MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: the ratio and root tests, and choosing a test (MATH 141)

This sheet is not a summary of sections 11.6 and 11.7 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on the ratio and root tests and on the choice of a test in MATH 141 at McGill University, and which precise gesture avoids each loss.

Every limit below is computed by hand, as on the exam, and every value quoted has been checked by an independent computation. The rule that runs through the whole sheet: read the form of the general term first, then compute one limit, then compare it with $1$ , and say what you do when it equals $1$ .

The thread of the chapter

The ratio and root tests compare a series with a GEOMETRIC one and read only the LIMIT $L$ : below $1$ it converges absolutely, above $1$ it diverges because the terms grow, and $L = 1$ says nothing. So the test is chosen by the FORM of the general term before computing, and an $L = 1$ always sends you to another test.

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

Ratio test
Root test
Absolute convergence
Conditional convergence
Inconclusive ratio test
Divergence test
Limit comparison test
Factorials in convergence tests
Strategy for testing series
Geometric tail estimate

The essentials

Two tests, one limit, three verdicts

• Ratio test: $L = \lim_{n\to\infty}\left\lvert\frac{a_{n+1}}{a_n}\right\rvert$ . Root test: $L = \lim_{n\to\infty}\sqrt[n]{\lvert a_n\rvert}$ . Same three conclusions for both.
• $L < 1$ : the series converges ABSOLUTELY, hence converges. No alternating series test is needed.
• $L > 1$ or $L = \infty$ : the series DIVERGES, signed or not, because $\lvert a_n\rvert$ eventually increases and cannot tend to $0$ .
• $L = 1$ : NO information. $\sum\frac{1}{n}$ and $\sum\frac{1}{n^2}$ both give $L = 1$ . Write inconclusive and name another test.
• The tests read the LIMIT only: a few first ratios above $1$ , or every ratio below $1$ , decide nothing.
• Both tests need the limit to EXIST. A ratio that oscillates without settling, as for $\frac{2 + (-1)^n}{3^n}$ , puts the ratio test out of play.

Three sequences of ratios settle on

\frac{1}{2}

1

and

\frac{3}{2}

: only where each one ends counts, not where it starts, and the middle one, on the dashed line, decides nothing.

Behind both tests is a comparison with a geometric series of ratio $r$ , $L < r < 1$ . When $L = 1$ no such $r$ exists, which is why the test is silent there and nowhere else.

Choosing by the form of the general term

• Factorials, products $1 \cdot 3 \cdots (2n-1)$ , constants to the power $n$ , recursions $a_{n+1} = f(n)\,a_n$ : RATIO test, the quotient collapses.
• The whole term raised to the power $n$ ( $b_n^{\,n}$ , $2^{n^2} = (2^n)^n$ , $(n!)^n$ ): ROOT test, the exponent disappears.
• Rational or algebraic in $n$ : $L = 1$ is GUARANTEED, do not compute it; limit comparison with $\frac{1}{n^{q-p}}$ .
• A polynomial factor never changes $L$ : $n^{100}$ contributes $\left(\frac{n+1}{n}\right)^{100} \to 1$ to the ratio and $\left(n^{1/n}\right)^{100} \to 1$ to the root.
• Signs that alternate: test $\sum\lvert a_n\rvert$ first; only if it diverges, try the alternating series test for conditional convergence.

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.

The forms, the limit they give, and the verdict

Most ratio and root questions reduce to one of these forms. The last column is the verdict; the red line is the form on which neither test can give one.

Form of the term	Limit to compute	Result
$\frac{c^n}{n!}$	ratio $\frac{c}{n+1} \to 0$	converges, for every $c$
Example: $\sum\frac{100^n}{n!}$ : ratio $\frac{100}{n+1} \to 0$ , converges although the terms grow until $n = 99$ .
$n^k c^n$ or $\frac{n^k}{c^n}$	ratio $\to c$ or $\frac{1}{c}$	the power $n^k$ is irrelevant
Example: $\sum\frac{n^3}{3^n}$ : ratio $\frac{1}{3}\left(1 + \frac{1}{n}\right)^3 \to \frac{1}{3}$ , converges.
$\frac{(n!)^2}{(2n)!}$	ratio $\frac{(n+1)^2}{(2n+2)(2n+1)}$	$\frac{1}{4}$ : converges
Example: Two new factors on the bottom: $\frac{n+1}{2(2n+1)} \to \frac{1}{4}$ .
$\frac{n!}{n^n}$	ratio $\left(\frac{n}{n+1}\right)^n$	$\frac{1}{e}$ : converges
Example: $\left(\frac{n}{n+1}\right)^n = \frac{1}{(1 + 1/n)^n} \to \frac{1}{e}$ , a $1^\infty$ form that is NOT $1$ .
$b_n^{\,n}$	root $\lvert b_n\rvert$	$\lim\lvert b_n\rvert$ against $1$
Example: $\sum\left(\frac{2n+3}{3n+1}\right)^n$ : root $\frac{2n+3}{3n+1} \to \frac{2}{3}$ , converges.
$\frac{P(n)}{Q(n)}$ , algebraic	ratio and root $\to 1$	no verdict inconclusive
Example: $\frac{\sqrt{n}}{n^2 + 1}$ : ratio $\to 1$ ; limit comparison with $\frac{1}{n^{3/2}}$ gives convergence. Same form, other result: $\sum\frac{1}{n}$ and $\sum\frac{1}{n^2}$ both give $L = 1$ ; the first diverges, the second converges. What to do: Do not compute the ratio. Compare with $\frac{1}{n^{q-p}}$ by a limit comparison, then apply the $p$ -series test.

The red line is the only one where the limit is not a verdict. Every other line ends on a number compared with $1$ , and the comparison is the sentence that scores.

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. Taking a ratio below 1 for every n as a limit below 1

the whole question

What not to write

“ $\frac{a_{n+1}}{a_n} = \frac{n}{n+1} < 1$ for every $n$ , so $\sum\frac{1}{n}$ converges by the ratio test.”

What to write

“ $\frac{n}{n+1} \to 1$ : the ratio test is inconclusive. $\sum\frac{1}{n}$ is the $p$ -series with $p = 1$ , which diverges.”

Why: The test needs a FIXED $r < 1$ above all the ratios from some rank on. Ratios that creep up to $1$ make the terms decrease more and more slowly, not geometrically.

2. Reading L = 1 as a verdict

the whole question

What not to write

“The ratio test gives $L = 1$ for $\sum\frac{1}{n^2}$ , so the series diverges.”

What to write

“ $L = 1$ : the ratio test is inconclusive. By the $p$ -series test with $p = 2 > 1$ , $\sum\frac{1}{n^2}$ converges.”

Both series give

L = 1

, yet the partial sums of

\frac{1}{n^2}

stay under the dashed line at

2

while those of

\frac{1}{n}

keep climbing.

Why: $L = 1$ happens for convergent and for divergent series alike. The only correct line after it is inconclusive, followed by the name of another test.

3. Adding one factor to a factorial instead of two

the whole question, since the verdict flips

What not to write

“ $\frac{(2n+2)!}{(2n)!} = 2n + 2$ , so for $\sum\frac{(n!)^2}{(2n)!}$ the ratio is $\frac{(n+1)^2}{2n+2} \to \infty$ : diverges.”

What to write

“ $(2n+2)! = (2n+2)(2n+1)(2n)!$ , so the ratio is $\frac{(n+1)^2}{(2n+2)(2n+1)} = \frac{n+1}{2(2n+1)} \to \frac{1}{4}$ : converges.”

Why: From $(2n)!$ to $(2n+2)!$ the argument rises by TWO, so two factors appear; from $(kn)!$ to $(kn+k)!$ , $k$ factors. Write the last factors out before cancelling.

4. Reading the form 1 to the infinity as 1

2 to 3 marks, the limit and the verdict

What not to write

“For $\sum\frac{n!}{n^n}$ , the ratio is $\left(\frac{n}{n+1}\right)^n$ , and $\frac{n}{n+1} \to 1$ , so $L = 1$ : inconclusive.”

What to write

“ $\left(\frac{n}{n+1}\right)^n = \frac{1}{(1 + 1/n)^n} \to \frac{1}{e} < 1$ : the series converges.”

Why: A base tending to $1$ raised to an exponent tending to infinity can tend to anything. The limit $\left(1 + \frac{x}{n}\right)^n \to e^x$ is the tool, and it is the one of the sequences chapter.

5. Looking for conditional convergence after L above 1

2 marks, and the time lost on a useless test

What not to write

“For $\sum\frac{(-4)^n}{3^n n^2}$ , $L = \frac{4}{3} > 1$ , so it is not absolutely convergent; the alternating series test will say if it converges.”

What to write

“ $L = \frac{4}{3} > 1$ , so $\lvert a_n\rvert$ eventually increases and $a_n \not\to 0$ : the series diverges.”

\lvert a_n\rvert = \frac{(4/3)^n}{n^2}

falls until

n = 7

, then climbs past

1

for ever: a long dip is not a limit of

0

Why: $L > 1$ is stronger than the divergence of $\sum\lvert a_n\rvert$ : it kills the divergence test's hypothesis $a_n \to 0$ , so the signed series diverges too. Conditional convergence is impossible.

6. Computing the ratio of a rational term

the whole question, and ten minutes

What not to write

“For $\sum\frac{n^2 + 1}{n^4 + 3}$ , after a page of algebra the ratio tends to $1$ , so the question cannot be decided.”

What to write

“Limit comparison with $\frac{1}{n^2}$ : $\frac{(n^2 + 1)n^2}{n^4 + 3} \to 1$ , and $\sum\frac{1}{n^2}$ converges, so the series converges.”

Why: For every rational or algebraic term the ratio and the root tend to $1$ . Recognising the form sends you to the comparison before any computation.

7. Calling a series conditionally convergent because the AST applies

1 to 2 marks, the classification

What not to write

“ $\sum(-1)^n\frac{n^2}{2^n}$ converges by the alternating series test, so it is conditionally convergent.”

What to write

“ $\left\lvert\frac{a_{n+1}}{a_n}\right\rvert = \frac{1}{2}\left(\frac{n+1}{n}\right)^2 \to \frac{1}{2} < 1$ : the series converges ABSOLUTELY.”

Why: Conditional means convergent AND not absolutely convergent. Test the absolute series first: when it converges, the alternating series test is not needed and conditional is false.

8. Taking cos n for an alternating sign

2 marks, the test is not applicable

What not to write

“ $\sum\frac{\cos n}{n^3}$ alternates, and $\frac{1}{n^3}$ decreases to $0$ , so it converges by the alternating series test.”

What to write

“ $\left\lvert\frac{\cos n}{n^3}\right\rvert \le \frac{1}{n^3}$ and $\sum\frac{1}{n^3}$ converges: the series converges absolutely by comparison.”

Why: $\cos(n\pi) = (-1)^n$ alternates; $\cos n$ does not ( $\cos 2$ , $\cos 3$ and $\cos 4$ are all negative). The ratio test is out too: $\frac{\lvert\cos(n+1)\rvert}{\lvert\cos n\rvert}$ has no limit.

9. Using L to compute the sum

the value, and the credibility of the rest of the answer

What not to write

“For $\sum\frac{n^2}{2^n}$ the ratio test gives $L = \frac{1}{2}$ , so the sum is $\frac{1}{1 - 1/2} = 2$ .”

What to write

“ $L = \frac{1}{2} < 1$ : the series converges. The test gives no value; already $s_3 = \frac{1}{2} + 1 + \frac{9}{8} = \frac{21}{8} > 2$ .”

Why: $L$ is the ratio of a geometric series used as a yardstick, not the ratio of the series itself. The value of $\sum\frac{n^2}{2^n}$ is $6$ , far from $2$ .

Which method to choose

Which test, by the FORM of the general term

Before computing anything, read the general term and go down this list; stop at the first line that fits

If the terms visibly do not tend to $0$ → divergence test: diverges, nothing else to do

Example: $\left(\frac{2n+1}{2n}\right)^n \to \sqrt{e} \ne 0$ : diverges
If a geometric, p-series or telescoping series, possibly in disguise → quote its known verdict, and its sum when asked

Example: $\frac{n!}{(n+2)!} = \frac{1}{n+1} - \frac{1}{n+2}$ : sum $\frac{1}{2}$
If factorials, products of $n$ factors, $c^n$ , or a recursion → ratio test; cancel the factorials before the limit

Example: $a_{n+1} = \frac{5n+1}{4n+3}a_n$ : $L = \frac{5}{4}$ , diverges
If the whole term raised to the power $n$ → root test; the exponent disappears

Example: $\frac{n^n}{2^{n^2}} = \left(\frac{n}{2^n}\right)^n$ : $L = 0$ , converges
If rational or algebraic in $n$ , or a bounded factor over a power → limit comparison or direct comparison with a $p$ -series; the ratio would give $1$

Example: $\frac{\arctan n}{n^2} < \frac{\pi}{2n^2}$ : converges
If signs that alternate → test $\sum\lvert a_n\rvert$ first; if it diverges, the alternating series test

Example: $\frac{(-1)^n}{\sqrt[3]{n}}$ : $p = \frac{1}{3}$ , then AST: conditional
If $f(n)$ with $f$ positive, decreasing and easy to integrate → integral test, hypotheses checked

Example: $\frac{1}{n(\ln n)^3}$ : $\int_2^\infty = \frac{1}{2(\ln 2)^2}$ , converges

When a test returns $L = 1$ , you have not failed: you have learned that the term is not geometric, and you move one line down. If the exact ratio is known and stays ABOVE $1$ , the terms increase and the series diverges.

How the answer is expected to be written

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

Writing a ratio or root test

When to use it: Any question that says determine whether the series converges or diverges, or classify it as absolutely convergent, conditionally convergent or divergent

1 Name the form of the term and the test it calls for, in one sentence.
2 Write the ratio $\left\lvert\frac{a_{n+1}}{a_n}\right\rvert$ as a product of two fractions, or the root $\sqrt[n]{\lvert a_n\rvert}$ , in full.
3 Cancel the factorials and powers, with the last factors written out.
4 Take the limit, naming the standard limit used ( $e$ , $n^{1/n} \to 1$ , degrees).
5 Compare $L$ with $1$ and conclude in a sentence: absolutely convergent, divergent, or inconclusive followed by another test.

Concluding sentence

“ $\left\lvert\frac{a_{n+1}}{a_n}\right\rvert = \frac{n+1}{2(2n+1)} \to \frac{1}{4} < 1$ , so by the ratio test $\sum\frac{(n!)^2}{(2n)!}$ converges absolutely.”

The trap: Writing the limit sign in front of the unsimplified ratio and then the value: the simplification is where the marker looks for the method mark.

Marking: Typically 1 mark for naming the test, 1 to 2 for the simplified ratio, 1 for the limit and 1 for the conclusion with the comparison to 1; after an inconclusive result, the second test carries the rest of the marks.

Check before you hand in

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

The polynomial-factor test

Powers of $n$ contribute exactly $1$ to $L$ . Cover them with your thumb: the constants to the power $n$ and the factorials that remain give $L$ .

$\frac{n^{100}}{(1.01)^n}$ : without $n^{100}$ it is $\frac{1}{1.01^n}$ , so $L = \frac{1}{1.01}$ .
The degree test

If no factorial and no constant to the power n appears, the term is algebraic and L = 1. Do not start the computation.

$\frac{n^2 + 1}{n^4 + 3}$ : algebraic, so limit comparison with $\frac{1}{n^2}$ straight away.
The side-of-1 test

When $L = 1$ and the exact ratio is known, check its side of $1$ for every $n$ . Always above $1$ means the terms increase: divergence. Always below $1$ proves nothing.

$\frac{4(n+1)^2}{(2n+2)(2n+1)} = \frac{2n+2}{2n+1} > 1$ : the terms of $\sum\frac{4^n(n!)^2}{(2n)!}$ increase from $2$ , diverges.
The absolute-first test

On a signed series, the word conditionally is only allowed after a proof that $\sum\lvert a_n\rvert$ diverges.

$\sum(-1)^n\frac{3^n}{n\,4^n}$ : $L = \frac{3}{4}$ for $\lvert a_n\rvert$ , so absolutely convergent, and the AST is never needed.

The typical problem, taken apart

A signed term with factorials: read, compute, classify

Classify $\sum_{n=1}^{\infty}(-1)^n\frac{(2n)!}{(n!)^2\,5^n}$ as absolutely convergent, conditionally convergent or divergent.

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

Step 1

Form: alternating signs, factorials and a constant to the power $n$ . So test $\sum\lvert a_n\rvert$ first, with the ratio test.

Why

Reading the form chooses the test and the order. Starting with the alternating series test would mean proving that $\frac{(2n)!}{(n!)^2 5^n}$ decreases, a much harder job, for a weaker conclusion.

Step 2

$\left\lvert\frac{a_{n+1}}{a_n}\right\rvert = \frac{(2n+2)!}{(2n)!} \cdot \frac{(n!)^2}{((n+1)!)^2} \cdot \frac{5^n}{5^{n+1}} = \frac{(2n+2)(2n+1)}{5(n+1)^2}$ .

Why

The ratio is written as a product of three quotients, one per ingredient, and each collapses: two new factors on top, $(n+1)^2$ on the bottom, a $5$ .

Step 3

Simplify: $\frac{2(n+1)(2n+1)}{5(n+1)^2} = \frac{2(2n+1)}{5(n+1)} = \frac{4 + 2/n}{5 + 5/n} \to \frac{4}{5}$ .

Why

The common factor $n + 1$ cancels BEFORE the limit; the limit is then a ratio of leading coefficients, readable at a glance.

Step 4

$L = \frac{4}{5} < 1$ : $\sum\lvert a_n\rvert$ converges by the ratio test, so the series converges ABSOLUTELY.

Why

Comparing $L$ with $1$ and naming absolute convergence is the conclusion mark. The alternating series test is never used.

Step 5

Check: $\frac{(2n)!}{(n!)^2}$ is a binomial coefficient of row $2n$ , at most the row sum $2^{2n} = 4^n$ , so $\lvert a_n\rvert \le \left(\frac{4}{5}\right)^n$ : a geometric bound of ratio $\frac{4}{5}$ , exactly $L$ .

Why

An independent comparison confirms both the verdict and the value of L. It also shows what the ratio test does in general: it finds the geometric series hiding underneath.

The conclusion, written out

“ $\left\lvert\frac{a_{n+1}}{a_n}\right\rvert = \frac{2(2n+1)}{5(n+1)} \to \frac{4}{5} < 1$ , so by the ratio test the series converges absolutely.”

The classic mistake on this problem: Writing $\frac{(2n+2)!}{(2n)!} = 2n + 2$ , which gives $\frac{2}{5(n+1)} \to 0$ : the right verdict for a wrong reason, and with $4^n$ instead of $5^n$ the same slip would turn an inconclusive $L = 1$ into a false convergence.

Learn by heart

• $L < 1$ : converges absolutely. $L > 1$ : diverges, signs or not. $L = 1$ : inconclusive, another test.
• Factorials, products, $c^n$ , recursions: ratio. The whole term to the power $n$ : root. Algebraic in $n$ : comparison.
• $(n+1)! = (n+1)\,n!$ ; $(2n+2)! = (2n+2)(2n+1)(2n)!$ .
• $\left(1 + \frac{1}{n}\right)^n \to e$ , $\left(\frac{n}{n+1}\right)^n \to \frac{1}{e}$ , $\sqrt[n]{n} \to 1$ , $\sqrt[n]{c} \to 1$ .
• The tests read the LIMIT: first terms and ratios below $1$ for every $n$ decide nothing.
• Signed series: $\sum\lvert a_n\rvert$ first. Conditional only if it diverges and the AST applies.
• The tests prove convergence; they never give the sum.

Frequently asked questions

When should I use the ratio test instead of the root test?

Look at the general term. If it contains factorials, products with a growing number of factors, a constant to the power n or a recursion, use the ratio test, because the quotient of two consecutive terms collapses. If the whole term is raised to the power n, use the root test, because the n-th root removes the exponent. When both apply, they give the same limit.

What does it mean when the ratio test gives exactly 1?

It means the test is inconclusive and says nothing about convergence. The harmonic series and the series of one over n squared both give 1, and one diverges while the other converges. Write that the test is inconclusive, then use another test: a comparison with a p-series, the divergence test, or the alternating series test.

If the ratio test gives a limit greater than 1, can the series still converge conditionally?

No. A limit greater than 1 means that from some point on each term is larger in absolute value than the one before, so the terms cannot tend to zero. By the divergence test the series itself diverges, whatever its signs. There is no need to try the alternating series test.

Why does the ratio test always fail on rational functions of n?

Because a polynomial evaluated at n plus one and at n have the same leading term, so their quotient tends to 1, for the numerator and for the denominator. The ratio therefore always tends to 1, which is the inconclusive case. For those series, compare with one over n to a power given by the difference of the degrees.

How do I choose a convergence test on a final exam?

Go through a fixed order. Check that the terms tend to zero. Recognise a geometric, p-series or telescoping series. Use the ratio test for factorials and exponentials, the root test for n-th powers, a comparison for algebraic terms, the absolute series and then the alternating series test for alternating signs, and the integral test for an easy integral.

Practise it

Corrected exercises: Ratio and root tests, and choosing a test, 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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Get in touch for a first session. The choice of a convergence test is the question that closes almost every MATH 141 final, and it is won by reading the form of the term before computing.

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Revision sheet: the ratio and root tests, and choosing a test (MATH 141)

The essentials

Two tests, one limit, three verdicts

Choosing by the form of the general term

The rules in table form

The forms, the limit they give, and the verdict

The mistakes that cost marks

1. Taking a ratio below 1 for every n as a limit below 1

2. Reading L = 1 as a verdict

3. Adding one factor to a factorial instead of two

4. Reading the form 1 to the infinity as 1

5. Looking for conditional convergence after L above 1

6. Computing the ratio of a rational term

7. Calling a series conditionally convergent because the AST applies

8. Taking cos n for an alternating sign

9. Using L to compute the sum

Which method to choose

Which test, by the FORM of the general term

How the answer is expected to be written

Writing a ratio or root test

Check before you hand in

The typical problem, taken apart

A signed term with factorials: read, compute, classify

Learn by heart

Frequently asked questions

When should I use the ratio test instead of the root test?

What does it mean when the ratio test gives exactly 1?

If the ratio test gives a limit greater than 1, can the series still converge conditionally?

Why does the ratio test always fail on rational functions of n?

How do I choose a convergence test on a final exam?

Corrected exercises: Ratio and root tests, and choosing a test, MATH 141 at McGill

See also

Looking for a MATH 141 tutor in Montreal?