredth entry and compare it with the claimed limit.
Why: The hundredth entry is 301 over 102, about 2.951. That sits within five hundredths of 3 and is still climbing, exactly how a sequence converging to 3 from below should behave.
Figure (svg): The same sequence plotted out to fourteen terms with a dashed line marking the limit at three.
Concept
Converging to a number does not mean reaching it. It means you can name any tolerance, however tight, and from some point onward every single remaining entry sits inside that tolerance.
\[ \lim_{n \to \infty} a_n = L \]
If no such number exists, the sequence diverges. Diverging covers two genuinely different behaviours, and it is worth keeping them apart in your head.
Diverges — Fails to converge. Either the entries grow past every bound, or they keep moving around without ever homing in on a single height.
Definition probe
Sort by long-run behaviour. Say the reason out loud before you place each one.
Sort into buckets
Sort each sequence by what it does in the long run.
Trap
The entries of this sequence never leave the stretch between negative one and one, so the sequence is fenced in. Being fenced in means it has to settle somewhere.
\[ a_n = (-1)^n \;: \quad -1, \, 1, \, -1, \, 1, \, \dots \]
Claimed limit: zero, since the entries balance each other out on average.
Being boxed in is not the same as homing in. The entries do stay between negative one and one, but they never get close to any one number and stay there.
\[ \left| a_n - 0 \right| = 1 \qquad \text{for every } n \]
Every single entry sits a full unit away from zero, so zero is not the limit. The same argument rules out every other candidate, because whatever number you propose, at least one of the two values sits at distance one or more from it. The sequence diverges by oscillation.
Bounded plus monotone is the combination that forces convergence. Bounded on its own is not enough, and an average is not a limit.
Worked example
The same move, one degree higher. The highest power in the denominator is now n squared.
\[ a_n = \frac{2n^2 + 3}{5n^2 - n} \]
Divide the top and the bottom by n squared.
Why: Choosing the highest power in the denominator guarantees the denominator settles on a nonzero number rather than collapsing, which would leave the limit undefined.
\[ a_n = \frac{\,2 + \frac{3}{n^2}\,}{\,5 - \frac{1}{n}\,} \]
Let n run to infinity and read each piece.
Why: Three over n squared and one over n both go to zero, so only the leading coefficients are left standing.
\[ \lim_{n \to \infty} a_n = \frac{2}{5} \]
Verify: test the tenth entry against the claim.
Why: The tenth entry is 203 over 490, about 0.414. The proposed limit is 0.4, so the gap is under two hundredths and still shrinking. Consistent.
Figure (svg): Sequence terms falling toward a dashed line at height zero point four.
Explain it to yourself
Two limits that look almost identical and come out completely differently.
\[ \lim_{n \to \infty} \frac{2n^2+3}{5n^2-n} = \frac{2}{5} \qquad \lim_{n \to \infty} \frac{2n+3}{5n^2-n} = 0 \]
Discussion prompt
Why does the second limit come out as zero when the first one does not?
Hint: Compare the degree of the top with the degree of the bottom.
Answer:
In the first, the top and the bottom grow at the same rate, so their ratio settles on the ratio of the leading coefficients. In the second, the bottom grows a whole power faster, so it outruns the top and drags the fraction to zero.
Compare degrees before anything else. Equal degrees give the ratio of leading coefficients, a bigger bottom gives zero, and a bigger top means the sequence diverges.
Concept
Sometimes you can prove a sequence converges without ever producing the number it converges to. That is worth having, because some limits are hard to name and easy to trap.
Monotone — Always heading the same way. Either every entry is at least as large as the one before it, or every entry is at least as small.
Bounded — Fenced in on both sides. There is a ceiling no entry rises above and a floor no entry drops below.
A sequence that is both monotone and bounded must converge. Picture a climber who never steps down and can never pass the ceiling: the climb is forced to flatten out against some height, even if nobody tells you which height.
Estimation
This one has no formula in n at all. Each entry is manufactured from the entry before it.
\[ a_1 = 1, \qquad a_{n+1} = \sqrt{2 + a_n} \]
Predict first
The entries run 1, then about 1.732, then about 1.932, then about 1.983. Roughly where do they stop?
Correct: At exactly 2
Why: The entries climb and every one of them stays under 2, so the sequence is monotone and bounded and therefore converges. Calling the limit L and taking limits on both sides of the recipe gives L equal to the square root of two plus L, so L squared equals L plus two, whose positive solution is exactly 2 rather than merely near it.
Worked example
The trick is to earn the right to assume convergence first, and then let the recipe pin the value down.
\[ a_1 = 1, \qquad a_{n+1} = \sqrt{2 + a_n} \]
Show the sequence increases and stays below 2.
Why: If an entry is under 2 then two plus it is under 4, so its square root is under 2. The first entry is 1, so by induction every entry is under 2, and each one is larger than its predecessor.
Conclude the limit exists by the monotone bounded rule, and give it the name L.
Why: Increasing and capped above forces convergence. Naming the limit is now legitimate rather than wishful, and that distinction is the whole content of this step.
\[ L = \sqrt{2 + L} \]
Square both sides and collect everything on one side.
Why: Both sides are non-negative here, so squaring cannot smuggle in a false solution that survives the final check.
\[ L^2 - L - 2 = 0 \quad \Longrightarrow \quad (L-2)(L+1) = 0 \]
Verify: test both roots against the sequence itself.
Why: Negative one is impossible because every entry is positive, so it is discarded on sight. Putting 2 back into the recipe gives the square root of four, which is 2, so the limit is self-consistent. The limit is exactly 2.
Figure (svg): Recursive sequence terms rising quickly and flattening against a dashed line at height two.
Counterexample
Both of these sequences increase. Only one of them converges.
\[ b_n = \frac{n}{n+1} \qquad \text{and} \qquad c_n = n \]
Discussion prompt
What does this pair tell you about the claim that every increasing sequence converges?
Hint: Look for the ceiling. Only one of them has one.
Answer:
The first increases and is capped by 1, so it converges to 1. The second increases with no cap at all, so it runs away. Increasing on its own guarantees nothing, so the claim is false and this pair is the counterexample.
The alternating sequence from the trap slide breaks the other half of the rule: it is bounded and still does not converge. You genuinely need both conditions at once, and each half has its own counterexample.
Comparison
Fill in the last column, then read straight down it.
Comparison matrix
| Sequence | Bounded | Monotone | Converges |
|---|---|---|---|
| n-th entry is one over n | yes | yes | yes, to 0 |
| n-th entry is negative one to the n | yes | no | no |
| n-th entry is n | no | yes | no |
| n-th entry is n over n plus one | yes | yes | yes, to 1 |
Only the rows carrying yes in both middle columns are guaranteed. That is the entire content of the monotone bounded rule, and the two no rows are exactly the counterexamples from the last slide.
Pattern
Four moves cover very nearly everything this course will hand you.
| shape you are looking at | what to reach for |
|---|---|
| ratio of polynomials | compare degrees |
| exponential or factorial | growth-rate ranking |
| trapped between two known sequences | the squeeze |
| recursive recipe | monotone bounded, then solve for the limit |
Check
Solve it on paper before you click.
Check your understanding
What is the limit of the sequence whose n-th entry is the quantity four n minus seven, divided by the quantity two n plus five?
Answer: B
Why: Top and bottom are both degree one, so the limit is the ratio of the leading coefficients, four over two, which is 2. The constants minus seven and plus five stop mattering once n is large.
Section
Turning an impossible sum into a possible limit
Concept
You cannot literally add infinitely many numbers. What you can do is add the first few, then a few more, and watch whether those running totals settle down. That is the definition, and it converts every question about an infinite sum into a question about a sequence.
\[ S_N = a_1 + a_2 + \dots + a_N = \sum_{n=1}^{N} a_n \]
Partial sum — The total of the first N terms. There is one for every N, so the partial sums form a sequence in their own right.
\[ \sum_{n=1}^{\infty} a_n = \lim_{N \to \infty} S_N \]
If that limit is a number, the series converges to it. If the limit does not exist, the series diverges and the infinite sum simply has no value. There is no third option.
Notation
Four pieces again, and one of them is not what it looks like.
Annotate
On: \( \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \)
Two different objects live inside this one symbol: the terms, and the running totals of the terms. Almost every mistake in this unit comes from confusing them.
Intuition
When you are asked whether a series converges, you are being asked about the running totals. The terms are only the ingredients, and ingredients do not decide the answer on their own.
| N | what has been added | running total |
|---|---|---|
| 1 | one half | 0.500 |
| 2 | and one sixth | 0.667 |
| 3 | and one twelfth | 0.750 |
| 4 | and one twentieth | 0.800 |
| 10 | and six more | 0.909 |
The right-hand column is climbing toward something just short of 1. That column, not the middle one, is the sequence whose limit decides the answer.
Picture it
Here are the harmonic terms beside the harmonic running totals. Both panels describe exactly the same series.
Figure (svg): Two side-by-side scatter plots: on the left the terms falling toward zero, on the right the running totals climbing steadily with no ceiling in sight.
Whenever you are stuck on a series, ask which of these two pictures you are actually reasoning about.
Worked example
When the general term splits into a difference of two nearly identical pieces, almost everything cancels and the partial sum becomes short enough to take a limit of.
\[ \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \]
Split the general term by partial fractions.
Why: Writing one over n times n plus one as a difference is what exposes the cancellation, and the cancellation is the only reason this sum is doable in closed form at all.
\[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \]
Write out the N-th partial sum and cancel the interior.
Why: Each negative piece is destroyed by the positive piece of the very next term, so only the first positive piece and the last negative piece survive the collapse.
\[ S_N = \left(1 - \tfrac{1}{2}\right) + \left(\tfrac{1}{2} - \tfrac{1}{3}\right) + \dots + \left(\tfrac{1}{N} - \tfrac{1}{N+1}\right) = 1 - \frac{1}{N+1} \]
Take the limit of the partial sums, now that they are a single short expression.
Why: One over N plus one shrinks to zero, so the running totals settle on 1.
\[ \sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \lim_{N \to \infty} \left(1 - \frac{1}{N+1}\right) = 1 \]
Verify: add the first four terms by hand and compare with the closed form.
Why: One half plus one sixth plus one twelfth plus one twentieth is four fifths, which is 0.8. The closed form at N equal to four gives one minus one fifth, also 0.8. The formula and the arithmetic agree exactly.
Figure (svg): Running totals of the telescoping series rising toward a dashed line at height one.
Invariant
Step through the cancellation one term at a time and note which pieces are never touched.
Step through it
Which two pieces are the only ones that ever survive the collapse?
Cancel inside the finite partial sum first, and only then take the limit. Reversing that order is how this technique goes wrong.
Trap
The terms of the harmonic series shrink to nothing. Adding numbers that are heading to zero cannot possibly build an infinite total, so the sum must be finite.
\[ \sum_{n=1}^{\infty} \frac{1}{n} = 1 + \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{4} + \dots \]
Claimed conclusion: converges, because the terms die out.
The terms really do shrink to zero, and the series diverges anyway. Shrinking terms are necessary for convergence and never sufficient for it.
\[ \lim_{n \to \infty} \frac{1}{n} = 0 \qquad \text{and yet} \qquad \sum_{n=1}^{\infty} \frac{1}{n} \text{ diverges} \]
The question is never whether the terms shrink. It is whether they shrink fast enough that the running totals stop climbing. Harmonic terms shrink too slowly, and the grouping argument later in this deck shows exactly how slowly.
Keep the two sequences apart at all times: the terms, and the partial sums. Only the second one is allowed to decide.
Concept
One family of series can be summed exactly, and it is the family you will meet most often. In it, every term is a fixed multiple of the one before it.
\[ \sum_{n=0}^{\infty} a r^n = a + ar + ar^2 + ar^3 + \dots \]
Common ratio — The number r that each term is multiplied by to produce the next one. Divide any term by its predecessor to find it, and check the answer does not depend on n.
\[ \sum_{n=0}^{\infty} a r^{n} = \frac{a}{1-r} \qquad \text{when } \left| r \right| < 1 \]
When the size of the ratio reaches 1 the terms stop shrinking and the series diverges. The closed form is only legal inside that window, and using it outside produces confident nonsense.
Faded example
Both gaps come straight from the definition. Name the first term and the ratio before you write anything.
Fill in the blanks
\sum_5^one third 5\left(\tfrac______\right)^n = \frac___}___}} = \tfrac______
Why: The numerator is the very first term of the series, the one you get by putting the starting index into the general term. The denominator is one minus the common ratio. Here the first term is 5 and the ratio is one third, so the total is 5 divided by two thirds, which is fifteen halves.
Worked example
A series that does not start at index zero, so the first-term rule has to do some real work.
\[ \sum_{n=1}^{\infty} \frac{2}{3^n} \]
Check that the ratio between consecutive terms is genuinely constant.
Why: Dividing the term at n plus one by the term at n gives one third every time, with no leftover n. That is what makes the series geometric and the closed form applicable.
\[ \frac{a_{n+1}}{a_n} = \frac{2/3^{n+1}}{2/3^{n}} = \frac{1}{3} \]
Identify the first term by putting the actual starting index into the general term.
Why: The formula wants the first term that is really present. The count starts at one here, not at zero, and assuming otherwise is the single most common error in this topic.
\[ a = \frac{2}{3^1} = \frac{2}{3}, \qquad r = \frac{1}{3} \]
Apply the closed form, since the size of the ratio is below one.
Why: One third sits comfortably inside the window where the partial sums settle, so the formula is legal here.
\[ S = \frac{a}{1-r} = \frac{2/3}{1 - 1/3} = \frac{2/3}{2/3} = 1 \]
Verify: add the first four terms and compare.
Why: Two thirds plus two ninths plus two twenty-sevenths plus two eighty-firsts is about 0.988. That is just under 1 and still creeping upward, which is what a series converging to 1 from below does.
Figure (svg): Running totals of the geometric series approaching a dashed line at height one.
Commit first
Read the ratio before you reach for any formula.
\[ \sum_{n=0}^{\infty} \left(\tfrac{5}{4}\right)^{n} \]
Predict first
Does this series converge, and if so to what?
Correct: No, it diverges
Why: The common ratio is five fourths, whose size is above one, so each term is larger than the one before it. The terms do not shrink to zero, the running totals climb without bound, and the closed form is simply unavailable. Applying it blindly hands you negative four, which is visibly nonsense for a sum of positive terms and is the best possible warning that you stepped outside the window.
Worked example
Repeating decimals are the most familiar infinite sums there are. You have been relying on their values since primary school without calling them series.
\[ 0.7777\dots = \frac{7}{10} + \frac{7}{100} + \frac{7}{1000} + \dots \]
Read off the first term and the common ratio.
Why: Each place value is one tenth of the one before it, so the ratio is one tenth and the first term is seven tenths.
\[ a = \frac{7}{10}, \qquad r = \frac{1}{10} \]
Apply the closed form.
Why: One tenth is comfortably inside the convergence window, so the formula is legal.
\[ S = \frac{7/10}{1 - 1/10} = \frac{7/10}{9/10} = \frac{7}{9} \]
Verify: divide seven by nine and compare with the decimal you started from.
Why: Long division gives 0.7777 continuing forever, which is exactly the number the series was built from. The geometric formula reproduced it precisely, which is a strong check on both the method and the arithmetic.
Figure (svg): Running totals of the repeating decimal series rising toward a dashed line at seven ninths.
Real world
A patient takes a fixed dose of a drug every day. The body clears a fixed fraction of whatever is present, so a quarter of each dose is still there when the next one arrives.
\[ \text{retained fraction } r = 0.25, \qquad \text{dose } a = 20 \text{ mg} \]
Discussion prompt
If this continues indefinitely, does the amount in the body grow without bound, and if not, what does it level off at?
Hint: Write down what is left of the first dose, the second and the third, then look for a common ratio.
Answer:
Just before each new dose, the drug still present is the total of what remains of every earlier dose. That is a geometric series with ratio one quarter.
\[ \frac{20}{1 - 0.25} = \frac{20}{0.75} \approx 26.7 \]
The level settles at about 26.7 milligrams rather than growing forever, because each dose contributes only a quarter as much as the one before it. This steady state is precisely why a dosing schedule can be repeated indefinitely without the patient accumulating a dangerous amount.
Sorting
For each geometric series, decide whether the closed form may be used at all.
Sort into buckets
Sort by whether the closed form applies.
The test is always on the size of the ratio, never on its sign. Nine tenths converges even though it takes a long time to get anywhere.
Pattern
Four questions, asked in this order, every single time.
| symptom | what it is telling you |
|---|---|
| the ratio still contains n | not geometric, reach for a different tool |
| size of the ratio is at least one | diverges, and there is no total to find |
| a negative answer for a sum of positive terms | you used the formula outside its window |
| an answer smaller than the first term | arithmetic slip, since the total must be at least the first term |
Check
Solve it on paper before you click.
Check your understanding
What is the value of the series whose terms are three times one quarter raised to the n, summed from n equal to zero to infinity?
Answer: C
Why: The first term is three, taken at n equal to zero where one quarter to the zero is one, and the ratio is one quarter. Three divided by one minus one quarter is three divided by three quarters, which is four.
Error analysis
Find the faulty step before you read the notes. The ratio is not the problem.
Annotate
On: \( \sum_{n=2}^{\infty} 6\left(\tfrac{1}{3}\right)^{n} \;\overset{?}{=}\; \frac{6}{1 - \tfrac{1}{3}} = 9 \)
Always compute the first term from the actual starting index. This is the single most common geometric slip, and it never announces itself.
Section
One that kills a series outright, and one reference family
Concept
There is one test worth running before any other, because it is cheap and it settles a surprising number of cases on the spot.
\[ \text{If } \lim_{n \to \infty} a_n \ne 0 \text{, then } \sum a_n \text{ diverges.} \]
The reason is not deep. If the terms do not die out, then every new term adds a chunk of roughly the same size forever, and running totals that keep gaining a fixed amount cannot possibly settle.
Read that arrow in one direction only. It tells you when a series diverges. It never, under any circumstances, tells you that a series converges.
Anomaly
Two series whose terms both shrink to zero, and which behave completely differently.
\[ \sum_{n=1}^{\infty} \frac{1}{n} \qquad \text{versus} \qquad \sum_{n=1}^{\infty} \frac{1}{n^2} \]
Predict first
The terms of both shrink to zero. What is actually true of the two totals?
Correct: The first diverges and the second converges
Why: Terms shrinking to zero is the entry requirement, not the finish line. One over n shrinks too slowly and its running totals climb without bound, while one over n squared shrinks fast enough that the totals settle, at about 1.645. The n-th term test cannot tell these two apart, which is exactly why the rest of the unit exists.
Worked example
The cheapest question you can ask about a series, asked properly and answered in three lines.
\[ \sum_{n=1}^{\infty} \frac{n}{2n+1} \]
Take the limit of the general term, not of the partial sums.
Why: The n-th term test looks only at the ingredients. Building partial sums here would be far more work for a verdict the terms already give away for free.
\[ \lim_{n \to \infty} \frac{n}{2n+1} = \lim_{n \to \infty} \frac{1}{2 + \frac{1}{n}} = \frac{1}{2} \]
Compare that limit with zero.
Why: One half is not zero, so the condition of the test is met and its conclusion becomes available immediately.
\[ \lim_{n \to \infty} a_n = \tfrac{1}{2} \ne 0 \;\Longrightarrow\; \sum_{n=1}^{\infty} \frac{n}{2n+1} \text{ diverges} \]
Verify: sanity-check the verdict with a rough total.
Why: Every term past the first few is close to one half, so a thousand terms contribute roughly five hundred and ten thousand terms roughly five thousand. The running totals grow without limit, which matches the verdict exactly.
Figure (svg): Sequence terms rising and levelling off at a dashed line at height one half, never approaching zero.
Trap
The terms of this series shrink to zero, so the n-th term test is satisfied, so the series converges.
\[ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}, \qquad \lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0 \]
Claimed conclusion: converges by the n-th term test.
The n-th term test has no convergence half. When the limit of the terms is zero the test is silent, and silence is not a verdict.
\[ \lim_{n \to \infty} a_n = 0 \;\Longrightarrow\; \text{no conclusion whatsoever} \]
This particular series is a p-series with p equal to one half, which is not above one, so it actually diverges. A student who read the test backwards would have arrived at exactly the wrong answer while feeling entirely justified.
Write the test on your formula sheet in this shape: nonzero limit means diverges, zero limit means try something else. Never write it as a two-way rule.
Two truths and a lie
Four statements about the n-th term test. Exactly one of them is the test itself.
Eliminate the wrong options
Rule out every statement that tries to extract a convergence verdict, and keep the survivor.
Survives elimination: b
Why: Only the second statement is the test itself. A nonzero limit for the terms genuinely forces divergence, because the running totals keep gaining a fixed amount forever. The other three all try to squeeze a convergence verdict out of a test that has none to give.
Intuition
The harmonic terms shrink, so the running totals climb ever more slowly. The only question is whether that slowdown is enough to stop the climb, and the answer turns out to be a clean no.
Group the terms into blocks whose lengths double: one term, then one, then two, then four, then eight. Inside each block, replace every term with the smallest one in that block. Doing so can only make the total smaller, so any bound proved for the shrunken version holds for the real one.
\[ \underbrace{\tfrac{1}{3} + \tfrac{1}{4}}_{\,>\, 2 \times \frac{1}{4} = \frac{1}{2}} \qquad \underbrace{\tfrac{1}{5} + \tfrac{1}{6} + \tfrac{1}{7} + \tfrac{1}{8}}_{\,>\, 4 \times \frac{1}{8} = \frac{1}{2}} \]
Every block beats one half. There are infinitely many blocks, so the total beats one half added infinitely often, which has no finite value. The climb never stops, it only slows.
Picture it
The same argument as a picture. Each bracket holds strictly more than the same fixed amount.
Figure (svg): Harmonic terms drawn as bars of decreasing height, with brackets grouping terms three and four, then five through eight, each bracket labelled as beating one half.
The bars getting shorter is exactly what makes this feel impossible. The brackets getting wider at the same rate is what makes it true.
Worked example
Turning that picture into an inequality you could actually defend on a written exam.
\[ S_{2^k} = 1 + \tfrac{1}{2} + \left(\tfrac{1}{3}+\tfrac{1}{4}\right) + \left(\tfrac{1}{5} + \dots + \tfrac{1}{8}\right) + \dots \]
Bound each doubling block from below by replacing every entry in it with that block's last and smallest entry.
Why: Shrinking every term in a block can only shrink the block's total, so whatever the shrunken version beats, the genuine version beats too.
\[ \left(\tfrac{1}{3}+\tfrac{1}{4}\right) > 2 \cdot \tfrac{1}{4} = \tfrac{1}{2} \]
Count the blocks out to the term at two to the k, and add up the guaranteed halves.
Why: Going that far produces k blocks, each worth more than one half, sitting on top of the leading 1.
\[ S_{2^k} > 1 + \frac{k}{2} \]
Verify: test the bound at k equal to four.
Why: The bound predicts the total of the first sixteen terms exceeds 3. Adding those sixteen terms directly gives about 3.381, which is indeed above 3. The bound is honest, and since it grows without limit as k does, the series diverges.
Figure (svg): Harmonic running totals climbing steadily with no ceiling out to twenty terms.
Ranking
Slowest growth of the running totals first.
Put in order
Why: The first converges outright, so its totals stop growing altogether. The harmonic totals grow like the natural logarithm, very slowly but without any bound. The square-root series grows like the square root of n, faster still. Adding one every time grows linearly, the fastest of the four by a wide margin.
Concept
One family is worth memorising outright, because almost every comparison you make for the rest of the course will be made against a member of it.
\[ \sum_{n=1}^{\infty} \frac{1}{n^p} \]
\[ \text{converges when } p > 1, \qquad \text{diverges when } p \le 1 \]
The harmonic series is the boundary case at p equal to one, and it falls on the diverging side. That single fact is why the boundary is worth remembering rather than re-deriving under exam pressure.
Matching
The exponent alone decides every one of these.
Match the pairs
Why: Three and one point one are both above one, so those two converge. The square root is the exponent one half, which is below one, so that diverges. An exponent of exactly one is the harmonic series, which sits on the boundary and falls on the diverging side.
Edge cases
All the interesting behaviour in this family is crammed right next to the boundary.
\[ \sum_{n=1}^{\infty} \frac{1}{n^{1.001}} \qquad \text{versus} \qquad \sum_{n=1}^{\infty} \frac{1}{n} \]
Discussion prompt
These two series have almost identical terms for any n you could actually write down. How can one converge and the other not?
Hint: Ask what happens in the tail, not in the first thousand terms.
Answer:
Convergence is decided entirely by the infinite tail, never by any finite stretch. For the first few million terms the two series really are nearly identical, and their running totals track each other closely.
Far enough out, the tiny extra exponent compounds. The first series accumulates only a finite amount in its tail and stops, while the harmonic tail keeps contributing forever. The first totals roughly a thousand, which is enormous but finite, and finite is the only thing that matters here.
The lesson generalises: no amount of numerical evidence from early terms can settle a convergence question. Only an argument about the tail can.
Pattern
Two tests so far, plus two exact-total families, and a fixed order that saves real work.
| shape of the general term | first thing to reach for |
|---|---|
| a constant multiple of a fixed ratio to the n | geometric |
| one over a fixed power of n | p-series |
| a fraction that does not shrink to zero | n-th term test |
| a product of two factors a fixed distance apart | partial fractions, then telescope |
Check
Solve it on paper before you click.
Check your understanding
Which of these series can be shown to diverge using the n-th term test?
Answer: B
Why: The terms of the second head toward 1 rather than zero, so the n-th term test applies directly and delivers divergence in one line. The other three all have terms shrinking to zero, so the test is silent about them whatever their actual behaviour turns out to be.
Section
Choosing the tool before doing the work
Elimination
One tool settles this in a single line. Rule out the three that would waste your time.
\[ \sum_{n=1}^{\infty} \frac{3^n}{4^n} \]
Eliminate the wrong options
Which single tool settles this series immediately?
Survives elimination: b
Why: Three to the n over four to the n is just three quarters to the n, which is geometric with ratio three quarters. The size of that ratio is below one, so the series converges, and the first term divided by one minus the ratio gives the exact total of three.
Discrimination
Sorting only. Do not compute anything, because the skill being practised here is recognition.
Sort into buckets
Sort each series by the first tool you would reach for.
Trade off
Fill the missing cells. Every tool buys you something and charges you something.
Comparison matrix
| Tool | Can prove convergence | Gives the exact total | What it costs |
|---|---|---|---|
| n-th term test | never | no | one limit |
| geometric | yes | yes | identify the first term and the ratio |
| telescoping | yes | yes | partial fractions plus a partial sum |
| p-series | yes | no | read one exponent |
Only two of these hand you an actual number. The other two give a verdict, which is usually all the question asked for anyway.
Fill the middle
Both ends are given. The gap in the middle is the partial-fractions split, and the gap on the right is the total.
Fill in the blanks
\sum_one half times the quantity one over n minus one over n plus two^three quarters \frac______ = \sum____^___ ___ \quad \Longrightarrow \quad S = ___
Why: Partial fractions give one half times the difference of one over n and one over n plus two. Because the gap between the two pieces is two rather than one, two leading terms survive the cancellation instead of one, so the total is one half times the quantity one plus one half, which is three quarters.
Reverse engineer
You are told the total and the ratio. Recover the first term.
Fill in the blanks
\sum_4^___ a r^n = 10, \qquad r = \tfrac______ \quad \Longrightarrow \quad a = ___
Why: The closed form says the total equals the first term divided by one minus the ratio. One minus three fifths is two fifths, so the first term is ten times two fifths, which is 4. Checking forwards: 4 divided by two fifths is 10, so the recovered value is right.
Socratic
This implication is true, and proving it is what makes the n-th term test legitimate rather than a rule of thumb.
\[ \sum_{n=1}^{\infty} a_n \text{ converges} \quad \Longrightarrow \quad \lim_{n \to \infty} a_n = 0 \]
Discussion prompt
Explain in your own words why the convergence of the total forces the terms down to zero.
Hint: Write the n-th term as the difference of two consecutive partial sums.
Answer:
If the running totals settle on a number, then two consecutive running totals must be getting arbitrarily close to each other, because both of them are getting arbitrarily close to the same limit.
\[ a_N = S_N - S_{N-1} \;\longrightarrow\; L - L = 0 \]
The gap between consecutive running totals is exactly the next term. So the terms are squeezed to zero. This is the contrapositive of the n-th term test, and writing it out this way is the cleanest reason to trust the test in the first place.
Explain it
A classmate produces the following and asks you to check it.
\[ \sum_{n=1}^{\infty} \left(\tfrac{1}{n} - \tfrac{1}{n+1}\right) = 1 \qquad \text{but} \qquad \sum_{n=1}^{\infty} \tfrac{1}{n} \;-\; \sum_{n=1}^{\infty} \tfrac{1}{n+1} \text{ is undefined} \]
Discussion prompt
Your classmate concludes that the telescoping total must therefore be undefined too. Where does the reasoning break?
Hint: Ask whether each of the two separated pieces converges on its own.
Answer:
Splitting one convergent series into a difference of two series is only legal when both pieces converge separately. Here both pieces diverge, so the split manufactures a meaningless expression out of a perfectly well-defined one.
The honest route is the one used earlier in this deck: build the N-th partial sum, cancel inside it while it is still finite, and take the limit only at the very end. That route never splits the series and never meets an undefined expression.
Rule of thumb worth keeping: cancel inside a finite partial sum, then take the limit. Never take the limit first and cancel afterwards.
Check
Solve it on paper before you click.
Check your understanding
Consider the series whose n-th term is the quantity two to the n plus three to the n, all divided by three to the n, summed from n equal to one. What does it do?
Answer: C
Why: Split the term into two thirds raised to the n, plus one. The first piece dies away but the constant 1 does not, so the terms head toward 1 rather than zero and the n-th term test delivers divergence immediately.
Connect it up
One page, in your own handwriting, that you will add to every time the course hands you a new test.
Draw it
Draw a decision tree for an unfamiliar series. Start from the general term. Put the n-th term test at the top as the first cheap check, then branch to geometric, p-series and telescoping according to the shape of the term. At each leaf write down what you actually get: a verdict only, or a verdict plus an exact total. Leave one branch open and label it for the comparison tests coming next.
Keep this at the front of your notes. A decision tree you built yourself is worth more than a printed table you did not.
Exit ticket
The single distinction this whole deck was built around.
Predict first
A series has terms that shrink to zero. What have you learned about the series?
Correct: Nothing yet - the cheap test is silent and more work is needed
Why: Terms shrinking to zero is the entry requirement for convergence, never a proof of it. The harmonic series clears that bar and diverges, while one over n squared clears it and converges. Geometric, p-series, telescoping or one of the comparison tests is needed to break the tie.
Recap
Six concrete abilities, and one distinction underneath all of them.
| what you are looking at | the move |
|---|---|
| a ratio of polynomials in n | compare degrees |
| a constant ratio between consecutive terms | geometric: first term over one minus the ratio |
| one over n to a fixed power | p-series: compare the exponent with one |
| terms that do not go to zero | n-th term test: diverges |
| terms that do go to zero | no verdict yet, keep working |
The distinction underneath all of it: the terms are one sequence, the partial sums are another, and only the second one ever decides.
OpenStax, Calculus Volume 2, Ch. 5 (Sequences and Series) Sections 5.1-5.3 — Sequences, infinite series, and the divergence test.
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