MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: alternating series, absolute and conditional convergence (MATH 141)

Q: How do I show that an alternating series converges?

Write each term as plus or minus a positive number b_n and check two things about b_n in writing: that it decreases, at least from some rank on, and that it tends to zero. If the decrease is not obvious, study the derivative of the matching function of x. Then name the Alternating Series Test in your conclusion.

Q: What is the difference between absolute and conditional convergence?

A series converges absolutely when the series of absolute values converges; it converges conditionally when it converges but the series of absolute values diverges. The alternating harmonic series, one minus one half plus one third and so on, is conditionally convergent: its sum is ln 2, but the harmonic series diverges.

Q: How many terms of an alternating series do I need for a given accuracy?

The error after n terms is at most the first term left out, b sub n plus 1. Solve b sub n plus 1 at most epsilon for the smallest n, checking the two integers around the threshold, then count the terms from where the index starts: a sum that starts at n equals zero has one more term than its last index.

Q: If the alternating series test fails, does the series diverge?

No. The test only gives sufficient conditions. If the terms do not tend to zero, the divergence test shows divergence. If they tend to zero but do not decrease, the series may converge, for instance absolutely by comparison, or diverge, and you must decide with another tool, such as the series of absolute values or the partial sums themselves.

Q: Is the error of an alternating series positive or negative?

The error, sum minus partial sum, has the sign of the first term left out. A partial sum that ends on an added term is above the sum, one that ends on a subtracted term is below it, so the sum always lies between two consecutive partial sums, which gives an exact interval without a calculator.

This sheet is not a summary of section 11.5 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on alternating series and on absolute against conditional convergence in MATH 141 at McGill University, and which precise gesture avoids each loss.

Everything is done by hand, as on the final: no calculator, bounds settled with squares, factorials and powers of ten, and every number below checked.

The thread of the chapter

Separate the SIGN from the SIZE: write each term as $(-1)^n b_n$ with $b_n > 0$ and ask every question of $b_n$ . The test checks that $b_n$ decreases to $0$ , the first omitted $b_{n+1}$ bounds the error, and absolute convergence is a question about $\sum b_n$ alone. A failed hypothesis is never a verdict.

This chapter is part of MATH 141, Calculus 2 (McGill)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Alternating series
Alternating series test
Alternating series estimation theorem
Remainder of a series
Absolute convergence
Conditional convergence
Alternating harmonic series
Divergence test
Monotone sequence
Convergence of p-series

The essentials

The test is about the size b_n, with three written checks

• Strip the sign first: $a_n = (-1)^{n-1} b_n$ or $(-1)^n b_n$ , with $b_n > 0$ . Name $b_n$ on the first line.
• Alternating Series Test: $b_n > 0$ , $b_{n+1} \le b_n$ from some rank on, and $\lim b_n = 0$ give the convergence of $\sum (-1)^{n-1} b_n$ .
• The decrease is a HYPOTHESIS, not a formality: $1 - \frac{1}{4} + \frac{1}{3} - \frac{1}{16} + \frac{1}{5} - \cdots$ alternates, has $b_n \to 0$ , and diverges.
• When the decrease is not obvious, pass to $f(x)$ with $f(n) = b_n$ and show $f'(x) < 0$ for $x \ge N$ . The same $f$ gives the limit by L'Hôpital's rule.
• Hidden alternating signs: $\cos(n\pi) = (-1)^n$ . Irregular signs such as $\sin n$ are NOT alternating: the test does not apply at all.

1 - \frac{1}{2} + \frac{1}{3} - \cdots

: odd sums stay above

\ln 2

, even sums below, and the gap between two consecutive sums,

b_{n+1}

, bounds the error.

The test proves convergence and nothing else. It never proves divergence, and it never says absolutely: both of those come from other tools.

The error, and the three verdicts

• Estimation theorem, same hypotheses: $|S - S_n| \le b_{n+1}$ , the FIRST TERM LEFT OUT, and $S$ lies between $S_n$ and $S_{n+1}$ .
• The sign of $S - S_n$ is the sign of that first omitted term, the opposite of the last term kept.
• Terms needed for accuracy $\varepsilon$ : solve $b_{n+1} \le \varepsilon$ for the smallest $n$ , then count the terms of $S_n$ from the starting index.
• Absolute convergence: $\sum |a_n| = \sum b_n$ converges, a chapter 19 question. It implies convergence.
• Conditional convergence: $\sum a_n$ converges, $\sum |a_n|$ diverges. Reference: $\sum \frac{(-1)^{n-1}}{n} = \ln 2$ .

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Absolute, conditional or divergent: read the two series side by side

For each series, the column $\sum |a_n|$ is answered with a chapter 19 test, the column $\sum a_n$ with the Alternating Series Test or the Divergence Test. The verdict follows from the pair. The red line is a combination that cannot happen.

Series	Sum of absolute values	Verdict
$\sum \frac{(-1)^n}{n^2}$	converges ( $p = 2$ )	absolutely convergent
Example: $\|S - S_3\| \le \frac{1}{16}$ , and no alternating test is even needed for the verdict.
$\sum \frac{\sin n}{n^2}$	converges (compare with $\frac{1}{n^2}$ )	absolutely convergent
Example: Signs $+, +, +, -, -, -, \ldots$ : not alternating, and $\frac{\|\sin n\|}{n^2} \le \frac{1}{n^2}$ settles it.
$\sum \frac{(-1)^{n-1}}{n}$	diverges (harmonic)	conditionally convergent
Example: Sum $\ln 2$ ; $999$ terms for an error at most $0.001$ .
$\sum \frac{(-1)^n}{\ln n}$ , $n \ge 2$	diverges ( $\frac{1}{\ln n} > \frac{1}{n}$ )	conditionally convergent
Example: $\frac{1}{\ln n}$ decreases to $0$ since $\ln$ increases to infinity.
$\sum (-1)^n \frac{n}{n+1}$	diverges	divergent
Example: $b_n = \frac{n}{n+1} \to 1$ : the Divergence Test decides, not the failure of the alternating test.
$\sum \|a_n\|$ converges but $\sum a_n$ diverges	converges	impossible cannot happen
Example: Absolute convergence implies convergence: $\sum \frac{(-1)^n}{n^2}$ cannot be made to diverge by its signs. What to do: If $\sum \|a_n\|$ converges, write absolutely convergent, hence convergent, and stop: no second test is needed.

Always fill the middle column FIRST. If it converges, the answer is absolutely and the work is over; only when it diverges does the alternating test have a job.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Checking only the limit of b_n

the whole question, since the verdict is reversed

What not to write

“The signs alternate and $b_n \to 0$ , so $1 - \frac{1}{4} + \frac{1}{3} - \frac{1}{16} + \frac{1}{5} - \cdots$ converges by the alternating series test.”

What to write

“ $b_3 = \frac{1}{3} > b_2 = \frac{1}{4}$ : $(b_n)$ is not decreasing and the test does not apply. $S_{2m} \ge \frac{1}{2} H_m - \frac{\pi^2}{24} \to \infty$ : the series diverges.”

Both series alternate and have

b_n \to 0

. The blue one, with

b_n

decreasing, settles at

\ln 2

; the red one, whose

b_n

jumps back up at every odd

n

, keeps climbing.

Why: The decrease is what makes the zigzag close. Without it, the odd terms $\frac{1}{n}$ can add up faster than the even terms $\frac{1}{n^2}$ take away, and the sums drift to infinity, as the figure shows.

2. Proving the decrease with the denominator grows

the mark for the decreasing hypothesis, often 2 marks

What not to write

“ $\frac{\sqrt{n}}{n+4}$ decreases because its denominator grows.”

What to write

“ $g(x) = \frac{\sqrt{x}}{x+4}$ has $g'(x) = \frac{4 - x}{2\sqrt{x}(x+4)^2} < 0$ for $x > 4$ , so $b_n$ decreases from $n = 4$ .”

Why: The numerator grows too, and this $b_n$ actually INCREASES until $n = 4$ : $b_3 < b_4$ reads $48 < 49$ after squaring, $b_4 > b_5$ reads $81 > 80$ . Only a sign study decides who wins.

3. Declaring divergence because the test fails

the whole question

What not to write

“ $b_n = \frac{2 + (-1)^n}{n^2}$ is not decreasing, so the alternating series test fails and $\sum (-1)^n b_n$ diverges.”

What to write

“The test does not apply. But $|a_n| \le \frac{3}{n^2}$ and $\sum \frac{3}{n^2}$ converges, so the series converges absolutely.”

Why: The test is one-way: its hypotheses are sufficient, not necessary. When one fails, look for absolute convergence, or for $a_n \not\to 0$ , which is the only thing that proves divergence quickly.

4. Bounding the error with the last term kept

1 mark, the answer being off by one

What not to write

“For $\sum \frac{(-1)^{n-1}}{n^2}$ , $|S - S_n| \le \frac{1}{n^2}$ , so $\frac{1}{n^2} \le 0.001$ and I need $32$ terms.”

What to write

“ $|S - S_n| \le b_{n+1} = \frac{1}{(n+1)^2} \le 0.001$ gives $(n+1)^2 \ge 1000$ ; $31^2 = 961$ , $32^2 = 1024$ , so $n + 1 = 32$ and $31$ terms.”

Why: The theorem bounds the error by the FIRST TERM LEFT OUT. Write that term explicitly before solving anything: the off-by-one disappears.

5. Counting the terms from the wrong index

1 mark

What not to write

“For $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}$ , $\frac{1}{8!} < 10^{-4}$ so $S_3$ is enough: three terms.”

What to write

“The sum starts at $n = 0$ , so $S_3 = 1 - \frac{1}{2} + \frac{1}{24} - \frac{1}{720} = \frac{389}{720}$ has FOUR terms, with error at most $\frac{1}{8!}$ .”

Why: The question asks for a number of terms, not for an index. From $n = 0$ , $S_N$ has $N + 1$ terms; from $n = 1$ , it has $N$ .

6. Giving the error the sign of the last term kept

1 mark, and every interval built on it

What not to write

“ $S_2 = 1 - \frac{1}{2}$ ends with a negative term, so $S_2$ is above the sum $\ln 2$ .”

What to write

“The first omitted term, $+\frac{1}{3}$ , is positive, so $S_2 = \frac{1}{2}$ is an underestimate: $\frac{1}{2} < \ln 2 < \frac{5}{6}$ .”

Why: $S$ lies between $S_n$ and $S_{n+1} = S_n + a_{n+1}$ , so $S - S_n$ has the sign of $a_{n+1}$ . A partial sum that ends on a subtraction has just jumped BELOW the limit.

7. Using the remainder bound on a series that is not alternating

1 to 2 marks, and a false guarantee

What not to write

“ $\sum_{n=1}^{\infty} \frac{1}{2^n}$ converges, so $|S - S_n| \le \frac{1}{2^{n+1}}$ .”

What to write

“The bound needs alternating signs and $b_n$ decreasing to $0$ . Here $S - S_n = \frac{1}{2^n}$ , twice the next term.”

Why: With positive terms nothing cancels: every omitted term adds to the error. For positive series the remainder is bounded with the tool of their own test, the integral bound of chapter 19 for instance.

8. Writing converges where the question asks how

1 mark per series in a classification question

What not to write

“ $\sum \frac{(-1)^n}{n^{3/2}}$ converges by the alternating series test.”

What to write

“ $\sum \frac{1}{n^{3/2}}$ is a $p$ -series with $p = \frac{3}{2} > 1$ , so the series converges absolutely, hence converges.”

Why: Classify means absolutely, conditionally or divergent. The alternating test proves the weakest of the three statements; start with $\sum |a_n|$ , which may end the question at once.

Which method to choose

Which tool decides, by the FORM of the terms

Look at the terms a_n before choosing anything: their limit, their signs, and the series of their absolute values

\sum \frac{(-1)^n}{n^p}

: divergent for

p \le 0

, conditionally convergent for

0 < p \le 1

, absolutely convergent for

p > 1

. A dot belongs to the region on its left.

If $a_n$ does not tend to $0$ (for instance $b_n \to 1$ or $b_n \to e$ ) → Divergence Test: the series diverges, stop

Example: $\sum (-1)^n \left(1 + \frac{1}{n}\right)^n$ , since $b_n \to e$
If $\sum |a_n|$ converges by a chapter 19 test → absolutely convergent, hence convergent, stop

Example: $\sum \frac{(-1)^n}{n^{3/2}}$ , or $\sum \frac{\sin n}{n^2}$ with irregular signs
If $\sum |a_n|$ diverges and the signs alternate, $b_n$ decreasing to $0$ → Alternating Series Test: conditionally convergent

Example: $\sum \frac{(-1)^{n+1} n}{n^2 + 1}$ , decrease by $h'(x) = \frac{1 - x^2}{(x^2+1)^2}$
If $\sum |a_n|$ diverges, the signs alternate, $b_n \to 0$ but $b_n$ does not decrease → the test is silent: study the partial sums directly

Example: $1 - \frac{1}{4} + \frac{1}{3} - \frac{1}{16} + \cdots$ diverges
If the signs are irregular and the absolute series diverges → out of reach of this chapter: say so and name what is missing

Example: $\sum \frac{\sin n}{n}$ is not decided by any test of MATH 141

The order matters: the limit of $a_n$ costs one line, the absolute series usually one comparison, and only then is the alternating test worth writing. For $\sum \frac{(-1)^n}{n^p}$ the whole tree is summed up in the figure.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing the alternating series test

When to use it: Any question that says show that the series converges, when the terms carry $(-1)^n$ , $(-1)^{n-1}$ or $\cos(n\pi)$

1 Write $a_n = (-1)^n b_n$ and give $b_n$ explicitly; say why $b_n > 0$ , from which $n$ .
2 Decrease: one line if it is obvious (a single growing denominator), otherwise $f(x)$ with $f(n) = b_n$ , $f'(x)$ computed and its sign discussed, and the rank $N$ from which it holds.
3 Limit: compute $\lim b_n$ , by L'Hôpital on $f(x)$ if needed; if it is not $0$ , stop and use the Divergence Test.
4 Conclude with the name of the test, and, if the question asks, go on to $\sum b_n$ for absolute convergence.
5 For an estimate, name the first term left out, $b_{n+1}$ , before solving for $n$ .

Concluding sentence

“ $b_n = \frac{\ln n}{n} > 0$ for $n \ge 2$ ; $f'(x) = \frac{1 - \ln x}{x^2} < 0$ for $x > e$ , so $(b_n)$ decreases from $n = 3$ ; $\lim b_n = 0$ . By the Alternating Series Test, $\sum (-1)^n \frac{\ln n}{n}$ converges.”

The trap: Stopping after the limit. The decrease is the hypothesis students skip and markers look for first.

Marking: Typically 1 mark for b_n named, 1 for the decrease justified, 1 for the limit, 1 for the conclusion with the test named, and 1 or 2 more for the absolute or conditional verdict.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The zigzag check

An interval for the sum must have one endpoint that ends on an addition and one that ends on a subtraction. Two partial sums on the same side prove nothing.

$\frac{115}{144} = S_4 < \frac{\pi^2}{12} < S_3 = \frac{31}{36}$ for $\sum \frac{(-1)^{n-1}}{n^2}$ .
Which term is left out?

Write the first omitted term with its index before any inequality: it gives both the size and the sign of the error.

For $S_4$ of $\sum \frac{(-1)^{n-1}}{n^3}$ , the first term left out is $+\frac{1}{125}$ : $S_4$ is low by at most $0.008$ .
Absolute series first

Before writing the alternating test, test the series of absolute values: if it converges, the question is over, and the stronger answer is on the page.

$\sum \frac{\sin n}{n^2}$ : $\frac{|\sin n|}{n^2} \le \frac{1}{n^2}$ , done in one line.
Plausibility of a term count

Slow $b_n$ gives huge counts: $\frac{1}{\sqrt{n}}$ needs about $\frac{1}{\varepsilon^2}$ terms, factorials only a handful. A count that contradicts the speed of $b_n$ hides a slip.

$\sum \frac{(-1)^{n-1}}{\sqrt{n}}$ within $0.01$ : $9999$ terms; $\sum \frac{(-1)^n}{(2n)!}$ within $10^{-4}$ : four terms.

The typical problem, taken apart

Classify, then count the terms for a given accuracy

Let $\sum_{n=1}^{\infty} (-1)^{n-1} \frac{\sqrt{n}}{n+1}$ . Decide whether it converges absolutely, conditionally, or not at all. Then find how many terms guarantee its sum with an error of at most $0.1$ .

No calculator. Every hypothesis must be justified as on a MATH 141 final.

Step 1

$b_n = \frac{\sqrt{n}}{n+1} > 0$ . Absolute series: $\frac{\sqrt{n}/(n+1)}{1/\sqrt{n}} = \frac{n}{n+1} \to 1$ , and $\sum \frac{1}{\sqrt{n}}$ diverges ( $p = \frac{1}{2}$ ), so $\sum b_n$ diverges by limit comparison.

Why

The absolute series comes first: if it had converged, the answer would already be absolutely convergent. Here it fails, so the alternating test now has a job.

Step 2

$f(x) = \frac{\sqrt{x}}{x+1}$ : $f'(x) = \frac{\frac{x+1}{2\sqrt{x}} - \sqrt{x}}{(x+1)^2} = \frac{1 - x}{2\sqrt{x}(x+1)^2} < 0$ for $x > 1$ . So $(b_n)$ decreases from $n = 1$ .

Why

Numerator and denominator both grow, so the decrease needs a proof. Multiplying top and bottom by $2\sqrt{x}$ puts the sign in the single factor $1 - x$ .

Step 3

$b_n = \frac{1/\sqrt{n}}{1 + 1/n} \to 0$ . By the Alternating Series Test the series converges, and since $\sum b_n$ diverges, it converges CONDITIONALLY.

Why

Three hypotheses, three lines, then the test named. The verdict combines both columns: convergent series, divergent absolute series.

Step 4

Error at most $b_{n+1}$ . With $m = n + 1$ : $\frac{\sqrt{m}}{m+1} \le \frac{1}{10}$ is equivalent, squaring positive numbers, to $100m \le (m+1)^2$ .

Why

Naming the first omitted term, $b_{n+1}$ , and renaming it $b_m$ keeps the index straight. Squaring removes the root without a calculator.

Step 5

$m = 97$ : $100 \cdot 97 = 9700 > 98^2 = 9604$ , fails. $m = 98$ : $100 \cdot 98 = 9800 \le 99^2 = 9801$ , holds. Since $b_m$ decreases, every larger $m$ works too. So $n + 1 = 98$ and $n = 97$ terms.

Why

Testing the two integers around the threshold is the proof that $97$ is the smallest count the theorem allows; the margin of $1$ in $9801$ shows why guessing fails.

The conclusion, written out

“The series converges conditionally: $\sum \frac{\sqrt{n}}{n+1}$ diverges by limit comparison with $\sum \frac{1}{\sqrt{n}}$ , and the Alternating Series Test applies since $b_n$ decreases to $0$ . By the estimation theorem, $S_{97}$ is within $b_{98} \le 0.1$ of the sum: $97$ terms.”

The classic mistake on this problem: Proving the decrease by the denominator grows, or solving $b_n \le 0.1$ instead of $b_{n+1} \le 0.1$ , which gives $98$ terms.

Learn by heart

• Name $b_n > 0$ first. The test needs $b_n$ decreasing (from some rank) AND $b_n \to 0$ .
• Not obvious? $f'(x) < 0$ for $x \ge N$ . Limits by L'Hôpital on $f(x)$ , never on $b_n$ .
• $|S - S_n| \le b_{n+1}$ , the first term left out, whose sign is the sign of the error.
• Count terms from the starting index: from $n = 0$ , $S_N$ has $N + 1$ terms.
• Absolute: $\sum |a_n|$ converges. Conditional: $\sum a_n$ converges, $\sum |a_n|$ diverges. Absolute implies convergent.
• $\sum \frac{(-1)^{n-1}}{n} = \ln 2$ , conditionally. $\sum \frac{(-1)^n}{n^p}$ : absolute $p > 1$ , conditional $0 < p \le 1$ , divergent $p \le 0$ .
• A failed hypothesis is NOT a verdict. Only $a_n \not\to 0$ proves divergence in one line.

Frequently asked questions

How do I show that an alternating series converges?

Write each term as plus or minus a positive number b_n and check two things about b_n in writing: that it decreases, at least from some rank on, and that it tends to zero. If the decrease is not obvious, study the derivative of the matching function of x. Then name the Alternating Series Test in your conclusion.

What is the difference between absolute and conditional convergence?

A series converges absolutely when the series of absolute values converges; it converges conditionally when it converges but the series of absolute values diverges. The alternating harmonic series, one minus one half plus one third and so on, is conditionally convergent: its sum is ln 2, but the harmonic series diverges.

How many terms of an alternating series do I need for a given accuracy?

The error after n terms is at most the first term left out, b sub n plus 1. Solve b sub n plus 1 at most epsilon for the smallest n, checking the two integers around the threshold, then count the terms from where the index starts: a sum that starts at n equals zero has one more term than its last index.

If the alternating series test fails, does the series diverge?

No. The test only gives sufficient conditions. If the terms do not tend to zero, the divergence test shows divergence. If they tend to zero but do not decrease, the series may converge, for instance absolutely by comparison, or diverge, and you must decide with another tool, such as the series of absolute values or the partial sums themselves.

Is the error of an alternating series positive or negative?

The error, sum minus partial sum, has the sign of the first term left out. A partial sum that ends on an added term is above the sum, one that ends on a subtracted term is below it, so the sum always lies between two consecutive partial sums, which gives an exact interval without a calculator.

Practise it

Corrected exercises: Alternating series, MATH 141 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Integral and comparison tests Next sheet Ratio and root tests

Looking for a MATH 141 tutor in Montreal?

Get in touch for a first session. Alternating series are where the convergence tests start to demand written hypotheses, and where most marks on the series part of the final are lost.

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Revision sheet: alternating series, absolute and conditional convergence (MATH 141)

The essentials

The test is about the size b_n, with three written checks

The error, and the three verdicts

The rules in table form

Absolute, conditional or divergent: read the two series side by side

The mistakes that cost marks

1. Checking only the limit of b_n

2. Proving the decrease with the denominator grows

3. Declaring divergence because the test fails

4. Bounding the error with the last term kept

5. Counting the terms from the wrong index

6. Giving the error the sign of the last term kept

7. Using the remainder bound on a series that is not alternating

8. Writing converges where the question asks how

Which method to choose

Which tool decides, by the FORM of the terms

How the answer is expected to be written

Writing the alternating series test

Check before you hand in

The typical problem, taken apart

Classify, then count the terms for a given accuracy

Learn by heart

Frequently asked questions

How do I show that an alternating series converges?

What is the difference between absolute and conditional convergence?

How many terms of an alternating series do I need for a given accuracy?

If the alternating series test fails, does the series diverge?

Is the error of an alternating series positive or negative?

Corrected exercises: Alternating series, MATH 141 at McGill

See also

Looking for a MATH 141 tutor in Montreal?