Integral Calculus 201-SN3-RE (201-NYB) • CEGEP, Montreal

Revision sheet: advanced convergence tests, 201-SN3-RE (201-NYB)

When the general term tends to zero, the ratio test returns $1$ and no simple equivalent shows up, four tools remain: condensation, the root test, Raabe's test and Abel's test. Each one answers a specific situation, and knowing which one is most of the work. This sheet is written for Integral Calculus 201-SN3-RE (formerly 201-NYB).

It gives the four tests with their exact hypotheses, and above all the pitfall common to all of them: equivalents and comparisons only hold when the sign is constant, which exam questions exploit systematically.

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The thread of the chapter

An asymptotic equivalent can only be used when the sign is CONSTANT. As soon as the terms change sign, two equivalent sequences can give series that behave differently, and this is the mistake that separates the papers on the final exam.

This chapter is part of 201-SN3-RE (201-NYB), Integral Calculus

The essentials

The four tests, with their exact hypotheses

• CONDENSATION: if $(a_{n})$ is POSITIVE and DECREASING, then $\sum a_{n}$ and $\sum 2^{n}a_{2^{n}}$ BEHAVE THE SAME WAY. Both hypotheses are essential.
• ROOT TEST: if $\sqrt[n]{\left|a_{n}\right|}\to L$ , the series converges if $L<1$ and diverges if $L>1$ . It concludes whenever the ratio test does, and sometimes more often.
• RAABE'S TEST: if $\dfrac{a_{n+1}}{a_{n}}=1-\dfrac{\alpha}{n}+o\!\left(\dfrac{1}{n}\right)$ with $a_{n}>0$ , the series converges if $\alpha>1$ and diverges if $\alpha<1$ .
• ABEL'S TEST (Dirichlet's test in most English textbooks): if $(a_{n})$ decreases to $0$ and the partial sums of $(b_{n})$ are BOUNDED, then $\sum a_{n}b_{n}$ converges. The alternating series test is the case $b_{n}=(-1)^{n}$ .
• CAUCHY PRODUCT: $c_{n}=\displaystyle\sum_{k=0}^{n}a_{k}b_{n-k}$ . It is guaranteed to converge to the product of the sums when both series converge ABSOLUTELY; by Mertens' theorem, absolute convergence of one of the two is already enough.

Raabe's test takes over exactly where the ratio test fails: when the ratio tends to $1$ , it looks at how FAST it gets there, and the coefficient $\alpha$ of that expansion decides.

Condensation, and the Bertrand family

• Condensation turns a slow series into a geometric series or a p-series: it is made for terms that contain logarithms.
• The p-series test in one line: for $a_{n}=\dfrac{1}{n^{p}}$ , you get $2^{n}a_{2^{n}}=\left(2^{1-p}\right)^{n}$ , a geometric series that converges if and only if $p>1$ .
• BERTRAND: for $a_{n}=\dfrac{1}{n(\ln n)^{p}}$ , you get $2^{n}a_{2^{n}}=\dfrac{1}{(n\ln 2)^{p}}$ , hence convergence if and only if $p>1$ .
• In particular $\displaystyle\sum\frac{1}{n\ln n}$ DIVERGES, even though its terms are smaller than those of the harmonic series.
• Being smaller than a divergent series proves nothing: only an upper bound by a CONVERGENT series concludes.

Grouped by powers of two, the terms of the harmonic series form blocks that each weigh at least

\dfrac{1}{2}

: the sum therefore grows without bound. This is exactly the idea that condensation generalizes.

Condensation requires the sequence to decrease: it therefore never applies as such to a sequence that oscillates, nor to a sequence whose first terms go up before coming down, unless you shift the starting point.

The sign pitfall, and telescoping

• Equivalents, the limit comparison test and the comparison test all require a CONSTANT SIGN, at least from some index on.
• Counterexample to know: $a_{n}=\dfrac{(-1)^{n}}{\sqrt{n}}$ and $b_{n}=a_{n}+\dfrac{1}{n}$ are equivalent, but $\sum a_{n}$ converges and $\sum b_{n}$ diverges.
• When the signs vary, there are only three ways forward: ABSOLUTE convergence, the alternating series test (LEIBNIZ), or ABEL'S test.
• TELESCOPING: if $a_{n}=u_{n}-u_{n+1}$ , then $\displaystyle\sum_{n=1}^{N}a_{n}=u_{1}-u_{N+1}$ . The sum is exact, not just the behaviour.
• Only TWO boundary terms remain: the first one kept and the first one dropped. Forgetting them makes the sum wrong, never the behaviour.

Telescoping is the only tool of the chapter that gives the SUM and not just the behaviour. You spot it from a partial fraction decomposition whose two pieces only differ by a shift of the index.

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. Using an equivalent on a series whose terms change sign

the whole question, 6 marks

What not to write

“ $\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{1}{n}$ is equivalent to $\dfrac{(-1)^{n}}{\sqrt{n}}$ , whose series converges: so the series converges.”

What to write

“Equivalence allows no conclusion when the sign varies. Here the first series converges and the second one DIVERGES, since their difference is the harmonic series.”

The two general terms are equivalent, but the partial sums of

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

level off (green) while those with

+\frac{1}{n}

keep climbing (red).

Why: Two equivalent sequences have a difference that is negligible compared with each of them, but the SERIES of that difference can perfectly well diverge. This is impossible when the sign is constant, and common otherwise.

2. Believing that a series with terms smaller than a divergent series converges

the whole question, 5 marks

What not to write

“ $\dfrac{1}{n\ln n}<\dfrac{1}{n}$ , so the series converges.”

What to write

“An upper bound by a DIVERGENT series proves nothing. Condensation shows that $\displaystyle\sum\frac{1}{n\ln n}$ diverges as well.”

The red curve always stays below that of

\dfrac{1}{n}

, and yet its series diverges too: being smaller than a divergent series tells you nothing at all.

Why: The comparison test only works one way: smaller than a CONVERGENT series gives convergence, larger than a DIVERGENT series gives divergence. The two other combinations say nothing.

3. Applying condensation without checking that the sequence decreases

the whole question, 5 marks

What not to write

“I apply condensation to $a_{n}=\dfrac{2+(-1)^{n}}{n}$ .”

What to write

“This sequence is not DECREASING: it oscillates between $\dfrac{1}{n}$ and $\dfrac{3}{n}$ . Condensation does not apply; here a direct comparison with $\dfrac{1}{n}$ and $\dfrac{3}{n}$ is enough to conclude that it diverges.”

a_{n}=\frac{2+(-1)^{n}}{n}

goes back up every other index: it is not decreasing, so condensation does not apply.

Why: Condensation relies on bounding each block by its first and last terms, which requires monotonicity. Without it, the bounds collapse and the theorem is false.

4. Invoking Raabe's test when the ratio does not tend to one

no marks lost, but ten minutes wasted on a useless expansion

What not to write

“The ratio tends to $\dfrac{1}{4}$ , so I apply Raabe's test to get more precision.”

What to write

“Raabe's test is ONLY useful in the undecided case of the ratio test, that is, when the ratio tends to $1$ . Here $\dfrac{1}{4}<1$ : the ratio test already concludes, and Raabe's test makes no sense.”

Why: Raabe's test examines how FAST the ratio approaches $1$ . If the ratio tends to something else, that expansion does not exist and writing $1-\dfrac{\alpha}{n}$ makes no sense.

5. Getting the direction of Raabe's test wrong

the whole question, 5 marks, with the conclusion reversed

What not to write

“ $\dfrac{a_{n+1}}{a_{n}}=1-\dfrac{1}{2n}+o\!\left(\dfrac{1}{n}\right)$ , so $\alpha=\dfrac{1}{2}$ and the series converges.”

What to write

“ $\alpha=\dfrac{1}{2}<1$ : the series DIVERGES. You need $\alpha>1$ for convergence, exactly as with the exponent of a p-series.”

Why: The threshold of Raabe's test is the same as that of the p-series test, and for the same reason: the coefficient $\alpha$ in the expansion of the ratio plays the role of the exponent in a comparison with $\dfrac{1}{n^{\alpha}}$ .

6. Forgetting the condition on the partial sums in Abel's test

the whole question, 5 marks

What not to write

“ $\dfrac{1}{n}$ decreases to $0$ , so $\displaystyle\sum\frac{b_{n}}{n}$ converges whatever the sequence $(b_{n})$ .”

What to write

“The PARTIAL SUMS of $(b_{n})$ must also be BOUNDED. With $b_{n}=1$ for every $n$ , they equal $n$ and are not bounded: you get the harmonic series back, which diverges.”

Why: This condition does all the work: it prevents $(b_{n})$ from piling up in one direction. For $b_{n}=(-1)^{n}$ or $b_{n}=\sin n$ , the partial sums stay bounded and the test applies.

7. Forming a Cauchy product without absolute convergence

3 marks on the reasoning question

What not to write

“Both series converge, so their Cauchy product converges to the product of the sums.”

What to write

“You need the ABSOLUTE convergence of at least one of the two. With two series that are only conditionally convergent, the Cauchy product can diverge, and the classic example is $\displaystyle\sum\frac{(-1)^{n}}{\sqrt{n}}$ multiplied by itself.”

Why: The product rearranges infinitely many terms, which is only allowed under absolute convergence. A conditionally convergent series, on the contrary, can be reordered to converge to any real number.

8. Forgetting the boundary terms of a telescoping sum

the whole question, 4 marks

What not to write

“ $\displaystyle\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+1}\right)=0$ , everything cancels.”

What to write

“TWO terms remain: $\displaystyle\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+1}\right)=1-\frac{1}{N+1}$ , which tends to $1$ .”

The whole sum reduces to TWO numbers: the first term and the one dropped at index

N+1

. The partial sums therefore climb toward

1

without ever reaching it.

Why: Telescoping cancels every term EXCEPT the first one kept and the last one dropped. A zero sum for a series with strictly positive terms is absurd anyway.

9. Believing that the ratio test and the root test are interchangeable

the whole question, 5 marks

What not to write

“The ratio $\dfrac{a_{n+1}}{a_{n}}$ has no limit, so nothing can be concluded.”

What to write

“The ROOT test can conclude where the ratio test fails. With $a_{n}=\dfrac{\left(2+(-1)^{n}\right)^{n}}{4^{n}}$ , the ratio swings between $0$ and infinity, but $\sqrt[n]{a_{n}}$ equals $\dfrac{3}{4}$ or $\dfrac{1}{4}$ , never more than three quarters, so the series converges (root test in its lim sup form, or simply comparison with a geometric series).”

Why: When the ratio has a limit, the root has the same one: the root test is therefore at least as powerful. It is strictly stronger as soon as the general term oscillates between two regimes.

10. Using the comparison test the wrong way round

the whole question, 5 marks

What not to write

“ $a_{n}\ge b_{n}$ and $\sum b_{n}$ converges, so $\sum a_{n}$ converges.”

What to write

“Comparison only concludes one way: SMALLER than a convergent series gives convergence, LARGER than a divergent series gives divergence.”

Why: A series larger than a convergent series can perfectly well diverge. The direction of the inequality and the behaviour of the reference series must match, otherwise the theorem does not apply.

11. Using a zero equivalent

the whole question, 5 marks

What not to write

“ $\dfrac{1}{n}-\sin\dfrac{1}{n}$ is equivalent to $0$ , so the series trivially converges.”

What to write

“No nonzero sequence is equivalent to $0$ . You need the higher-order expansion: $\sin u=u-\dfrac{u^{3}}{6}+\ldots$ , so the difference is equivalent to $\dfrac{1}{6n^{3}}$ , and the series converges by the p-series test with $p=3$ .”

Why: Saying that a quantity is equivalent to zero makes no sense, since equivalence is defined through a quotient. It is the signal that you must go one step further in the expansion.

Which method to choose

Which advanced test, and when

What failed before, and the form of the general term

If the term contains a LOGARITHM and the sequence is positive and decreasing → CONDENSATION: $\sum a_{n}$ and $\sum 2^{n}a_{2^{n}}$ behave the same way

Example: $\displaystyle\sum\frac{1}{n(\ln n)^{p}}$ converges if and only if $p>1$
If the general term is an $n$ th POWER, or oscillates between two regimes → ROOT test, more powerful than the ratio test

Example: $a_{n}=\dfrac{\left(2+(-1)^{n}\right)^{n}}{4^{n}}$
If the ratio test gives exactly $1$ and the term contains factorials → RAABE'S TEST: expand the ratio as $1-\dfrac{\alpha}{n}$ and compare $\alpha$ with $1$

Example: $\dfrac{a_{n+1}}{a_{n}}=\dfrac{2n+1}{2n+2}$ gives $\alpha=\dfrac{1}{2}$ , hence divergence
If the term is a PRODUCT of a decreasing sequence and an oscillating sequence → ABEL'S TEST, after checking that the partial sums of the second sequence are bounded

Example: $\displaystyle\sum\frac{\sin n}{n}$ converges
If the term splits into partial fractions shifted by one index → TELESCOPING, which gives the exact sum and not just the behaviour

Example: $\displaystyle\sum\frac{1}{n(n+1)}=1$
If the signs vary without regular alternation → ABSOLUTE convergence or Abel's test, never an equivalent

this is exactly the case where the limit comparison test is illegitimate

These tests do not replace the four from the course: they come after them. The right strategy is always to try an equivalent and the p-series test first, and to bring out these tools only when they fail.

What a test requires before it is applied

The hypotheses to check explicitly, or lose the rigour mark

If you use an EQUIVALENT or a limit comparison → check the CONSTANT SIGN from some index on

Example: forbidden on $\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{1}{n}$
If you use CONDENSATION → check POSITIVITY and that the sequence DECREASES

Example: forbidden on a sequence that oscillates between $\dfrac{1}{n}$ and $\dfrac{3}{n}$
If you use RAABE'S TEST → check that the terms are positive AND that the ratio tends to $1$

Example: pointless if the ratio tends to $\dfrac{1}{4}$
If you use ABEL'S TEST → check that the first sequence decreases to $0$ AND that the partial sums of the second one are bounded

Example: $\left|\displaystyle\sum_{k=1}^{n}\sin k\right|$ is bounded
If you form a CAUCHY PRODUCT → check the ABSOLUTE convergence of at least one of the two series

without it, rearranging infinitely many terms is not allowed

Each unchecked hypothesis costs a rigour mark, and sometimes the whole question when it is actually violated. Writing them down takes three lines and earns more than the computation itself.

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.

Handling a family by condensation

When to use it: The general term contains a logarithm and depends on a parameter, and the question asks for a discussion according to that parameter.

1 Check that the sequence is POSITIVE and DECREASING from some index on, and justify it by studying a function.
2 Write the condensed term $2^{n}a_{2^{n}}$ by replacing $n$ with $2^{n}$ , and simplify it completely.
3 Recognize the series obtained: geometric, a p-series, or one to condense again.
4 Conclude on its behaviour, then transfer the conclusion to the original series with the condensation theorem.
5 Discuss according to the parameter, giving the critical value explicitly.

Concluding sentence

“The sequence $\dfrac{1}{n(\ln n)^{p}}$ is positive and decreasing for $n\ge 3$ ; the condensed term equals $\dfrac{2^{n}}{2^{n}(n\ln 2)^{p}}=\dfrac{1}{(\ln 2)^{p}}\cdot\dfrac{1}{n^{p}}$ , a p-series that converges if and only if $p>1$ : the same therefore holds for the original series.”

The trap: Substituting $2^{n}$ into $a_{n}$ without multiplying by $2^{n}$ . The theorem is about $2^{n}a_{2^{n}}$ , and that factor is what makes up for the indices becoming sparse.

Marking: Typically 2 marks for the checked hypotheses, 3 for the simplified condensed term, 2 for recognizing the series, 1 for the transfer and the discussion.

Justifying that an equivalent is legitimate

When to use it: You want to replace the general term with an equivalent in order to conclude by comparison.

1 Determine the SIGN of the general term, and the index from which it is constant.
2 If the sign varies, do not use an equivalent: switch to absolute convergence, the alternating series test or Abel's test, and say so.
3 If the sign is constant, write the equivalent and state the neighbourhood considered, here $n\to+\infty$ .
4 Check that the quotient tends to a FINITE and NONZERO limit, then state the limit comparison test.
5 Conclude with the p-series test or the geometric series, depending on the form obtained.

Concluding sentence

“The general term is positive from $n=2$ on, which allows the comparison; it is equivalent to $\dfrac{1}{2n^{2}}$ , whose series converges by the p-series test with $p=2$ : the series converges.”

The trap: Writing “equivalent to $0$ ” when the first order cancels out. No nonzero sequence is equivalent to zero: it is the signal that you need an expansion one order further.

Check before you hand in

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

The sign before any equivalent

Go back over your paper and check that a sentence justifies the constant sign before each comparison or equivalent.

An equivalent applied to a term containing a factor $(-1)^{n}$ : the manipulation is illegitimate.
The hypotheses of each test written down

Each advanced test has two or three hypotheses. Count them on your paper and check that none was passed over in silence.

A condensation that never mentions that the sequence decreases: a hypothesis is missing.
Raabe's threshold consistent with the p-series threshold

The threshold is the same, $\alpha>1$ for convergence. Reasoning that makes the series converge for $\alpha<1$ contradicts the comparison the test is built on.

An $\alpha=\dfrac{1}{2}$ declared convergent: it is the analogue of $\displaystyle\sum\frac{1}{\sqrt{n}}$ , which diverges.
The boundary terms of a telescoping sum

After simplification, exactly two terms must remain. A zero sum for a series with positive terms is impossible.

A partial sum announced as zero when all the terms are positive.
The direction of the comparison test

Check that the direction of the inequality and the behaviour of the reference series match: smaller and convergent, or larger and divergent.

An upper bound by a divergent series used to conclude convergence: the theorem says nothing in that case.

The typical problem, taken apart

The Bertrand family, handled by condensation

Discuss, according to the real number $p$ , the behaviour of the series $\displaystyle\sum_{n\ge 2}\frac{1}{n(\ln n)^{p}}$ .

Pay particular attention to the case $p=1$ , and explain why comparison with the harmonic series does not settle it.

Step 1

General term: $a_{n}=\dfrac{1}{n(\ln n)^{p}}$ , defined and strictly POSITIVE for $n\ge 2$ . It tends to $0$ , so the divergence test is inconclusive.

Why

Positivity is the first hypothesis of condensation, and every comparison depends on it as well. One line mentioning it is expected by the marking scheme.

Step 2

Decrease: the function $x\mapsto x(\ln x)^{p}$ is increasing for $x\ge 3$ as soon as $p\ge 0$ , so $a_{n}$ decreases. For $p<0$ , the series is larger than the harmonic series and diverges right away.

Why

Set aside the trivial case $p<0$ at once: the logarithm moves to the numerator and the series is then larger than the harmonic series, hence divergent with no further work.

Step 3

Condensation: $2^{n}a_{2^{n}}=2^{n}\cdot\dfrac{1}{2^{n}\left(\ln 2^{n}\right)^{p}}=\dfrac{1}{\left(n\ln 2\right)^{p}}=\dfrac{1}{(\ln 2)^{p}}\cdot\dfrac{1}{n^{p}}$ .

Why

The factor $2^{n}$ of the theorem exactly cancels the $2^{n}$ in the denominator: this simplification is what makes condensation so effective on terms in $\dfrac{1}{n}$ .

Step 4

Up to the constant $\dfrac{1}{(\ln 2)^{p}}$ , the condensed series is the p-series $\displaystyle\sum\frac{1}{n^{p}}$ : it converges if and only if $p>1$ .

Why

A nonzero constant factor never changes the behaviour of a series. You can therefore set it aside in a few words and fall back on a well-known case.

Step 5

By the condensation theorem, the original series BEHAVES THE SAME WAY: it converges if and only if $p>1$ .

Why

The transfer is immediate once both hypotheses are checked. That is the whole point of the theorem: it reduces a hard family to a reference family.

Step 6

Case $p=1$ : the series $\displaystyle\sum\frac{1}{n\ln n}$ DIVERGES, since the condensed series is then $\dfrac{1}{\ln 2}\displaystyle\sum\frac{1}{n}$ , the harmonic series.

Why

This is the result the question is after. It is counterintuitive, because the terms are clearly smaller than those of the harmonic series, and yet the sum still grows without bound.

Step 7

Why comparison fails: $\dfrac{1}{n\ln n}<\dfrac{1}{n}$ , but being SMALLER than a DIVERGENT series allows no conclusion. The comparison test only works the other way.

Why

This answers the second question, and it is worth as many marks as the computation. The direction of the inequality and the behaviour of the reference series must match.

Step 8

Numerical check: the sum of the first twenty thousand terms for $p=1$ is about $3.09$ , and it keeps growing like $\ln\ln n$ , hence without bound, but so slowly that no finite computation shows it.

Why

The numerical check illustrates the difficulty of the chapter: a series can diverge so slowly that no partial sum reveals it. Only the proof settles the question.

The conclusion, written out

“The Bertrand series $\displaystyle\sum_{n\ge 2}\frac{1}{n(\ln n)^{p}}$ converges if and only if $p>1$ . In particular, for $p=1$ , it DIVERGES, even though its terms are strictly smaller than those of the harmonic series: comparison does not conclude in that direction.”

The classic mistake on this problem: Concluding convergence for $p=1$ by writing $\dfrac{1}{n\ln n}<\dfrac{1}{n}$ . The bound is true, but it is a bound by a divergent series: it tells you nothing. Only an upper bound by a CONVERGENT series concludes.

Learn by heart

• An EQUIVALENT can only be used when the SIGN is CONSTANT. Otherwise: absolute convergence, Leibniz or Abel.
• CONDENSATION: a POSITIVE and DECREASING sequence; then $\sum a_{n}$ and $\sum 2^{n}a_{2^{n}}$ behave the same way.
• BERTRAND: $\displaystyle\sum\frac{1}{n(\ln n)^{p}}$ converges if and only if $p>1$ ; in particular $\displaystyle\sum\frac{1}{n\ln n}$ DIVERGES.
• ROOT TEST: concludes whenever the ratio test does, and also when the term oscillates between two regimes.
• RAABE: ratio $=1-\dfrac{\alpha}{n}+o\!\left(\dfrac{1}{n}\right)$ ; converges if $\alpha>1$ , exactly like a p-series.
• ABEL: $(a_{n})$ decreases to $0$ AND the partial sums of $(b_{n})$ are BOUNDED.
• TELESCOPING: TWO boundary terms remain, never zero.
• Comparison: SMALLER than a CONVERGENT series, or LARGER than a DIVERGENT one. The other two cases say nothing.

Frequently asked questions

Why is an equivalent not enough when the signs vary?

Because two equivalent sequences have a difference that is negligible compared with each of them, but the series of that difference can diverge. With a constant sign this is impossible. With varying signs, the classic example is the sum of a convergent alternating series and the harmonic series: the two general terms are equivalent, and yet one series converges while the other diverges.

What is the Cauchy condensation test used for?

It turns a slow series, typically one with a logarithm, into a known series. It requires the sequence to be positive and decreasing. You then replace the series with the series of the terms whose index is a power of two, multiplied by two to the power n. It gives the p-series test in one line, and the whole Bertrand family just as quickly.

Does the series of one over n times the log of n converge?

No, it diverges, even though its terms are clearly smaller than those of the harmonic series. Condensation turns it into exactly the harmonic series, up to a constant. It is the counterexample that reminds you that an upper bound by a divergent series tells you nothing.

When should you use Raabe's test?

Only when the ratio test fails, that is, when the ratio of two consecutive terms tends to one. You then expand that ratio in the form one minus alpha over n, and the coefficient alpha decides: convergence if it is greater than one, divergence if it is smaller. It is exactly the threshold of the p-series test.

Is the root test more powerful than the ratio test?

Yes. When the ratio has a limit, the root has the same one, so whatever the first test concludes, the second one concludes too. But the root test also concludes when the general term oscillates between two regimes, a case where the ratio has no limit at all. It is never weaker.

What is left after a telescoping sum cancels?

Exactly two terms: the first term kept and the first term dropped. Everything else cancels in pairs. A partial sum equal to zero for a series with strictly positive terms is therefore impossible, and that is the fastest check on this kind of computation.

Practise it

Corrected exercises: Advanced tests: root, condensation, Raabe, Abel, 201-SN3-RE (201-NYB)

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.

15 corrected exercises
150 points
225 minutes

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-calc2-advanced-convergence-tests. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: advanced convergence tests, 201-SN3-RE (201-NYB)

The essentials

The four tests, with their exact hypotheses

Condensation, and the Bertrand family

The sign pitfall, and telescoping

The mistakes that cost marks

1. Using an equivalent on a series whose terms change sign

2. Believing that a series with terms smaller than a divergent series converges

3. Applying condensation without checking that the sequence decreases

4. Invoking Raabe's test when the ratio does not tend to one

5. Getting the direction of Raabe's test wrong

6. Forgetting the condition on the partial sums in Abel's test

7. Forming a Cauchy product without absolute convergence

8. Forgetting the boundary terms of a telescoping sum

9. Believing that the ratio test and the root test are interchangeable

10. Using the comparison test the wrong way round

11. Using a zero equivalent

Which method to choose

Which advanced test, and when

What a test requires before it is applied

How the answer is expected to be written

Handling a family by condensation

Justifying that an equivalent is legitimate

Check before you hand in

The typical problem, taken apart

The Bertrand family, handled by condensation

Learn by heart

Frequently asked questions

Why is an equivalent not enough when the signs vary?

What is the Cauchy condensation test used for?

Does the series of one over n times the log of n converge?

When should you use Raabe's test?

Is the root test more powerful than the ratio test?

What is left after a telescoping sum cancels?

Corrected exercises: Advanced tests: root, condensation, Raabe, Abel, 201-SN3-RE (201-NYB)

See also

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