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

Revision sheet: series with sin, cos, exp and ln terms, 201-SN3-RE (201-NYB)

A general term in $\sin\dfrac{1}{n}$ , $e^{1/n}-1$ or $\ln\left(1+\dfrac{1}{n}\right)$ is never studied directly: replace it by its equivalent, and the resulting series is almost always a p-series whose behaviour is known.

This sheet for 201-SN3-RE (formerly 201-NYB) gives the five equivalents to know by heart, the test to choose according to the form of the term, the three conditions of the Alternating Series Test, and the distinction between absolute and conditional convergence, which is often worth half the marks of the question.

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The thread of the chapter

Facing a sine, an exponential or a logarithm whose argument tends to zero, the first move is ALWAYS the same: replace it by its equivalent. All the rest of the chapter is only the choice of the test to apply next.

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

The essentials

The five equivalents, and the Limit Comparison Test

• At $0$ : $\sin u\sim u$ , $\tan u\sim u$ , $1-\cos u\sim\dfrac{u^{2}}{2}$ , $e^{u}-1\sim u$ , $\ln(1+u)\sim u$ . They are applied with $u=\dfrac{1}{n}$ , which does tend to $0$ .
• Limit COMPARISON: if $\dfrac{a_{n}}{b_{n}}\to L$ finite and NONZERO, with $a_{n}$ and $b_{n}$ positive, the two series BEHAVE THE SAME WAY.
• P-SERIES: $\displaystyle\sum\frac{1}{n^{p}}$ converges if and only if $p>1$ . The case $p=1$ , the harmonic series, DIVERGES.
• The complete chain is therefore: equivalent, then Limit Comparison Test, then p-series. Three lines are enough to handle half of the exam questions.
• NECESSARY condition: if $a_{n}\not\to 0$ , the series diverges. The converse is FALSE, and that is the whole difficulty of the chapter.

Near

0

\sin x

and

x

are almost indistinguishable: that is the equivalent

\sin u\sim u

. Away from

0

, they separate: it is applied only to an argument that tends to

0

An equivalent never applies to a SUM or a DIFFERENCE whose first-order terms cancel: $\ln(1+u)-u$ is not equivalent to $0$ , but to $-\dfrac{u^{2}}{2}$ . In that case you need an expansion to a higher order.

The tests, and what they do not say

• RATIO TEST (d'Alembert): if $\dfrac{a_{n+1}}{a_{n}}\to L$ , the series converges if $L<1$ , diverges if $L>1$ , and the test is INCONCLUSIVE if $L=1$ .
• It is made for FACTORIALS and $n$ th POWERS; it systematically fails on rational functions, where $L=1$ .
• An essential limit: $\left(1+\dfrac{1}{n}\right)^{n}\to e$ , hence $\dfrac{n^{n}}{(n+1)^{n}}\to\dfrac{1}{e}$ . This is what makes the thresholds in $\dfrac{k}{e}$ appear.
• ALTERNATING SERIES, Leibniz test: three conditions, the term must ALTERNATE, $|a_{n}|$ must DECREASE, and $|a_{n}|$ must TEND TO ZERO. All three, not one.
• Under these conditions, the remainder satisfies $\left|R_{N}\right|\le\left|a_{N+1}\right|$ : the first OMITTED term, not the last one kept.

The sequence climbs toward

e

without ever reaching it: it is what gives

\dfrac{n^{n}}{(n+1)^{n}}\to\dfrac{1}{e}

, and therefore the ratio

\dfrac{1}{e}<1

that makes

\displaystyle\sum\frac{n!}{n^{n}}

converge.

When the ratio test returns $1$ , it says nothing at all: neither convergence nor divergence. Go back to an equivalent and the p-series test, which is almost always possible.

Absolute, conditional, or divergent

• ABSOLUTELY convergent: $\sum\left|a_{n}\right|$ converges. It is the strongest case, and it implies that $\sum a_{n}$ converges.
• CONDITIONALLY convergent: $\sum a_{n}$ converges but $\sum\left|a_{n}\right|$ DIVERGES. It is the typical case of alternating series whose terms decrease slowly.
• The question “absolutely or conditionally” is handled in TWO steps: first the series of absolute values, then, if it diverges, the Leibniz test.
• A convergent series of POSITIVE terms is automatically absolutely convergent: the distinction only makes sense for terms of varying sign.
• The textbook example: $\displaystyle\sum\frac{(-1)^{n}}{n}$ converges, but $\displaystyle\sum\frac{1}{n}$ diverges: it is conditionally convergent.

The partial sums BRACKET the limit, alternating from one side to the other: that is exactly why the error made by stopping at index

N

never exceeds the first omitted term.

The order of the two steps matters: starting with the absolute values lets you conclude at once, if they converge, without even looking at the signs.

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. Concluding convergence because the general term tends to zero

the whole question, 5 marks

What not to write

“ $\dfrac{1}{n}\to 0$ , so $\displaystyle\sum\frac{1}{n}$ converges.”

What to write

“A general term tending to zero is a NECESSARY condition, not a sufficient one. The harmonic series is the counterexample: it diverges even though its term tends to zero.”

Why: This is the founding mistake of the chapter. The Test for Divergence only ever concludes DIVERGENCE, when the limit is not zero; it never concludes convergence.

2. Concluding despite a limit equal to one in the ratio test

the whole question, 5 marks

What not to write

“ $\dfrac{a_{n+1}}{a_{n}}\to 1$ , so the series diverges.”

What to write

“The test is INCONCLUSIVE when $L=1$ : change method. $\displaystyle\sum\frac{1}{n}$ and $\displaystyle\sum\frac{1}{n^{2}}$ both give $L=1$ , and one diverges while the other converges.”

Why: The case $L=1$ is exactly the one where the test is blind, and it is also the case of every rational function. On these series, go through an equivalent and the p-series test.

3. Applying an equivalent to a difference whose first-order terms cancel

the whole question, 6 marks

What not to write

“ $\ln\left(1+\dfrac{1}{n}\right)-\dfrac{1}{n}\sim\dfrac{1}{n}-\dfrac{1}{n}=0$ , so the series trivially converges.”

What to write

“The first-order terms cancel: the next term is needed. $\ln(1+u)=u-\dfrac{u^{2}}{2}+\ldots$ , so the difference is equivalent to $-\dfrac{1}{2n^{2}}$ , and the series converges by the p-series test with $p=2$ .”

Why: Equivalents cannot be subtracted: two equivalent quantities can have a difference of any order at all. As soon as a subtraction cancels the leading term, a further expansion is needed.

4. Checking only one of the three conditions of the Alternating Series Test

2 marks for rigour, and sometimes the whole question

What not to write

“The term tends to zero and the series alternates, so it converges.”

What to write

“ $\left|a_{n}\right|$ must ALSO be DECREASING, at least from some index on. Without this condition, the test does not apply.”

Why: The decrease is what guarantees that the partial sums close in on each other. Without it, an alternating series whose terms tend to zero in absolute value can perfectly well diverge, and the counterexample comes up regularly on exams.

5. Confusing convergence with absolute convergence

3 marks, and half the question

What not to write

“ $\displaystyle\sum\frac{(-1)^{n}}{n}$ converges, so it converges absolutely.”

What to write

“ $\displaystyle\sum\left|\frac{(-1)^{n}}{n}\right|=\sum\frac{1}{n}$ DIVERGES: the series is CONDITIONALLY convergent, not absolutely convergent.”

Why: The implication only goes one way: absolute convergence implies convergence, never the reverse. The question “absolutely or conditionally” therefore expects two studies, not one.

6. Using the wrong index in the remainder bound

2 marks, and an index one too large

What not to write

“I stop at index $N$ , so $\left|R_{N}\right|\le\left|a_{N}\right|$ .”

What to write

“ $\left|R_{N}\right|\le\left|a_{N+1}\right|$ : it is the first OMITTED term that bounds the error, not the last term kept.”

Why: The remainder starts at index $N+1$ : it is bounded by this first omitted term. Using $a_{N}$ gives a bound that is correct but weaker, and above all a wrong index when you look for the smallest $N$ .

7. Believing that decreasing to zero is enough for a series of positive terms to converge

the whole question, 5 marks

What not to write

“ $\dfrac{\ln n}{n}$ decreases to zero, so $\displaystyle\sum\frac{\ln n}{n}$ converges.”

What to write

“ $\dfrac{\ln n}{n}\ge\dfrac{1}{n}$ as soon as $n\ge 3$ : the series is LARGER than the harmonic series, so it DIVERGES.”

From

n=3

on, the curve of the logarithm divided by

n

lies ABOVE that of

\dfrac{1}{n}

and never comes back below it: the series is larger than the harmonic series, so it diverges.

Why: The logarithm grows too slowly to make up for anything: $\dfrac{\ln n}{n}$ tends to zero more slowly than $\dfrac{1}{n}$ , so the series diverges even more surely than the harmonic series.

8. Using the Limit Comparison Test on terms of varying sign

the whole question, 5 marks, with the conclusion reversed

What not to write

“ $\dfrac{(-1)^{n}}{n}$ and $\dfrac{1}{n}$ have a ratio equal to $\pm 1$ , so the two series behave the same way.”

What to write

“The Limit Comparison Test requires POSITIVE terms, and the ratio must have a LIMIT, which is not the case here. For an alternating series, use the Leibniz test.”

Why: Without the positivity hypothesis, cancellations between terms of opposite signs make the comparison meaningless: the two series of the example in fact behave differently.

9. Forgetting that the ratio test fails on rational functions

no marks lost, but ten minutes

What not to write

“ $\displaystyle\sum\frac{n^{2}+1}{n^{4}+3}$ : I apply the ratio test.”

What to write

“The ratio tends to $1$ : the test is useless. Use the equivalent $\dfrac{n^{2}}{n^{4}}=\dfrac{1}{n^{2}}$ , a convergent p-series.”

Why: For any rational function, the ratio of two consecutive terms tends to $1$ : it is structural. The ratio test is only useful in front of a factorial or an $n$ th power.

10. Applying an equivalent to an argument that does not tend to zero

the whole question, 5 marks

What not to write

“ $\sin n\sim n$ , so $\displaystyle\sum\frac{\sin n}{n^{2}}\sim\sum\frac{1}{n}$ .”

What to write

“The standard equivalents hold when the argument tends to ZERO. Here $n$ tends to infinity and $\sin n$ oscillates: bound it by $\dfrac{1}{n^{2}}$ and conclude ABSOLUTE convergence.”

Why: $\sin u\sim u$ is an approximation near $0$ , and nothing else. Facing an unbounded argument, the only way is to bound the absolute value by a convergent series.

11. Ignoring the index from which a hypothesis holds

2 marks for rigour

What not to write

“ $\dfrac{\ln n}{n}$ is not decreasing at $n=2$ , so the Leibniz test does not apply.”

What to write

“It is enough for it to decrease FROM SOME INDEX ON: here $n\ge 3$ , since the derivative of $\dfrac{\ln x}{x}$ is negative as soon as $x>e$ . Whether a series converges does not depend on its first terms.”

Why: Changing a FINITE number of terms never changes whether a series converges. All the hypotheses of the tests are therefore understood from some index on, and it is enough to state that index.

Which method to choose

Which test, according to the form of the general term

What you see in $a_{n}$ , in this order

If $a_{n}$ does NOT tend to $0$ → the series DIVERGES, and it is over in one line

Example: $\displaystyle\sum\frac{n}{n+1}$ diverges outright
If $a_{n}$ contains a FACTORIAL or an $n$ th power → RATIO TEST, using $\left(1+\dfrac{1}{n}\right)^{n}\to e$ if needed

Example: $\displaystyle\sum\frac{n!}{n^{n}}$ converges, with $L=\frac{1}{e}$
If $a_{n}$ contains $\sin$ , $\tan$ , $\ln(1+\cdot)$ or $e^{\cdot}-1$ with an argument tending to $0$ → EQUIVALENT, then Limit Comparison Test, then p-series

Example: $\sin\dfrac{1}{n}\sim\dfrac{1}{n}$ , so the series diverges
If $a_{n}$ is a RATIONAL function → equivalent to the ratio of the leading terms, then p-series; the ratio test is useless

Example: $\dfrac{n^{2}+1}{n^{4}+3}\sim\dfrac{1}{n^{2}}$ , convergent
If $a_{n}$ changes sign regularly → study $\sum\left|a_{n}\right|$ FIRST; if it diverges, apply LEIBNIZ

Example: $\displaystyle\sum\frac{(-1)^{n}\ln n}{n}$ is conditionally convergent
If $a_{n}$ oscillates without alternating, like $\sin n$ → BOUND the absolute value by a convergent series

$\left|\dfrac{\sin n}{n^{2}}\right|\le\dfrac{1}{n^{2}}$ gives absolute convergence

The first test takes five seconds and closes a third of the questions. Move on to the other tests only after checking that the general term does tend to zero.

Absolutely convergent, conditionally convergent or divergent

The order of the two studies, which is not arbitrary

If $\sum\left|a_{n}\right|$ CONVERGES → the series is ABSOLUTELY convergent, hence convergent. No other study needed

Example: $\displaystyle\sum\frac{(-1)^{n}}{n^{2}}$
If $\sum\left|a_{n}\right|$ DIVERGES and the series alternates with $|a_{n}|$ decreasing to $0$ → the series is CONDITIONALLY convergent

Example: $\displaystyle\sum\frac{(-1)^{n}}{n}$ , with sum $-\ln 2$
If $\sum\left|a_{n}\right|$ diverges and $a_{n}\not\to 0$ → the series DIVERGES

Example: $\displaystyle\sum(-1)^{n}\frac{n}{n+1}$
If $\sum\left|a_{n}\right|$ diverges, the series alternates, but $|a_{n}|$ does not decrease → the Leibniz test does not apply: the partial sums must be studied directly

a rare case at CEGEP, but it appears in reasoning questions
If the terms are all POSITIVE → the distinction does not arise: convergent means absolutely convergent

saying so in one sentence saves a useless study

Starting with the absolute values is always the right order: if they converge, the question is closed, and if they diverge, you already know the answer will be “conditionally” or “divergent”.

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.

Studying an alternating series from start to finish

When to use it: The general term contains a factor $(-1)^{n}$ or changes sign regularly.

1 First study $\sum\left|a_{n}\right|$ : if it converges, conclude ABSOLUTE convergence and stop there.
2 If it diverges, state the THREE conditions of the Leibniz test and check them one by one: alternation, decrease of $|a_{n}|$ , zero limit.
3 Justify the decrease by studying a function, stating the index from which it holds.
4 Conclude convergence, and therefore CONDITIONAL convergence since the series of absolute values diverges.
5 If the question asks for it, bound the remainder by $\left|a_{N+1}\right|$ and solve the inequality to find the index.

Concluding sentence

“The series of absolute values, $\displaystyle\sum\frac{\ln n}{n}$ , is larger than the harmonic series, so it diverges; the Leibniz test applies because $\dfrac{\ln n}{n}$ decreases to $0$ for $n\ge 3$ : the series is CONDITIONALLY CONVERGENT.”

The trap: Skipping the study of the absolute values and concluding directly with the Leibniz test. You then get convergence, but not the answer to the question asked, which almost always concerns the KIND of convergence.

Marking: Typically 2 marks for the study of the absolute values, 3 for the three conditions checked, 1 for the conclusion of conditional convergence, 2 for the remainder bound.

Concluding with an equivalent and a comparison

When to use it: The general term contains a transcendental function whose argument tends to zero.

1 Check that the argument does tend to $0$ , and say so: it is the condition for using equivalents.
2 Replace by the standard equivalent, writing explicitly which one is used.
3 Check that the terms are POSITIVE, at least from some index on, which allows the Limit Comparison Test.
4 State the comparison: the ratio tends to a finite nonzero limit, so the two series behave the same way.
5 Conclude with the p-series test, giving the value of $p$ explicitly.

Concluding sentence

“Since $\dfrac{1}{n}\to 0$ , we have $1-\cos\dfrac{1}{n}\sim\dfrac{1}{2n^{2}}$ ; the terms being positive, the Limit Comparison Test applies, and the p-series with exponent $2$ converges: the series converges.”

The trap: Using an equivalent without checking positivity. The Limit Comparison Test is a theorem about series of positive terms, and forgetting this hypothesis costs the rigour mark every time.

Check before you hand in

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

Does the general term tend to zero

Before any test, compute the limit of the general term. If it is not zero, the series diverges and the question is over.

A term in $\dfrac{n}{n+1}$ that tends to $1$ : no need to go any further.
The p-series exponent stated explicitly

After an equivalent, write the value of $p$ and conclude. A $p$ equal to $1$ means divergence, a $p$ strictly greater than $1$ means convergence.

An equivalent in $\dfrac{1}{n}$ followed by a conclusion of convergence: $p=1$ diverges.
The three Leibniz conditions written one by one

Reread your answer and count: alternation, decrease, zero limit. Three lines, not one.

A conclusion of convergence with no mention of the decrease: the test has not been applied.
The index of the remainder

Check that the bound uses $a_{N+1}$ , the first omitted term, and not $a_{N}$ .

An index found equal to $648$ instead of $647$ : the index of the remainder is off by one.
Consistency between the two studies

A series cannot be both absolutely convergent and conditionally convergent. Check that the two conclusions agree.

A series of absolute values declared convergent and a conclusion of conditional convergence: the two statements contradict each other.

The typical problem, taken apart

An alternating series with a logarithm, from its behaviour to the index of the remainder

Determine whether $\displaystyle\sum_{n\ge 2}\frac{(-1)^{n}\ln n}{n}$ converges or diverges, and if it converges, whether absolutely or only conditionally.

Then determine the index $N$ from which the partial sum approximates the sum of the series to within $0.01$ .

Finally, show with the ratio test that $\displaystyle\sum\frac{n!}{n^{n}}$ converges.

\left|a_{n}\right|=\frac{\ln n}{n}

rises up to

x=e

then decreases to

0

: the hypotheses of the Alternating Series Test hold from index

3

on, which is enough.

Step 1

General term: $\left|a_{n}\right|=\dfrac{\ln n}{n}$ , which tends to $0$ since the logarithm grows more slowly than $n$ . The necessary condition is met and you can continue.

Why

This first test proves nothing about convergence, but its failure would have stopped everything. Mentioning it in one line is expected by the marking scheme.

Step 2

Absolute values: for $n\ge 3$ , $\ln n\ge 1$ , so $\dfrac{\ln n}{n}\ge\dfrac{1}{n}$ . The series $\displaystyle\sum\frac{\ln n}{n}$ is LARGER than the harmonic series, which diverges: so it diverges too.

Why

Always start with the absolute values: if they converged, the question would be closed here. Their divergence already announces that the final answer will be “conditionally convergent” or “divergent”.

Step 3

Leibniz test, condition 1: the series does ALTERNATE, thanks to the factor $(-1)^{n}$ .

Why

The three conditions are written separately and numbered. The marking scheme gives one mark per condition, and a conclusion without all three is worth none.

Step 4

Condition 2, decrease: the derivative of $f(x)=\dfrac{\ln x}{x}$ is $\dfrac{1-\ln x}{x^{2}}$ , negative as soon as $x>e$ . So $\left|a_{n}\right|$ decreases for $n\ge 3$ .

Why

The decrease is proved by studying a function, not by an impression. Index $3$ is enough: changing a finite number of terms does not change whether a series converges.

Step 5

Condition 3, zero limit: $\dfrac{\ln n}{n}\to 0$ , already established. The three conditions being met, the series CONVERGES.

Why

The three conditions are independent and all necessary. It is the second one, the decrease, that is almost always forgotten.

Step 6

Conclusion on the behaviour: the series converges but the series of absolute values diverges, so it is CONDITIONALLY CONVERGENT.

Why

This is the answer to the question asked. An answer that stops at “converges” loses half the marks, since the question explicitly asks for the kind of convergence.

Step 7

Remainder: $\left|R_{N}\right|\le\left|a_{N+1}\right|=\dfrac{\ln(N+1)}{N+1}$ . We look for $\dfrac{\ln(N+1)}{N+1}<0.01$ . The function $\dfrac{\ln x}{x}$ goes below $0.01$ from $x=648$ on, so $N=647$ works.

Why

It is the first OMITTED term that bounds the error, hence the index $N+1$ . Using $a_{N}$ would give $N=648$ , one too many, which is precisely the trap of the exercise.

Step 8

Second part, the ratio test on $\displaystyle\sum\frac{n!}{n^{n}}$ : $\dfrac{a_{n+1}}{a_{n}}=\dfrac{(n+1)!}{(n+1)^{n+1}}\cdot\dfrac{n^{n}}{n!}=\dfrac{n^{n}}{(n+1)^{n}}=\dfrac{1}{\left(1+\frac{1}{n}\right)^{n}}\to\dfrac{1}{e}$ .

Why

A factorial AND an $n$ th power point to the ratio test without hesitation. The limit $\left(1+\dfrac{1}{n}\right)^{n}\to e$ is the one you need to know to get there.

Step 9

Since $\dfrac{1}{e}\approx 0.368<1$ , the series CONVERGES. Numerical check: the first ten terms already give $1.879$ , and the full sum is about $1.88$ .

Why

The conclusion is read directly from the comparison with $1$ . The numerical check confirms that the series converges quickly, which is consistent with a ratio clearly less than $1$ .

The conclusion, written out

“The series $\displaystyle\sum_{n\ge 2}\frac{(-1)^{n}\ln n}{n}$ is CONDITIONALLY CONVERGENT: it converges by the Alternating Series Test, while the series of its absolute values is larger than the harmonic series and diverges. You must go up to index $N=647$ to guarantee an error less than $0.01$ . Finally, $\displaystyle\sum\frac{n!}{n^{n}}$ converges, its ratio tending to $\frac{1}{e}$ .”

The classic mistake on this problem: Concluding that the series converges absolutely because $\dfrac{\ln n}{n}$ tends to zero. Decreasing to zero says nothing about the convergence of a series of positive terms: $\dfrac{1}{n}$ is the universal counterexample, and $\dfrac{\ln n}{n}$ is even larger.

Learn by heart

• $a_{n}\to 0$ is NECESSARY, never sufficient: the harmonic series is the counterexample.
• Equivalents at $0$ : $\sin u\sim u$ , $\tan u\sim u$ , $1-\cos u\sim\dfrac{u^{2}}{2}$ , $e^{u}-1\sim u$ , $\ln(1+u)\sim u$ .
• p-series: $\displaystyle\sum\frac{1}{n^{p}}$ converges if and only if $p>1$ ; $p=1$ DIVERGES.
• Ratio test: $L<1$ converges, $L>1$ diverges, $L=1$ is INCONCLUSIVE. Useless on a rational function.
• $\left(1+\dfrac{1}{n}\right)^{n}\to e$ , hence the thresholds in $\dfrac{k}{e}$ .
• Leibniz: THREE conditions, alternation, decrease of $|a_{n}|$ , zero limit.
• Remainder of an alternating series: $\left|R_{N}\right|\le\left|a_{N+1}\right|$ , the first OMITTED term.
• Conditionally convergent: the series converges, the series of absolute values diverges. Study the absolute values FIRST.

Frequently asked questions

If the general term tends to zero, does the series converge?

No. It is a necessary condition but not a sufficient one. The harmonic series, whose general term is one over n, is the counterexample: its term does tend to zero and yet it diverges. The Test for Divergence can only conclude divergence, when the limit is not zero.

What should you do when the ratio test gives one?

Change method, because the test concludes neither one way nor the other. This is always the case for rational functions. Go through an equivalent of the general term instead, then a comparison with a p-series, whose behaviour depends on the exponent.

What are the conditions of the Alternating Series Test?

There are three: the series must alternate, the absolute value of the general term must decrease, and it must tend to zero. All three are necessary. The decrease, often forgotten, is proved by studying a function, and it is enough for it to hold from some index on.

What is the difference between absolute and conditional convergence?

A series is absolutely convergent if the series of its absolute values converges; it is then convergent. It is conditionally convergent if it converges while the series of its absolute values diverges. The right method is to study the absolute values first: if they converge, the question is closed.

How do you bound the remainder of an alternating series?

By the absolute value of the first omitted term, that is, the term of index N plus one if you stop at index N. It follows directly from the way the partial sums close in, bracketing the limit by alternating from one side to the other. Using the term of index N gives an index that is off by one.

Can you use an equivalent on a difference?

Not if the first-order terms cancel. Two equivalent quantities can have a difference of any order at all, so the equivalent no longer says anything. You then need a Taylor expansion taken one order further, which brings out the next term and gives the true order of magnitude.

Practise it

Corrected exercises: Series with sin, cos, exp and ln terms, 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

Do the exercises

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Revision sheet: series with sin, cos, exp and ln terms, 201-SN3-RE (201-NYB)

The essentials

The five equivalents, and the Limit Comparison Test

The tests, and what they do not say

Absolute, conditional, or divergent

The mistakes that cost marks

1. Concluding convergence because the general term tends to zero

2. Concluding despite a limit equal to one in the ratio test

3. Applying an equivalent to a difference whose first-order terms cancel

4. Checking only one of the three conditions of the Alternating Series Test

5. Confusing convergence with absolute convergence

6. Using the wrong index in the remainder bound

7. Believing that decreasing to zero is enough for a series of positive terms to converge

8. Using the Limit Comparison Test on terms of varying sign

9. Forgetting that the ratio test fails on rational functions

10. Applying an equivalent to an argument that does not tend to zero

11. Ignoring the index from which a hypothesis holds

Which method to choose

Which test, according to the form of the general term

Absolutely convergent, conditionally convergent or divergent

How the answer is expected to be written

Studying an alternating series from start to finish

Concluding with an equivalent and a comparison

Check before you hand in

The typical problem, taken apart

An alternating series with a logarithm, from its behaviour to the index of the remainder

Learn by heart

Frequently asked questions

If the general term tends to zero, does the series converge?

What should you do when the ratio test gives one?

What are the conditions of the Alternating Series Test?

What is the difference between absolute and conditional convergence?

How do you bound the remainder of an alternating series?

Can you use an equivalent on a difference?

Corrected exercises: Series with sin, cos, exp and ln terms, 201-SN3-RE (201-NYB)

See also

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