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

Revision sheet: infinite series, 201-SN3-RE (201-NYB)

This sheet does not recite the list of tests: it is in your notes, and knowing it is not enough. It gives the order in which to try them, the form of the general term that points to each one, and the cases where a test does NOT conclude, which is the most profitable information in the chapter for 201-SN3-RE (formerly 201-NYB).

It is written for CEGEP students in the Science program (Sciences de la nature). Once the sheet is read, the corrected exercise set on the same chapter puts every reflex to the test.

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The thread of the chapter

A convergence test is not chosen at random: you read it off the FORM of the general term. And no test, ever, gives the SUM of a series: knowing whether it converges and knowing what it equals are two separate questions, and the second has a simple answer in only two cases.

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

The essentials

What a series is, and what converging means

• $\sum_{n=1}^{\infty}u_{n}$ is the LIMIT of the partial sums $S_{n}=u_{1}+\cdots+u_{n}$ .
• The series CONVERGES if $\left(S_{n}\right)$ has a finite limit; that limit is the SUM.
• A series that converges necessarily has $u_{n}\to 0$ . The converse is FALSE.
• Only two families have a sum that is simple to compute: GEOMETRIC and TELESCOPING series.
• Everywhere else, you decide WHETHER the series converges, never its value.

Two series whose terms tend to

0

: the partial sums of the geometric series hug

1

, those of the harmonic series keep rising with no ceiling.

A question that asks whether a series converges expects “convergent” or “divergent”; a question that asks for “the sum” expects a number, and the series is almost always geometric or telescoping.

The reference series, to know by heart

• GEOMETRIC $\sum_{n=0}^{\infty}ar^{n}$ : converges if and only if $\left|r\right|<1$ , and its sum is then $\frac{a}{1-r}$ , where $a$ is the FIRST term written.
• p-SERIES $\sum\frac{1}{n^{p}}$ : converges if and only if $p>1$ .
• HARMONIC $\sum\frac{1}{n}$ : the case $p=1$ , it DIVERGES.
• ALTERNATING $\sum\frac{(-1)^{n}}{n}$ : converges, but not absolutely.
• TELESCOPING $\sum\left(a_{n}-a_{n+1}\right)$ : the partial sum reduces to $a_{1}-a_{n+1}$ .

Partial sums of

\sum\frac{1}{n^{p}}

: for

p=2

they settle below

\frac{\pi^{2}}{6}

, for

p=1

and

p=\frac{1}{2}

they rise without end. The boundary is

p=1

These five references serve as benchmarks for almost every other series: they are what you compare with, never an invented series.

The tests, and what each one does not say

• DIVERGENCE TEST: if $u_{n}\not\to 0$ , the series DIVERGES. If $u_{n}\to 0$ , it does NOT conclude.
• COMPARISON: for series with positive terms, smaller than a convergent series gives convergent, larger than a divergent series gives divergent.
• INTEGRAL: if $f$ is continuous, positive and DECREASING, $\sum f(n)$ and $\int_{1}^{+\infty}f$ both converge or both diverge.
• RATIO: if $\lim\left|\frac{u_{n+1}}{u_{n}}\right|=L<1$ it converges absolutely, if $L>1$ it diverges, if $L=1$ the test does NOT conclude.
• ALTERNATING SERIES: converges if $\left|u_{n}\right|$ DECREASES and tends to $0$ . Both conditions.

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, and it is the most frequent trap of the chapter

What not to write

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

What to write

“The divergence test concludes in one direction only: $u_{n}\to 0$ proves nothing, and the harmonic series diverges.”

Both sequences of terms tend to

0

and look alike to the eye: nothing in this picture says that one generates a divergent series and the other a convergent one.

Why: $\frac{1}{n}$ and $\frac{1}{n^{2}}$ both tend to $0$ , yet the first series diverges and the second converges: the limit of the general term does not tell them apart. This test serves only to prove a DIVERGENCE.

2. Getting the first term of a geometric series wrong

the whole question

What not to write

$\sum_{n=2}^{\infty}\left(\frac{1}{3}\right)^{n}=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}$

What to write

The first term written is $\left(\frac{1}{3}\right)^{2}=\frac{1}{9}$ , so the sum is $\frac{1/9}{1-1/3}=\frac{1}{6}$ .

Why: In $\frac{a}{1-r}$ , $a$ is the value of the FIRST term actually summed, not the coefficient in front of the power. The check is immediate: the sum of a series with positive terms must be larger than its first term and of the same order of magnitude.

3. Using the ratio test on a p-series

the whole question

What not to write

“ $\sum\frac{1}{n^{2}}$ : the ratio tends to $1$ , so the series diverges.”

What to write

“The ratio tends to $1$ : the test DOES NOT CONCLUDE. I use the p-series test, $p=2>1$ , so it converges.”

Why: $L=1$ is the failure case of the ratio test, and it is exactly what happens to every p-series. The ratio test is made for factorials and $n$ th powers, not for rational functions of $n$ .

4. Comparing in the wrong direction

the whole question on convergence

What not to write

“ $\frac{1}{n^{2}+1}\le\frac{1}{n^{2}}$ and $\sum\frac{1}{n^{2}}$ converges, so $\sum\frac{1}{n^{2}+1}$ diverges.”

What to write

“Bounded above by a convergent series, so CONVERGENT.”

Each term

\frac{1}{n^{2}+1}

(green) lies below

\frac{1}{n^{2}}

(blue): smaller than a convergent series, the series converges.

Why: The direction is the whole content of the test: smaller than a convergent series is convergent, larger than a divergent series is divergent. Stating it with these words rather than with symbols keeps you from reversing it under exam pressure.

5. Applying the integral test without checking that the function decreases

the whole question

What not to write

“ $\sum\frac{\sin^{2}n}{n}$ : I integrate $\frac{\sin^{2}x}{x}$ .”

What to write

“The function is not DECREASING: the integral test does not apply. Bounding by $\frac{1}{n}$ proves nothing; I group two terms: $\frac{\sin^{2}n}{n}+\frac{\sin^{2}(n+1)}{n+1}\geq\frac{1-\cos 1}{n+1}$ , and the series DIVERGES like the harmonic series.”

\frac{\sin^{2}x}{x}

rises again between consecutive zeros: it is not decreasing, and bounding by the integral is worthless.

Why: The test rests on bounding the sum with inscribed and circumscribed rectangles, which only makes sense if the function decreases. Without that hypothesis, the area and the sum have no reason left to compare, and the argument collapses.

6. Declaring an alternating series convergent without checking that the terms decrease

1 to 2 marks, and the whole question in the borderline cases

What not to write

“The terms alternate and tend to $0$ , so the series converges.”

What to write

“ $\left|u_{n}\right|$ must ALSO be decreasing from some index on.”

Why: The alternating series test has two hypotheses, and forgetting the decrease is the most frequent omission. The decrease has to be proved, often by differentiating the associated function, and that computation is exactly what the marking scheme expects.

7. Confusing convergence with absolute convergence

1 to 2 marks

What not to write

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

What to write

“It converges, but not ABSOLUTELY: the series of absolute values is the harmonic series, which diverges.”

The sums of

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

close in on

\ln 2

, those of

\sum\frac{1}{n}

rise without end: convergent, but not absolutely.

Why: Absolute convergence implies convergence, never the reverse. A series that converges without converging absolutely is called conditionally convergent, and its behaviour is fragile: reordering its terms can change its sum, which is impossible for an absolutely convergent series.

8. Answering with a number when the question asked whether the series converges

1 to 2 marks, and the value given cannot even be proved at CEGEP

What not to write

“ $\sum\frac{1}{n^{2}}$ converges and is about $1.64$ , so I write $1.64$ .”

What to write

“The question asks WHETHER it converges: I answer convergent, citing the p-series test with $p=2>1$ .”

Why: Deciding whether a series converges and computing its sum are two separate questions, and the second has an elementary answer only for geometric and telescoping series. The sum $\frac{\pi^{2}}{6}$ requires tools that are not in the course.

Which method to choose

Which test, depending on the form of the general term

Try the tests in this order, stopping at the first one that concludes. The divergence test always goes first because it costs three seconds.

If $u_{n}$ does not tend to $0$ → the series DIVERGES, you are done

Example: $\sum\frac{n}{n+1}$ diverges because the term tends to $1$
If $u_{n}=ar^{n}$ → GEOMETRIC series: converges if $\left|r\right|<1$ , and the sum is $\frac{a}{1-r}$

Example: $\sum_{n=1}^{\infty}\frac{2}{3^{n}}=\frac{2/3}{1-1/3}=1$
If $u_{n}=\frac{1}{n^{p}}$ , or a rational function of $n$ → p-SERIES test on the dominant degree, or comparison with $\frac{1}{n^{p}}$

Example: $\frac{n+1}{n^{3}+2}$ behaves like $\frac{1}{n^{2}}$ , so it converges
If $u_{n}$ contains a FACTORIAL or an $n$ th power → RATIO test, which simplifies these expressions

Example: $\sum\frac{2^{n}}{n!}$ : the ratio tends to $0$ , so convergence
If $u_{n}=f(n)$ with $f$ positive, continuous and decreasing, and $\int f$ computable → INTEGRAL test

Example: $\sum\frac{1}{n\ln n}$ diverges because $\int\frac{dx}{x\ln x}=\ln\ln x$ diverges
If the signs ALTERNATE → alternating series test: $\left|u_{n}\right|$ decreasing AND limit zero

Example: $\sum\frac{(-1)^{n}}{\sqrt{n}}$ converges

When a test does not conclude, that does not mean the series diverges: it means you must try another one. Writing “the test is inconclusive” is worth half a mark.

Convergent, absolutely convergent, or conditionally convergent

For a series whose terms change sign, first study the series of ABSOLUTE VALUES.

If $\sum\left|u_{n}\right|$ converges → the series converges ABSOLUTELY, hence it converges

Example: $\sum\frac{(-1)^{n}}{n^{2}}$
If $\sum\left|u_{n}\right|$ diverges but the alternating series test applies → the series is CONDITIONALLY CONVERGENT

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

Example: $\sum(-1)^{n}$

The word “absolutely” changes the expected answer: a series can converge without converging absolutely, and the question always distinguishes the two.

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.

Whether a series converges, written up for the grader

When to use it: Any question asking you to determine whether a series converges or diverges.

1 Test the general term: “ $u_{n}\to 0$ , the divergence test is inconclusive.”
2 NAME the test you use and check its hypotheses explicitly.
3 Carry out the computation of the test, limit or integral, in full.
4 Compare the result with the threshold of the test.
5 Conclude with the word CONVERGES or DIVERGES, citing the test.

Concluding sentence

Since the function $f(x)=\frac{1}{x\ln x}$ is continuous, positive and decreasing on $\left[2,+\infty\right)$ , and $\int_{2}^{+\infty}\frac{dx}{x\ln x}$ diverges, the integral test shows that $\sum_{n=2}^{\infty}\frac{1}{n\ln n}$ DIVERGES.

The trap: Not checking the hypotheses of the test. The integral test requires continuous, positive AND decreasing: all three are written down, and the marking scheme often sets aside a full mark for them.

Marking: Usually 1 mark for the divergence test, 1 mark for the hypotheses checked, 2 marks for the computation, 1 mark for the conclusion naming the test.

Check before you hand in

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

Compute a few partial sums

Add up the first ten or twenty terms on your calculator: a convergent series shows sums that settle down, a divergent one sums that keep growing.

$\sum\frac{1}{n^{2}}$ gives $1.54$ at the tenth term and then $1.63$ at the hundredth: it settles down ✓. The harmonic series goes from $2.93$ to $5.19$ : it is still climbing.
Compare with the dominant behaviour

Replace the general term with its equivalent in $\frac{1}{n^{p}}$ for large $n$ , and apply the p-series test to that equivalent.

$\frac{n+1}{n^{3}+2}$ behaves like $\frac{n}{n^{3}}=\frac{1}{n^{2}}$ : $p=2>1$ , so convergence, which the detailed computation must confirm.
Check a geometric sum against its first term

The sum of a geometric series with positive terms is larger than its first term and of the same order of magnitude. A ratio of more than ten signals an error on $a$ .

First term $\frac{1}{9}$ and sum $\frac{1}{6}$ : consistent ✓. A sum of $\frac{3}{2}$ for a first term of $\frac{1}{9}$ is not.
Check that the test used can conclude

Before writing the conclusion, check that the test is not in its failure case: ratio equal to $1$ , general term tending to $0$ , function not decreasing.

On every p-series, the ratio tends to $1$ : this test is useless there, and writing so keeps you from a wrong conclusion.
Reread the question: convergence or sum

Spot the key word in the question. “Determine whether it converges”, “study the convergence” expect a word; “compute the sum” expects a number, and then the series is almost always geometric or telescoping.

Answering $1.64$ to a question about convergence earns nothing, even if the number is right.

The typical problem, taken apart

The integral test on a borderline series

Determine whether $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}$ converges or diverges.

The rectangles of height

f(n)

placed to the right of each integer stick out above the decreasing curve: this comparison, and it alone, allows you to replace the sum with an integral.

Step 1

General term: $\frac{1}{n\ln n}\to 0$ as $n\to+\infty$ . The divergence test DOES NOT CONCLUDE.

Why

You write it anyway, because it costs three seconds and half a mark is attached to it. Above all, it keeps you from concluding convergence on that basis alone, which is the error of the chapter.

Step 2

The p-series test does not apply directly: $\frac{1}{n\ln n}$ is smaller than $\frac{1}{n}$ but larger than $\frac{1}{n^{p}}$ for every $p>1$ , once $n$ is large enough. Comparison is therefore undecided.

Why

Noting that a test fails is part of the answer. Here the series is squeezed strictly between the divergent $\frac{1}{n}$ and all the convergent $\frac{1}{n^{p}}$ : the typical borderline case.

Step 3

Hypotheses of the integral test: $f(x)=\frac{1}{x\ln x}$ is CONTINUOUS on $\left[2,+\infty\right)$ , POSITIVE since $x>0$ and $\ln x>0$ for $x\ge 2$ , and DECREASING since $x\ln x$ is increasing.

Why

The three hypotheses are written one by one, each with its reason. This is the block the marking scheme grades, and it is also what distinguishes a legitimate application from a mechanical one.

Step 4

$\int_{2}^{+\infty}\frac{dx}{x\ln x}$ : let $u=\ln x$ , so $du=\frac{dx}{x}$ , and the integral becomes $\int_{\ln 2}^{+\infty}\frac{du}{u}=\lim_{T\to+\infty}\left[\ln u\right]_{\ln 2}^{T}=+\infty$ .

Why

The substitution $u=\ln x$ is the one the form $\frac{u'}{u}$ calls for. The limit is written with a variable bound, like any improper integral: this line is marked separately.

Step 5

The integral DIVERGES, so the series $\sum_{n=2}^{\infty}\frac{1}{n\ln n}$ diverges as well.

Why

The test passes on the behaviour in both directions: the sum and the integral behave the same way. It gives NO information on the value, on the other hand, which is to be expected since there is none here.

Step 6

Numerical check: the partial sums are about $2.32$ at $n=100$ , $2.73$ at $n=1000$ and $3.02$ at $n=10\,000$ . They are still rising, very slowly, like $\ln\ln n$ ✓.

Why

Growth in $\ln\ln n$ is so slow that no partial sum can prove the divergence: that is exactly why the theoretical test is indispensable, and why a conclusion drawn from the first sums would be wrong.

The conclusion, written out

Since $f(x)=\frac{1}{x\ln x}$ is continuous, positive and decreasing on $\left[2,+\infty\right)$ , and its integral from $2$ to infinity diverges, the integral test shows that $\sum_{n=2}^{\infty}\frac{1}{n\ln n}$ DIVERGES.

The classic mistake on this problem: Concluding convergence because $\frac{1}{n\ln n}$ is smaller than $\frac{1}{n}$ . Being smaller than a DIVERGENT series proves nothing: comparison passes on convergence only downward, from a convergent series.

Learn by heart

• $u_{n}\to 0$ is NECESSARY, never sufficient. This test serves only to prove a divergence.
• Geometric: converges if $\left|r\right|<1$ , sum $\frac{a}{1-r}$ where $a$ is the FIRST term summed.
• p-series: $\sum\frac{1}{n^{p}}$ converges if and only if $p>1$ . The harmonic series diverges.
• Ratio: $L<1$ converges, $L>1$ diverges, $L=1$ DOES NOT CONCLUDE. Useless on p-series.
• Integral: continuous, positive, DECREASING. The three hypotheses are written down.
• Alternating: $\left|u_{n}\right|$ decreasing AND limit zero. Both.
• Comparison: smaller than a convergent series converges, larger than a divergent series diverges.
• Convergence and sum are two questions: the sum can be computed only for geometric and telescoping series.

Frequently asked questions

Does a series whose terms tend to zero converge?

No, not necessarily. It is a necessary condition but not a sufficient one. The harmonic series is the counterexample: its terms tend to zero and its partial sums grow without limit. The divergence test serves only to prove a divergence, when the term does not tend to zero.

How do you choose the right convergence test?

You read the form of the general term. A constant power points to a geometric series, a rational function calls for the p-series test or a comparison, a factorial calls for the ratio test, a decreasing function you can integrate calls for the integral test, and alternating signs call for the alternating series test.

Why is the ratio test inconclusive for one over n squared?

Because the ratio of two consecutive terms tends to one, which is exactly the failure case of the test. This happens to every p-series, whose terms decrease too slowly for the ratio to move away from one. You then have to switch to the p-series test itself.

What is the difference between convergence and absolute convergence?

A series converges absolutely if the series of its absolute values converges. That implies convergence, but not the reverse. The alternating series of the reciprocals of the integers converges without converging absolutely: it is called conditionally convergent, and its sum can change if its terms are reordered.

How do you compute the sum of a series?

It can be done simply in only two cases. A geometric series equals its first term divided by one minus the common ratio. A telescoping series collapses because the terms cancel in pairs in the partial sum. Everywhere else, you only decide whether the series converges, not its value.

Practise it

Corrected exercises: Infinite series, 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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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-calc2-infinite-series. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: infinite series, 201-SN3-RE (201-NYB)

The essentials

What a series is, and what converging means

The reference series, to know by heart

The tests, and what each one does not say

The mistakes that cost marks

1. Concluding convergence because the general term tends to zero

2. Getting the first term of a geometric series wrong

3. Using the ratio test on a p-series

4. Comparing in the wrong direction

5. Applying the integral test without checking that the function decreases

6. Declaring an alternating series convergent without checking that the terms decrease

7. Confusing convergence with absolute convergence

8. Answering with a number when the question asked whether the series converges

Which method to choose

Which test, depending on the form of the general term

Convergent, absolutely convergent, or conditionally convergent

How the answer is expected to be written

Whether a series converges, written up for the grader

Check before you hand in

The typical problem, taken apart

The integral test on a borderline series

Learn by heart

Frequently asked questions

Does a series whose terms tend to zero converge?

How do you choose the right convergence test?

Why is the ratio test inconclusive for one over n squared?

What is the difference between convergence and absolute convergence?

How do you compute the sum of a series?

Corrected exercises: Infinite series, 201-SN3-RE (201-NYB)

See also

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