6 concepts19 worked examples25 exercises3 exam-level4 figures
What are you here for?
02Series: partial sums, the geometric and telescoping series, and the Test for Divergence
Drop a ball from a height of $2$ m. Each time it lands it comes back to exactly three quarters of the height it fell from, then falls again, and it keeps doing this forever. A tape measure that adds up every rise and every fall records a longer and longer number as the bounces pile up. The bounces never stop, so it is tempting to say the tape reading runs off to infinity — yet the ball plainly comes to rest in a few seconds over a patch of floor you could cover with your hand.
By the end of this section you can add those infinitely many bounces to one exact number, $14$ m, and state precisely when an endless sum settles on a value and when it genuinely blows up.
In 60 seconds
An infinite series adds up to the number its running totals close in on: name the type first (geometric, telescoping, or neither), apply that type's method to the partial sum $s_n$, and take $n\to\infty$.
always the first check; if the terms do not shrink to zero the sum cannot settle
Three most common mistakes
Reading $a_n\to 0$ as a guarantee of convergence. It is only a necessary condition; the harmonic series $\sum 1/n$ has $a_n\to 0$ and still diverges.
Using $\dfrac{a}{1-r}$ without checking $|r|<1$. When $|r|\ge 1$ that fraction is a meaningless number for a series that has no sum at all.
Taking the first term $a$ to be the wrong value when the sum starts at $n=0$ instead of $n=1$, or when the terms are written as $ar^{n}$ instead of $ar^{n-1}$.
On this course the Quiz and Homework components carry $10\%$ each. The three moves in this section — sum a geometric series, collapse a telescoping one, and reject a series by the Test for Divergence — are exactly the short single-method questions those two components are built from.
How much time do you have?
10 minutes
The two moves that clear most quiz questions: the geometric sum with its |r|<1 check, and the one-line test that rejects a series whose terms do not go to zero.
card, c-geometric-series, c-divergence-test, formula card
45 minutes
Every type you can be handed: what a sum even means, the geometric and telescoping methods, the one series everybody trips over (harmonic), the rejection test, and one full ladder from a worked solution to a bare one.
card, c-series-as-partial-sums, c-geometric-series, c-telescoping, c-harmonic, c-divergence-test, faded ladder, formula card
full read
Where an infinite sum comes from, why each formula is allowed, and enough mixed practice that you decide the method yourself instead of being told it.
hook, prerequisites, c-series-as-partial-sums, c-geometric-series, c-telescoping, c-harmonic, c-divergence-test, c-algebra-of-series, contrast pairs, faded ladder, exam example, practice A to D
By the end of this section
Define the sum of an infinite series as the limit of its sequence of partial sums, and read off convergence or divergence from that limit.
Sum a geometric series with the formula $a/(1-r)$ after checking $|r|<1$, and turn a repeating decimal or a bouncing distance into such a series.
Collapse a telescoping series by splitting its term into a difference and cancelling the middle, then take the limit of what survives.
Explain why the harmonic series diverges even though its terms shrink to zero, using the grouping argument.
Apply the Test for Divergence to reject a series whose terms do not tend to zero, and state exactly why the test says nothing when they do.
Combine convergent series with the linearity rules, and recognise when a sum of a convergent and a divergent series must diverge.
Syllabus coverage
11.2
Series: partial sums and convergence, the geometric series, telescoping sums, the harmonic series, the Test for Divergence, and the algebra of convergent series
The single book section for this week is spread over six blocks below: the definition through partial sums, the geometric series, telescoping sums, the harmonic series, the Test for Divergence, and the linearity rules for combining convergent series.
covered
Recall first
Limit of a sequence
$\lim_{n\to\infty} c_n = L$ means the terms $c_n$ close in on $L$ as $n$ grows; if no single such $L$ exists the sequence diverges.
The sum of a series is defined as one particular sequence limit, so every convergence question here is a sequence limit in disguise.
Limit of a power
$\lim_{n\to\infty} r^{\,n} = 0$ when $|r|<1$, while $|r^{\,n}|\to\infty$ when $|r|>1$.
This single fact is what makes the geometric series for small ratios and blow up for large ones.
Partial fractions
A quotient like $\dfrac{1}{n(n+1)}$ can be rewritten as a difference $\dfrac{1}{n}-\dfrac{1}{n+1}$ by solving for the unknown numerators.
Splitting a term into a difference is the whole trick that makes a telescoping series collapse.
Sequence versus its running totals
From a sequence $a_n$ we build a second sequence $s_n=a_1+\cdots+a_n$; these are different objects with possibly different fates.
The previous section studied $a_n$; this section studies $s_n$, and confusing the two is the single most common error below.
Try it yourself first (3 questions)
1§02.0 — do shrinking terms force a finite sum?●○○○○
Before the section, an honest guess. A classmate argues: even though the terms of $1+\tfrac12+\tfrac13+\tfrac14+\cdots$ get as small as we like, the running totals still climb past every bound, so the sum is infinite.
Given
Claim: even though $\dfrac1n\to 0$, the running totals of $\displaystyle\sum_{n=1}^{\infty}\dfrac1n$ grow past every bound, so the sum is infinite.
Find
(a) True or false? A guess is fine; the section settles it.
Hint 1/4
Ask whether small terms alone can stop a running total from growing.
Hint 2/4
Terms going to zero is necessary for a finite sum, but the question is whether it is enough.
Hint 3/4
Think of adding many terms each near, say, $1/1000$: a thousand of them still add to about $1$.
Hint 4/4
Shrinking terms do not by themselves force a finite total; the harmonic totals do run off, so the claim is true.
two terms already beat one half, and later blocks do the same
Answer $$\boxed{\text{True}}$$
Check
The full grouping argument is developed in the harmonic-series block below.
2§02.0 — a running total you can already read●○○○○
You have met sequence limits already. Here is a sequence of running totals with a formula, so you can find the sum without knowing the individual terms.
Given
The running total of the first $n$ terms is $s_n=\dfrac{2n}{n+1}$.
Find
(a) What number do the running totals approach as n grows?
Hint 1/4
The sum of the series is just the limit of the running totals you are given.
Hint 2/4
For a quotient of polynomials of equal degree, the limit is the ratio of leading coefficients.
Hint 3/4
Here $s_n=\dfrac{2n}{n+1}$, and dividing top and bottom by $n$ gives $\dfrac{2}{1+1/n}$.
divide numerator and denominator by the highest power, then let 1/n vanish
Answer $$\boxed{2}$$
Check
Whenever you are handed a formula for the running totals, the sum is one limit away.
3§02.0 — is a geometric-looking sum always finite?●○○○○
One more guess. A geometric-looking sum multiplies each term by a fixed number to get the next. Does such a sum always land on a finite value?
Given
Claim: $\displaystyle\sum_{n=1}^{\infty} 2^{\,n-1}=1+2+4+8+\cdots$ adds up to a finite number.
Find
(a) True or false?
Hint 1/4
Ask whether the running totals here settle or keep growing.
Hint 2/4
A geometric series only settles when the fixed multiplier has size below $1$.
Hint 3/4
Here the multiplier is $2$, so each term is bigger than the last: $1,2,4,8,\dots$
Hint 4/4
The totals $1,3,7,15,\dots$ run off to infinity, so the sum is not finite.
Show solutionLook at the running totals
$$s_1=1,\ s_2=3,\ s_3=7,\ s_4=15$$
each total is one less than the next power of two, so they increase without bound
Answer $$\boxed{\text{False}}$$
Check
Geometric sums settle only when the ratio is small; this is the content of the geometric block.
Notation
symbol
reads as
means
watch out
$\displaystyle\sum_{n=1}^{\infty} a_n$
the sum from n equals 1 to infinity of a sub n
the limit of the running totals of the terms a_1, a_2, a_3 and so on
It is a single number (or the word diverges), not a list of terms; the letter $n$ is a dummy and never appears in the answer.
$s_n$
s sub n, the n-th partial sum
the finite total of just the first n terms, a_1 through a_n
This is an ordinary finite sum; the infinite series is $\lim_{n\to\infty} s_n$, a different object from $s_n$ itself.
$a_n$
a sub n, the n-th term
one single term of the series, the thing being added at step n
Do not confuse $a_n$ (a term) with $s_n$ (a running total); a whole section of mistakes lives in that confusion.
$r$
r, the
the fixed number each term of a geometric series is multiplied by to reach the next
Compute it as $r=a_{n+1}/a_n$ and confirm it is the same for every $n$; if it is not constant the series is not geometric.
$\text{converges}$ / $\text{diverges}$
converges or diverges
the running totals do, or do not, close in on a single finite number
Diverges covers three different behaviours: running off to $\pm\infty$, or oscillating with no limit at all.
Conventions used here
What convergence of a series means on this page.
A series converges when its sequence of partial sums $s_n$ has a finite limit, and that limit is called the sum. It diverges otherwise. We never treat $\sum a_n$ as a literal act of adding infinitely many numbers; it is always shorthand for $\lim_{n\to\infty} s_n$. So the question is never can you add them all up, but does the running total settle.
The whole subject is built on one definition, and half of the errors below come from reasoning about the terms $a_n$ when the definition only ever talks about the totals $s_n$.
How we report a divergent series.
When the partial sums run off in one direction we write $\sum a_n=\infty$ or $-\infty$, because that says more than the bare word diverges. When they oscillate with no limit, as for $\sum(-1)^n$, we simply say the series diverges. Either way there is no finite sum, and writing a number down for a divergent series is wrong even if the number is $\infty$ treated as an ordinary value.
Both phrasings are standard, and mixing them inside one answer is what loses the mark that the divergence discussion is meant to earn.
Where the index starts.
A geometric series can be written $\sum_{n=1}^{\infty} ar^{\,n-1}$ (first term $a$) or $\sum_{n=0}^{\infty} ar^{\,n}$ (also first term $a$) or $\sum_{n=1}^{\infty} ar^{\,n}$ (first term $ar$). The letter $a$ in the formula $a/(1-r)$ is always the actual first term you are adding, not the constant sitting in front of the power. Always read off the first term by plugging in the starting index.
The single most common geometric-series slip is using the wrong $a$ because the starting index or the exponent was read carelessly.
Adding or dropping finitely many terms.
Changing, inserting or deleting a finite number of terms never changes whether a series converges or diverges, though it does change the sum. So $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=5}^{\infty} a_n$ live or die together; only the value differs by the finitely many terms you dropped.
Convergence is a statement about the tail, and the tail is untouched by any finite edit at the front.
2.1A series is the limit of its partial sums
Turns an endless sum into a sequence limit by watching the running totals $s_n$ instead of adding all the terms at once.
The previous section handed us a way to take the limit of a sequence; now we point that same tool at the running totals of an infinite list of numbers.
Solvable with what we have
Add any finite list: $3+1+4+1+5=14$.
Take the limit of a sequence, from the previous section.
Not solvable yet
Literally add infinitely many numbers, one after another, because that process never ends.
Decide by eye whether $1+\tfrac12+\tfrac14+\cdots$ is a real number or nonsense.
Try to just keep adding. For $1+\tfrac12+\tfrac14+\tfrac18+\cdots$ the running totals are $1,\ 1.5,\ 1.75,\ 1.875,\ 1.9375,\dots$ and they press up toward $2$. For $1+2+3+4+\cdots$ the running totals $1,3,6,10,\dots$ just keep climbing. The adding never finishes in either case, so how can one have an answer and the other not?
Why it fails
There is no last term to reach, so adding is the wrong picture. What separates the two examples is not the adding but the running totals: one sequence of totals has a limit and the other does not. That observation is the whole definition.
DefinitionDefinition: sum of an infinite series
Conditions
the terms $a_1,a_2,a_3,\dots$ are given
the partial sums are $s_n=a_1+a_2+\cdots+a_n$
$$\boxed{\ \sum_{n=1}^{\infty} a_n=\lim_{n\to\infty} s_n\quad\text{when this limit exists (finite).}\ }$$
The sum of the whole series is the number the running totals close in on. If those running totals settle on a finite value the series converges to it; if they run off or oscillate, the series diverges and has no sum.
The running totals of 1/2 + 1/4 + 1/8 + ... climb 0.5, 0.75, 0.875, 0.9375 and press up against the line s = 1 without landing on it. That line is the sum.
Looks like this, but is not
The series $1-1+1-1+1-\cdots$ has terms that stay small and bounded, so it looks like it ought to have a sum, perhaps $0$ by pairing, or $\tfrac12$ by a clever average.
The partial sums are $1,0,1,0,1,0,\dots$, which never close in on a single number. Boundedness is not convergence; the definition demands a limit of the totals, and this sequence of totals has none, so the series diverges.
Summing 1/2 + 1/4 + 1/8 + ... by a closed form for the totals
Show that $\displaystyle\sum_{n=1}^{\infty}\frac1{2^{\,n}}$ converges and find its sum.
Given
$a_n=\dfrac1{2^{\,n}}$, so the terms are $\tfrac12,\tfrac14,\tfrac18,\dots$
both are sequences indexed by n and both are written with subscripts, so they blur together
wrong$$a_3\stackrel{?}{=}s_3$$
right$$a_3=s_3-s_2$$
2.2The geometric series and the a over one minus r formula
Sums any series whose terms share a constant ratio $r$, giving $a/(1-r)$ the moment $|r|<1$.
The doubling series above blew up because its ratio was $2$; the halving series settled because its ratio was $\tfrac12$. That contrast is a theorem waiting to be written.
TheoremTheorem: the geometric series
Conditions
each term is a fixed multiple of the previous one: $a_{n+1}=r\,a_n$
$a$ is the first term actually added
$$\boxed{\ \sum_{n=1}^{\infty} ar^{\,n-1}=\frac{a}{1-r}\ \text{ if }|r|<1;\quad\text{diverges if }|r|\ge 1.\ }$$
If every term is r times the one before and r has size below one, the endless sum equals the first term divided by one minus the ratio. If r has size one or more, the terms do not shrink and the series has no sum.
Cut a square of area 1 in half, then halve what is left again and again. The pieces have areas 1/2, 1/4, 1/8, 1/16, ... and together they are the whole square, so the sum of those areas is 1.
Looks like this, but is not
The series $\displaystyle\sum_{n=1}^{\infty}\frac{n}{2^{\,n}}=\frac12+\frac24+\frac38+\cdots$ looks geometric because a power of two sits in every denominator.
The ratio of consecutive terms is $\dfrac{a_{n+1}}{a_n}=\dfrac{n+1}{2n}$, which changes with $n$ (it is $1$, then $\tfrac34$, then $\tfrac23,\dots$). A geometric series needs the same ratio every time, so $a/(1-r)$ does not apply here; the extra factor $n$ disqualifies it.
Summing 3 + 3/4 + 3/16 + ... with the geometric formula
Sums a non-geometric series by splitting each term into a difference so consecutive pieces cancel and $s_n$ collapses.
Not every series is geometric, but some still have a partial sum you can write in closed form, because the terms are secretly differences that eat each other.
MethodMethod: telescoping a series
Conditions
the term can be written as a difference $a_n=b_n-b_{n+1}$
If each term is one piece minus the next piece, then in the running total every interior piece is added once and subtracted once, so only the very first piece and the very last piece survive. The sum is the first piece minus the limit of the last.
Writing out the partial sum of 1/(n(n+1)) shows every interior fraction paired with its own negative. All the middle pieces cancel and only the first and last survive: s_n = 1 - 1/(n+1).
Looks like this, but is not
The series $\displaystyle\sum_{n=1}^{\infty}\frac1{n^2}$ has a tidy term and it is tempting to hunt for a telescoping split $\frac1{n^2}=b_n-b_{n+1}$.
No elementary $b_n$ makes $b_n-b_{n+1}=\frac1{n^2}$, so this series does not telescope; its convergence needs a different tool (the tests in the next section). Telescoping is a lucky structure, not a universal method.
2.4The harmonic series diverges, even though its terms vanish
Shows by grouping that $\sum 1/n$ has no sum, the standard warning that $a_n\to 0$ is not enough for convergence.
We have summed several series exactly; here is a famous one whose terms shrink to zero and that still has no sum, and it is the reason the next test has to be worded so carefully.
TheoremTheorem: the harmonic series diverges
Conditions
the terms are $a_n=\dfrac1n$, which do tend to $0$
The sum of the reciprocals of the whole numbers grows without bound. The individual terms get as small as you like, but there are so many of them that the running total still climbs past every ceiling.
Sample the harmonic partial sums at 1, 2, 4, 8, 16 terms: they reach 1, 1.5, 2.08, 2.72, 3.38. Every time the term count doubles the total gains at least 1/2, and half added infinitely often has no ceiling.
Looks like this, but is not
Because the terms $\tfrac1n$ march down to $0$, the running total looks like it should flatten out and settle, exactly as $\sum\tfrac1{2^n}$ did.
Shrinking terms only stop the total from jumping; they do not stop it from creeping upward forever. The grouping shows the creep never runs out: every doubling of the term count buys another $\tfrac12$, and you can double as often as you please.
Why halving every term does not rescue the harmonic series
Does $\displaystyle\sum_{n=1}^{\infty}\frac{1}{2n}=\frac12+\frac14+\frac16+\cdots$ converge?
right$$a_n\to 0\ \text{is necessary, not sufficient: } \sum\tfrac1n=\infty$$
2.5The Test for Divergence
Rejects any series whose terms fail to reach $0$; silent, by design, whenever the terms do reach $0$.
The harmonic series showed that shrinking terms are not enough for a sum. Turn that around and you get the fastest test there is: if the terms do not even shrink to zero, the series is doomed.
TheoremTheorem: Test for Divergence
Conditions
$a_n$ is the general term of the series $\sum a_n$
$$\boxed{\ \text{If }\lim_{n\to\infty}a_n\neq 0\text{ or does not exist, then }\sum_{n=1}^{\infty} a_n\text{ diverges.}\ }$$
If the terms do not settle down to zero, the running total keeps getting kicked by pieces of a fixed size and cannot possibly converge. The test only ever proves divergence; it never proves convergence.
Looks like this, but is not
It is tempting to read the test as a two-way rule and conclude that $a_n\to 0$ proves $\sum a_n$ converges.
That direction is exactly what the harmonic series refutes: $\tfrac1n\to 0$ and yet $\sum\tfrac1n=\infty$. When $a_n\to 0$ the Test for Divergence returns no verdict at all, and you must reach for a different test.
Rejecting a series whose terms head to a nonzero limit
$$\lim_{n\to\infty}(-1)^{\,n}\ \text{does not exist}$$
the terms flip between -1 and +1 and never settle
Apply the Test for Divergence
$$\text{limit not }0\ (\text{in fact DNE})\ \Rightarrow\ \text{diverges}$$
the partial sums bounce -1, 0, -1, 0 and never converge
Answer $$\boxed{\text{Diverges}}$$
Check
Check by partial sums: they read -1, 0, -1, 0, ..., which has no single limit, matching the verdict.
A limit of the terms that fails to exist counts as not zero for this test, so oscillating terms are an immediate rejection.
Checkpoint
§02.5 — can this test ever prove the harmonic series diverges?●●●○○
Careful reading of what the test can and cannot do. A classmate applies the Test for Divergence to $\sum\tfrac1n$ and reports that, because $\lim\tfrac1n=0$, the test returns no verdict here.
Given
Claim: applied to $\displaystyle\sum_{n=1}^{\infty}\frac1n$, the Test for Divergence is inconclusive, because $\lim_{n\to\infty}\tfrac1n=0$.
Find
(a) True or false, and say what the test actually returns here.
Hint 1/4
Ask what the test needs to see in order to say anything.
Hint 2/4
The Test for Divergence fires only when $\lim a_n\neq 0$; if $\lim a_n=0$ it is silent.
Hint 3/4
Here $\lim_{n\to\infty}\tfrac1n=0$, so the test returns no verdict at all.
Hint 4/4
So the claim is true: the harmonic series does diverge, but by grouping, not by this test — here the test itself is silent.
Show solutionCheck the trigger of the test
$$\lim_{n\to\infty}\frac1n=0$$
the test only concludes divergence when this limit is nonzero or missing
right$$a_n\to 0\ \Rightarrow\ \text{test says nothing; try another test}$$
⚠ Claiming the Test for Divergence proves convergence
having one fast test, students hope it is a full decision procedure rather than a one-sided rejection rule
wrong$$\lim a_n=0\ \stackrel{?}{\Rightarrow}\ \text{Test for Divergence proves it converges}$$
right$$\text{The test can only ever prove DIVERGENCE, never convergence.}$$
2.6Combining series: linearity and its one trap
Adds and scales convergent series term by term, and flags the case where a convergent plus a divergent series must diverge.
Once you can sum the standard types, real problems hand you a mix; the rules for adding and scaling series let you take them one recognizable piece at a time.
You may pull a constant out of a convergent series and you may add two convergent series term by term, in each case getting the sum you would expect. The catch is that both pieces must already converge for the term-by-term rule to be legal.
Looks like this, but is not
The rule $\sum(a_n+b_n)=\sum a_n+\sum b_n$ looks like it should hold no matter what, so one might split $\sum\big(\tfrac1n-\tfrac1n\big)$ into $\sum\tfrac1n-\sum\tfrac1n$.
The left side is $\sum 0=0$, but the right side is $\infty-\infty$, which is meaningless. The rule needs both series to converge; here neither does, so splitting is illegal even though the combined series is perfectly well behaved.
Summing a mixture of a geometric and a telescoping series
if the sum converged, subtracting the convergent geometric part would make the harmonic series converge, which it does not
Answer $$\boxed{\text{Diverges}}$$
Check
Check the logic by contradiction: a finite total minus a finite total is finite, but the harmonic total is not, so the combined total cannot be finite.
A single divergent ingredient poisons the whole sum; you cannot average it away with a convergent one.
Checkpoint
§02.6 — scale and add two convergent series●●●○○
Use linearity to evaluate a combination of two geometric series you can each sum by formula.
Partial sums 3, 3.75, 3.9375 already sit just under 4.
2 · you write the reasoning
An easier one, and this time you write the reasons. Compute $$\sum_{n=1}^{\infty}\left(\tfrac12\right)^{\,n-1}$$ and, in the empty column, say why each line is allowed.
reasoning
The ratio between consecutive terms is the constant 1/2, and its size is below 1, so this is a convergent geometric series and the formula applies.
reasoning
Apply the geometric partial-sum formula with first term a = 1 and ratio r = 1/2.
reasoning
As n grows, (1/2)^n tends to 0, so the only moving part disappears and the total is 2.
3 · find the buried error
Harder, and the work is done for you, badly. Two of the four steps are wrong. Find both. The problem: sum the telescoping series $$\sum_{n=1}^{\infty}\frac{1}{n(n+2)}.$$
the two buried errors (2)
⚠ step 2
The partial-fraction split is missing its coefficient. Solving $1=A(n+2)+Bn$ gives $A=\tfrac12,\ B=-\tfrac12$, so $\dfrac{1}{n(n+2)}=\dfrac12\!\left(\dfrac1n-\dfrac1{n+2}\right)$, with a factor of $\tfrac12$.
when the two factors differ by more than 1, the numerators are not plus and minus 1, but the eye expects the clean 1/n - 1/(n+2) pattern from the differ-by-1 case.
right
Keep the $\tfrac12$: $\dfrac{1}{n(n+2)}=\dfrac12\!\left(\dfrac1n-\dfrac1{n+2}\right)$.
⚠ step 3
With a gap of 2, two terms survive at each end, not one. The collapse leaves $s_n=\tfrac12\!\left(1+\tfrac12-\tfrac1{n+1}-\tfrac1{n+2}\right)$, because 1 and 1/2 both fail to find a partner.
the differ-by-1 telescope leaves a single front survivor, and that habit gets copied to the gap-2 case where in fact two front terms and two back terms are left over.
at n=1 the first term is 3*(1/5)=3/5, not 3; ratio 1/5
Add the two sums
$$2+\frac34=\frac{11}{4}$$
linearity permits adding the two convergent totals
Answer $$\boxed{\dfrac{11}{4}}$$
Check
Independent check: the telescoping part is exactly 2, and the geometric part 0.6 + 0.12 + 0.024 + ... clearly totals about 0.75, so 2.75 = 11/4 is right.
Two standard sums plus one addition; the only trap is reading the geometric first term as 3 instead of 3/5.
Exam mixtures are almost always linearity over pieces you already know: split, sum each, add.
Practice
A · concept 4 questions
1§02.4 — shrinking terms and convergence●●○○○
A study partner writes down a rule of thumb and wants you to confirm it before the quiz.
Given
Rule of thumb: if $\lim_{n\to\infty}a_n=0$, then $\sum a_n$ converges.
Find
(a) True or false? If false, name a counterexample.
Hint 1/4
Ask whether the terms going to zero has ever been shown to be enough on its own.
Hint 2/4
The condition $a_n\to 0$ is necessary for convergence but not sufficient.
Hint 3/4
The harmonic series $\sum\tfrac1n$ has $a_n\to 0$ yet diverges, which kills the rule.
Hint 4/4
So the statement is false, with $\sum\tfrac1n$ as the counterexample.
Close the page and write out, from memory: the definition of the sum of a series in terms of partial sums; the geometric formula with its condition and how you read off the first term; how a telescoping series collapses; why the harmonic series diverges; and exactly what the Test for Divergence can and cannot conclude. Then reopen and compare; every gap is a block to reread, and none of this is scored.
State that the sum of a series is the limit of its partial sums, and recover a term as s_n minus s_(n-1)?
c-series-as-partial-sums
Sum a geometric series after checking |r| < 1, reading off the correct first term even when the index starts at 0?
c-geometric-series
Split a term into a difference and collapse the telescoping partial sum, keeping any coefficient and every surviving boundary term?
c-telescoping
Explain, by grouping, why the harmonic series diverges despite its terms tending to zero?
c-harmonic
Apply the Test for Divergence to reject a series, and say why it is silent when the terms tend to zero?
c-divergence-test
Combine convergent series by linearity, and recognise that a convergent plus a divergent series diverges?
c-algebra-of-series
Glossary (10 terms)
seriesseri
The sum of the terms of an infinite sequence, defined as the limit of its partial sums.
partial sumkısmi toplam
The finite total of the first n terms of a series, written s_n.
convergeyakınsamak
To have partial sums that close in on a single finite number, which is then the sum.
divergeıraksamak
To fail to converge, either by running off to plus or minus infinity or by oscillating with no limit.
geometric seriesgeometrik seri
A series in which each term is a fixed multiple r of the previous one.
common ratioortak oran
The fixed number r that multiplies one term of a geometric series to give the next.
telescoping seriesteleskopik seri
A series whose term is a difference b_n minus b_(n+1), so the partial sums collapse to first minus last.
harmonic seriesharmonik seri
The series of reciprocals of the whole numbers, the sum of 1 over n, which diverges.
Test for Divergenceıraksaklık testi
The rule that a series diverges whenever its terms do not tend to zero.
toplam indisi
The counting letter n in a sum, a dummy symbol that does not appear in the final value.
What comes next
§03 · Convergence tests: the Integral, Comparison and Alternating Series Tests
We can now sum a geometric or telescoping series exactly and reject a series whose terms miss zero, but the Test for Divergence goes silent the moment the terms do tend to zero. The next section fills that silence with real tests, the Integral Test, the Comparison Tests and the Alternating Series Test, which decide the many series this one leaves open.
Sources
James Stewart, Calculus, 9th edition, section 11.2 (Series) The definition, the geometric-series theorem, the telescoping and harmonic examples and the Test for Divergence follow this text, the required book for the course.
MATH 102 course plan: weekly schedule and assessment weights Week 2 covers section 11.2; the Quiz and Homework weights of 10 percent each come from the grading scheme.
Grouping (Oresme) argument for the divergence of the harmonic series The block-doubling lower bound s_(2^k) > 1 + k/2 is the standard elementary proof and is reproduced in the harmonic block.