MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: the integral test and the comparison tests (MATH 141)

Q: How do I know when I can use the integral test on a series?

You need a function f that is continuous, positive and decreasing on some interval from N to infinity, with f(n) equal to the n-th term. The decrease only has to start at some N: ln x over x increases up to x = e, so you apply the test from n = 3. If f oscillates, like a function with sin or cos in it, use a comparison instead.

Q: Does the integral test give the sum of the series?

No. It only says whether the series converges. The integral is the area under the curve and the series is a sum of rectangles, so they differ: the series of 1 over n cubed is more than its first term 1, while the integral from 1 to infinity is one half. What the integral does give is a bracket, the sum lying between the integral from 1 and the first term plus that integral.

Q: Why does my comparison not prove anything even though the inequality is true?

Because it points the wrong way. A series smaller than a divergent one can converge or diverge, and so can a series larger than a convergent one. To prove convergence you need your terms below a convergent series; to prove divergence, above a divergent one. Usually you fix it by giving up a constant, for example 1 over 3n instead of 1 over 2n.

Q: How do I choose the series to compare with in the limit comparison test?

Keep only the dominant term of the numerator and of the denominator and simplify: that is your benchmark, usually a p-series. For two n squared plus 3n over the square root of n to the seventh plus one, the benchmark is 1 over n to the three halves. If the limit of the ratio comes out as zero or infinity on the wrong side, move the exponent closer to the term.

Q: Is the series of 1 over n to the power 1 plus 1 over n a convergent p-series?

No, and it diverges. A p-series needs a constant exponent, and here the exponent tends to 1. Compared with the harmonic series, the ratio is n to the power minus 1 over n, which tends to 1, so by the limit comparison test both series diverge together, even though every single exponent is greater than 1.

This sheet is not a summary of sections 11.3 and 11.4 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on the integral test and the comparison tests in MATH 141 at McGill University, and which precise gesture avoids each loss.

Every number below is done by hand, as on the exam, and every series quoted has positive terms: the alternating series, the ratio test and the root test belong to the next chapters.

The thread of the chapter

Every test of the chapter judges a series against something whose fate is already known, an integral or a benchmark series, and it is a theorem: its hypotheses are written before the verdict, a comparison only decides below a convergent series or above a divergent one, and the integral gives the fate of the series, never its value.

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

Integral test
Convergence of p-series
Direct comparison test
Limit comparison test
Integral test remainder estimate
Harmonic series
Geometric series as a benchmark
Eventually decreasing function
Dominant term of a sequence
Improper integral

The essentials

The integral test: three hypotheses, one verdict, no value

• If $f$ is continuous, positive and DECREASING on $[N, \infty)$ and $f(n) = a_n$ , then $\sum_{n \ge N} a_n$ and $\int_N^{\infty} f(x)\,dx$ both converge or both diverge.
• Decreasing only has to hold from some $N$ on: $\frac{\ln x}{x}$ increases up to $x = e$ , so the test is applied from $N = 3$ . Finitely many terms never change convergence.
• Decrease without a derivative: a product of POSITIVE decreasing functions decreases, and so does the reciprocal of a product of positive increasing ones.
• The integral is NOT the sum. Rectangles give the bracket $\int_1^{\infty} f \le S \le a_1 + \int_1^{\infty} f$ , and after $n$ terms $\int_{n+1}^{\infty} f \le R_n \le \int_n^{\infty} f$ .
• $p$ -series: $\sum \frac{1}{n^p}$ converges exactly when $p > 1$ ; for $p \le 0$ the integral test does not apply and the divergence test does.

The outlined rectangles, heights

a_1

a_5

, stick out above

y = \frac{1}{x^2}

; the shaded ones, heights

a_2

a_6

, sit under it: the sum is squeezed between two integrals, and equals neither.

The picture rebuilds every inequality of the chapter in ten seconds: a rectangle whose height is the value at the LEFT end sticks out above a decreasing curve, one whose height is the value at the RIGHT end fits under it.

The comparison tests: which direction concludes

• Positive terms only. $a_n \le b_n$ for $n \ge N$ and $\sum b_n$ converges: $\sum a_n$ converges. $a_n \ge b_n$ for $n \ge N$ and $\sum b_n$ diverges: $\sum a_n$ diverges.
• Smaller than a divergent series, or larger than a convergent one: NO conclusion, whatever the inequality looks like.
• Limit comparison: $\frac{a_n}{b_n} \to c$ with $0 < c < \infty$ gives the same fate. Build $b_n$ from the dominant term of the numerator and of the denominator.
• Limit $0$ concludes only with $\sum b_n$ convergent, limit $\infty$ only with $\sum b_n$ divergent.
• A logarithm is dominated by every positive power: $\frac{(\ln n)^q}{n^p}$ behaves like $\frac{1}{n^p}$ unless $p = 1$ , where the integral test decides.

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.

What a comparison can conclude, and what it cannot

Read a line as: when the terms of $\sum a_n$ and of the benchmark $\sum b_n$ are positive and related as in the first column, and the benchmark is as in the second, the series is as in the third. The red lines are the ones students write and markers refuse.

Relation between the terms	Benchmark	Verdict for the series
$a_n \le b_n$	$\sum b_n$ converges	converges
Example: $\frac{1 + \cos n}{n^{3/2}} \le \frac{2}{n^{3/2}}$ and $p = \frac{3}{2} > 1$ : converges.
$a_n \le b_n$	$\sum b_n$ diverges	no conclusion no conclusion
Example: $\frac{1}{n^2} \le \frac{1}{n}$ and $\sum \frac{1}{n^2}$ converges, while $\frac{1}{2n + \sqrt{n}} \le \frac{1}{2n}$ and $\sum \frac{1}{2n + \sqrt{n}}$ diverges. What to do: Bound BELOW instead: $2n + \sqrt{n} \le 3n$ gives $\frac{1}{2n + \sqrt{n}} \ge \frac{1}{3n}$ .
$a_n \ge b_n$	$\sum b_n$ diverges	diverges
Example: $\frac{\ln n}{\sqrt{n}} \ge \frac{1}{\sqrt{n}}$ for $n \ge 3$ and $p = \frac{1}{2}$ : diverges.
$a_n \ge b_n$	$\sum b_n$ converges	no conclusion no conclusion
Example: $\frac{1}{n} \ge \frac{1}{n^2}$ and $\sum \frac{1}{n}$ diverges, while $\frac{1}{n^2 - 5} \ge \frac{1}{n^2}$ and $\sum_{n \ge 3} \frac{1}{n^2 - 5}$ converges. What to do: Give up a constant: $\frac{1}{n^2 - 5} \le \frac{2}{n^2}$ for $n \ge 4$ , or take the limit of the ratio.
$\frac{a_n}{b_n} \to c$ , $0 < c < \infty$	either	same as $\sum b_n$
Example: $\frac{\sin(1/n)}{1/n} \to 1$ and $\sum \frac{1}{n}$ diverges: $\sum \sin\frac{1}{n}$ diverges.
$\frac{a_n}{b_n} \to 0$	$\sum b_n$ converges	converges
Example: $\frac{\ln n / n^2}{1 / n^{3/2}} = \frac{\ln n}{n^{1/2}} \to 0$ and $p = \frac{3}{2}$ : converges.
$\frac{a_n}{b_n} \to 0$	$\sum b_n$ diverges	no conclusion no conclusion
Example: Against $\frac{1}{n}$ both $\frac{\ln n}{n^2}$ and $\frac{1}{n \ln n}$ give the limit $0$ ; the first series converges, the second diverges. What to do: Move the benchmark closer to the term: an exponent strictly between the two, such as $\frac{1}{n^{3/2}}$ for $\frac{\ln n}{n^2}$ .
$\frac{a_n}{b_n} \to \infty$	$\sum b_n$ diverges	diverges
Example: $\frac{\ln n / n}{1 / n} = \ln n \to \infty$ and $\sum \frac{1}{n}$ diverges: $\sum \frac{\ln n}{n}$ diverges.
$\frac{a_n}{b_n} \to \infty$	$\sum b_n$ converges	no conclusion no conclusion
Example: Against $\frac{1}{n^2}$ both $\frac{1}{n}$ and $\frac{\ln n}{n^2}$ give the limit $\infty$ ; the first series diverges, the second converges. What to do: Lower the benchmark exponent just enough to absorb the extra factor, here $\frac{1}{n^{3/2}}$ .

The four blue lines are the only verdicts a comparison can give. Every red line has a repair, and it is always the same move: change the benchmark so that the inequality, or the limit, lands on a blue line.

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. Comparing in the direction that proves nothing

most of the question: the verdict alone earns one mark out of four

What not to write

“ $\frac{1}{2n + \sqrt{n}} < \frac{1}{2n}$ and $\sum \frac{1}{2n}$ diverges, so the series diverges.”

What to write

“For $n \ge 1$ , $\sqrt{n} \le n$ , so $\frac{1}{2n + \sqrt{n}} \ge \frac{1}{3n}$ ; $\sum \frac{1}{3n}$ diverges, so the series diverges by comparison.”

Why: Smaller than a divergent series is compatible with both verdicts: $\frac{1}{n^2} < \frac{1}{n}$ too. Divergence needs a LOWER bound, convergence an UPPER one; a constant factor like $\frac{1}{3}$ never matters.

2. Reading the value of the integral as the sum of the series

1 to 2 marks, and the credibility of the whole answer: the first term alone is 1

What not to write

“By the integral test, $\sum_{n=1}^{\infty} \frac{1}{n^3} = \int_1^{\infty} \frac{dx}{x^3} = \frac{1}{2}$ .”

What to write

“The integral converges, so the series converges by the integral test, and $\frac{1}{2} \le \sum_{n=1}^{\infty} \frac{1}{n^3} \le 1 + \frac{1}{2}$ .”

Why: The series is an area of rectangles that stick out above the curve, the integral the area under it. The test transfers the FATE, never the value; the value is only bracketed, $\int_1^{\infty} f \le S \le a_1 + \int_1^{\infty} f$ .

3. Applying the integral test without checking that f decreases

1 to 2 marks for the missing hypothesis, the whole question when it actually fails

What not to write

“ $f(x) = \frac{\ln x}{x}$ is continuous and positive on $[1, \infty)$ and $\int_1^{\infty} f$ diverges, so the series diverges.”

What to write

“ $f'(x) = \frac{1 - \ln x}{x^2} < 0$ for $x > e$ , so $f$ decreases on $[3, \infty)$ ; $\int_3^{\infty} f$ diverges, so $\sum_{n \ge 3} \frac{\ln n}{n}$ diverges, and so does the whole series.”

The series

\sum f(n)

is a sum of zeros (red dots), while the shaded area under

\frac{\sin^2(\pi x)}{x}

is at least

\frac{1}{2}\sum \frac{1}{k + 1}

: infinite. Continuous and positive is not enough.

Why: Without the decrease, the values at the integers say nothing about the area between them. The figure is the counterexample: $f(x) = \frac{\sin^2(\pi x)}{x}$ is continuous and positive, every $f(n)$ is $0$ , and yet the integral diverges.

4. Calling a series with a varying exponent a p-series

the whole question: the verdict is reversed

What not to write

“ $\sum \frac{1}{n^{1 + 1/n}}$ converges: it is a $p$ -series with $p = 1 + \frac{1}{n} > 1$ .”

What to write

“ $\frac{a_n}{1/n} = n^{-1/n} \to 1$ , and $\sum \frac{1}{n}$ diverges, so the series diverges by limit comparison.”

Why: In the $p$ -series test the exponent is a CONSTANT. An exponent that tends to $1$ puts the series on the border, where only a computation decides: here $n^{1/n} \to 1$ , and with $1 + \frac{1}{\ln n}$ the factor is exactly $n^{1/\ln n} = e$ .

5. Taking a limit comparison that lands on 0 or infinity on the wrong side

the whole question: the verdict is reversed

What not to write

“ $\frac{\ln n / n^2}{1 / n^2} = \ln n \to \infty$ , so $\sum \frac{\ln n}{n^2}$ diverges.”

What to write

“ $\frac{\ln n / n^2}{1 / n^{3/2}} = \frac{\ln n}{\sqrt{n}} \to 0$ and $\sum \frac{1}{n^{3/2}}$ converges, so $\sum \frac{\ln n}{n^2}$ converges.”

Why: Limit $\infty$ concludes only against a DIVERGENT benchmark, limit $0$ only against a CONVERGENT one. When the limit lands on the wrong side, the benchmark was too far from the term: move its exponent between the two, where the logarithm is absorbed.

6. Swapping n and n + 1 in the remainder bounds

1 to 2 marks, and an interval that is empty

What not to write

“For $\sum \frac{1}{n^2}$ , $\int_3^{\infty} \frac{dx}{x^2} \le R_3 \le \int_4^{\infty} \frac{dx}{x^2}$ , so $\frac{1}{3} \le R_3 \le \frac{1}{4}$ .”

What to write

“ $\int_4^{\infty} \frac{dx}{x^2} \le R_3 \le \int_3^{\infty} \frac{dx}{x^2}$ , so $\frac{1}{4} \le R_3 \le \frac{1}{3}$ and $\frac{29}{18} \le S \le \frac{61}{36}$ .”

Why: $R_3 = a_4 + a_5 + \cdots$ : placed on $[3, 4], [4, 5], \dots$ these rectangles use the RIGHT end and fit under the curve from $3$ , which gives the upper bound. An interval whose left end exceeds its right end is the sign the bounds were swapped.

7. Comparing a series whose terms are not positive

the whole question

What not to write

“ $-\frac{1}{n} \le \frac{1}{n^2}$ and $\sum \frac{1}{n^2}$ converges, so $\sum \left(-\frac{1}{n}\right)$ converges.”

What to write

“The comparison test needs $a_n \ge 0$ . Here $\sum \left(-\frac{1}{n}\right) = -\sum \frac{1}{n}$ , which diverges.”

Why: Below a convergent series there is room for anything that runs off to $-\infty$ . Positivity is the first hypothesis of both comparison tests: write it, with its range of $n$ , before any inequality.

8. Confusing a p-series with a geometric series

1 mark per series, and a reversed verdict for the disguised one

What not to write

“ $\sum \frac{1}{2^n}$ and $\sum \frac{1}{n^2}$ are the same kind of series, and so are $\sum 2^{-\ln n}$ and $\sum 2^{-n}$ .”

What to write

“ $\sum \frac{1}{n^2}$ is a $p$ -series ( $n$ in the base, $p = 2$ ); $\sum \frac{1}{2^n}$ is geometric ( $n$ in the exponent, $r = \frac{1}{2}$ ); $2^{-\ln n} = n^{-\ln 2}$ is a $p$ -series with $p = \ln 2 < 1$ , so it diverges.”

Why: Where $n$ sits decides the family: in the base, a $p$ -series; alone in the exponent, a geometric series. Rewrite with $a^{\ln n} = n^{\ln a}$ before choosing, since $\ln 2 < 1 < \ln 3$ separates the two verdicts.

Which method to choose

Which test, by the FORM of the term

Before writing anything, look at the shape of $a_n$ and check that its terms are positive

If $a_n$ is a quotient of powers and roots of $n$ → limit comparison with the ratio of the dominant terms

Example: $\frac{2n^2 + 3n}{\sqrt{n^7 + 1}}$ against $\frac{1}{n^{3/2}}$ , limit $2$
If a bounded oscillating factor, $\sin^2 n$ , $1 + \cos n$ → direct comparison, bounding the factor; the integral test is out, f does not decrease

Example: $\frac{1 + \cos n}{n^{3/2}} \le \frac{2}{n^{3/2}}$
If $a_n = f(n)$ with $f$ easy to integrate by a substitution → integral test, hypotheses first

Example: $\frac{e^{1/n}}{n^2}$ : $\int_1^{\infty} \frac{e^{1/x}}{x^2}\,dx = e - 1$
If $n$ alone in an exponent, $2^n$ , $3^n$ → compare with a geometric series

Example: $\frac{2^n}{3^n + n} < \left(\frac{2}{3}\right)^n$
If a function of $\frac{1}{n}$ : $\sin$ , $1 - \cos$ , $\ln(1 + \cdot)$ , $e^{\cdot} - 1$ → limit comparison with the power given by the MATH 140 limit

Example: $1 - \cos\frac{1}{n}$ against $\frac{1}{n^2}$ , limit $\frac{1}{2}$
If a power of $n$ times a power of $\ln n$ → exponent $p \ne 1$ : limit comparison with a slightly shifted power; $p = 1$ : integral test

Example: $\frac{(\ln n)^5}{n^{6/5}}$ against $\frac{1}{n^{11/10}}$ ; $\frac{1}{n \ln n}$ by the integral
If the question asks how close $s_n$ is to $S$ → remainder estimate, which only the integral test provides

Example: $\frac{1}{101} < R_{100} < \frac{1}{100}$ for $\sum \frac{1}{n^2}$

Two branches can both apply: $\frac{\ln n}{n}$ yields to the integral test and to a comparison with $\frac{1}{n}$ . Choose the one whose hypotheses are quickest to write; the verdict is the same.

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.

An integral test answer that earns full marks

When to use it: Any question that says use the integral test, or any series $\sum f(n)$ whose $f$ has an antiderivative you can find

1 Name the function: $f(x) = \dots$ , with $f(n) = a_n$ .
2 Check the three hypotheses on an explicit interval $[N, \infty)$ : continuous, positive, decreasing, with $f'(x) \le 0$ or the product argument.
3 Write the improper integral as a limit: $\int_N^{\infty} f = \lim_{t \to \infty} \int_N^t f$ , compute, and say whether the limit is finite.
4 Conclude with the name of the test and both objects: the integral converges or diverges, therefore the series does.
5 Never write the value of the integral as the sum; if a value is asked for, bracket it with the remainder estimate.

Concluding sentence

“ $f$ is continuous, positive and decreasing on $[3, \infty)$ and $\int_3^{\infty} \frac{\ln x}{x}\,dx$ diverges, so $\sum \frac{\ln n}{n}$ diverges by the integral test.”

The trap: Stopping at continuous and positive. The decrease is the hypothesis that fails in exam questions, and the one markers look for first.

Marking: Typically 1 mark for the hypotheses, 2 for the integral with its limit, 1 for the conclusion naming the test.

A comparison answer that earns full marks

When to use it: Any series whose term resembles a p-series or a geometric series without being one

1 State that the terms are positive, with the range of $n$ .
2 Choose the benchmark $b_n$ from the dominant terms, and say what it does: $p$ -series with $p = \frac{3}{2} > 1$ , geometric with $|r| < 1$ , harmonic.
3 Direct version: write the inequality with its range, $a_n \le b_n$ for $n \ge N$ , in the direction that concludes. Limit version: compute $\lim \frac{a_n}{b_n}$ , factoring the dominant power out.
4 Conclude with the name of the test.

Concluding sentence

“For $n \ge 4$ , $0 < \frac{1}{n^2 - 5} \le \frac{2}{n^2}$ , and $\sum \frac{2}{n^2}$ converges ( $p = 2 > 1$ ), so $\sum \frac{1}{n^2 - 5}$ converges by the comparison test.”

The trap: Writing the inequality for all $n \ge 1$ when it holds only from $N = 4$ : a false line on the page costs the mark even with the right verdict.

Marking: Typically 1 mark for positivity and the benchmark, 2 for the inequality or the limit, 1 for the conclusion.

Check before you hand in

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

The direction check

Before concluding a comparison, read your inequality aloud: below a convergent series, or above a divergent one? Any other combination means the verdict is not yet proved.

$\frac{1}{2n + \sqrt{n}} < \frac{1}{2n}$ is below a DIVERGENT series: not proved. $\ge \frac{1}{3n}$ is above one: proved.
The sum is larger than its first term

With positive terms, $S > a_1$ . If the number you are about to write as the sum is smaller than $a_1$ , you have written the integral.

$\sum \frac{1}{n^3} = \frac{1}{2}$ is impossible: $a_1 = 1$ .
The exponent after the algebra

Simplify the term to $\frac{1}{n^p}$ before reading $p$ , counting the power of $n$ in the numerator.

$\frac{n}{(n^2 + 1)^p}$ behaves like $\frac{1}{n^{2p - 1}}$ , not $\frac{1}{n^{2p}}$ : it converges for $p > 1$ , not $p > \frac{1}{2}$ .
The bracket must be an interval

After a remainder estimate, check that the lower bound is below the upper bound, and that the partial sum lies below both.

$s_3 = \frac{49}{36} \le \frac{29}{18} \le S \le \frac{61}{36}$ for $\sum \frac{1}{n^2}$ .

The typical problem, taken apart

A series you cannot sum, bracketed exactly

Show that $\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$ converges, and use $s_2$ to bracket its sum. How many terms guarantee an error below $\frac{1}{100}$ ?

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

The shaded area from

2

to infinity is

\arctan\frac{1}{2}

; the boxes inside it, heights

a_3, a_4, \dots

, add up to

R_2

, which is therefore smaller.

Step 1

$f(x) = \frac{1}{1 + x^2}$ is continuous and positive on $[1, \infty)$ , and $f'(x) = -\frac{2x}{(1 + x^2)^2} < 0$ there: the integral test applies from $N = 1$ .

Why

The three hypotheses come first, on a named interval. Here they are quick, which is no reason to skip them: they carry a mark.

Step 2

$\int_n^{\infty} \frac{dx}{1 + x^2} = \lim_{t \to \infty} (\arctan t - \arctan n) = \frac{\pi}{2} - \arctan n = \arctan\frac{1}{n}$ for $n > 0$ . It is finite, so the series CONVERGES by the integral test.

Why

The identity $\frac{\pi}{2} - \arctan n = \arctan\frac{1}{n}$ holds for $n > 0$ : the two angles are complementary in a right triangle with legs $1$ and $n$ . It turns every bound into a clean exact number.

Step 3

$s_2 = \frac{1}{2} + \frac{1}{5} = \frac{7}{10}$ , and $\int_3^{\infty} f \le R_2 \le \int_2^{\infty} f$ gives $\arctan\frac{1}{3} \le R_2 \le \arctan\frac{1}{2}$ .

Why

The upper bound starts at $n = 2$ , the lower at $n + 1 = 3$ : the figure shows why, the boxes of heights $a_3, a_4, \dots$ fit under the curve from $2$ .

Step 4

$S = s_2 + R_2$ , so $\frac{7}{10} + \arctan\frac{1}{3} \le S \le \frac{7}{10} + \arctan\frac{1}{2}$ .

Why

An exact bracket, as the course wants without a calculator. The integral from $1$ , $\frac{\pi}{4}$ , is NOT the sum: it is only a lower bound, and a poor one.

Step 5

$R_n \le \arctan\frac{1}{n} < \frac{1}{n}$ , since $\arctan t < t$ for $t > 0$ . So $n = 100$ terms give $R_{100} < \frac{1}{100}$ .

Why

When the exact bound is awkward to invert, a simpler upper bound for it is enough: it answers what suffices. It does not say that fewer terms fail; that needs the lower bound.

The conclusion, written out

“By the integral test the series converges, and $\frac{7}{10} + \arctan\frac{1}{3} \le S \le \frac{7}{10} + \arctan\frac{1}{2}$ ; $100$ terms guarantee an error below $\frac{1}{100}$ .”

The classic mistake on this problem: Writing $S = \frac{\pi}{4}$ , the value of $\int_1^{\infty} f$ ; or bracketing with $\arctan\frac{1}{2} \le R_2 \le \arctan\frac{1}{3}$ , an empty interval since $\arctan$ increases.

Learn by heart

• Integral test: $f$ continuous, positive, DECREASING on $[N, \infty)$ , $f(n) = a_n$ . Same fate, never the same value.
• $\sum \frac{1}{n^p}$ converges exactly when $p > 1$ , with $p$ a CONSTANT.
• $\int_{n+1}^{\infty} f \le R_n \le \int_n^{\infty} f$ , and $\int_1^{\infty} f \le S \le a_1 + \int_1^{\infty} f$ .
• Comparison concludes below a CONVERGENT series or above a DIVERGENT one. Positive terms, range of $n$ written.
• Limit comparison: $\frac{a_n}{b_n} \to c \in (0, \infty)$ , same fate. $b_n$ from the dominant terms.
• Limit $0$ : only with $\sum b_n$ convergent. Limit $\infty$ : only with $\sum b_n$ divergent.
• $\frac{(\ln n)^q}{n^p}$ follows $p$ ; at $p = 1$ the integral test decides, and it converges only for $q < -1$ .
• As $t \to 0$ : $\sin t \sim t$ , $1 - \cos t \sim \frac{t^2}{2}$ , $\ln(1 + t) \sim t$ , $e^t - 1 \sim t$ .

Frequently asked questions

How do I know when I can use the integral test on a series?

You need a function f that is continuous, positive and decreasing on some interval from N to infinity, with f(n) equal to the n-th term. The decrease only has to start at some N: ln x over x increases up to x = e, so you apply the test from n = 3. If f oscillates, like a function with sin or cos in it, use a comparison instead.

Does the integral test give the sum of the series?

No. It only says whether the series converges. The integral is the area under the curve and the series is a sum of rectangles, so they differ: the series of 1 over n cubed is more than its first term 1, while the integral from 1 to infinity is one half. What the integral does give is a bracket, the sum lying between the integral from 1 and the first term plus that integral.

Why does my comparison not prove anything even though the inequality is true?

Because it points the wrong way. A series smaller than a divergent one can converge or diverge, and so can a series larger than a convergent one. To prove convergence you need your terms below a convergent series; to prove divergence, above a divergent one. Usually you fix it by giving up a constant, for example 1 over 3n instead of 1 over 2n.

How do I choose the series to compare with in the limit comparison test?

Keep only the dominant term of the numerator and of the denominator and simplify: that is your benchmark, usually a p-series. For two n squared plus 3n over the square root of n to the seventh plus one, the benchmark is 1 over n to the three halves. If the limit of the ratio comes out as zero or infinity on the wrong side, move the exponent closer to the term.

Is the series of 1 over n to the power 1 plus 1 over n a convergent p-series?

No, and it diverges. A p-series needs a constant exponent, and here the exponent tends to 1. Compared with the harmonic series, the ratio is n to the power minus 1 over n, which tends to 1, so by the limit comparison test both series diverge together, even though every single exponent is greater than 1.

Practise it

Corrected exercises: The integral test and the comparison tests, 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

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Revision sheet: the integral test and the comparison tests (MATH 141)

The essentials

The integral test: three hypotheses, one verdict, no value

The comparison tests: which direction concludes

The rules in table form

What a comparison can conclude, and what it cannot

The mistakes that cost marks

1. Comparing in the direction that proves nothing

2. Reading the value of the integral as the sum of the series

3. Applying the integral test without checking that f decreases

4. Calling a series with a varying exponent a p-series

5. Taking a limit comparison that lands on 0 or infinity on the wrong side

6. Swapping n and n + 1 in the remainder bounds

7. Comparing a series whose terms are not positive

8. Confusing a p-series with a geometric series

Which method to choose

Which test, by the FORM of the term

How the answer is expected to be written

An integral test answer that earns full marks

A comparison answer that earns full marks

Check before you hand in

The typical problem, taken apart

A series you cannot sum, bracketed exactly

Learn by heart

Frequently asked questions

How do I know when I can use the integral test on a series?

Does the integral test give the sum of the series?

Why does my comparison not prove anything even though the inequality is true?

How do I choose the series to compare with in the limit comparison test?

Is the series of 1 over n to the power 1 plus 1 over n a convergent p-series?

Corrected exercises: The integral test and the comparison tests, MATH 141 at McGill

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Looking for a MATH 141 tutor in Montreal?