MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: sequences (MATH 141)

Q: Can I use L'Hôpital's rule on a sequence?

Not directly, because a sequence is only defined at the integers and has no derivative. Define the function f of a real variable x that matches the sequence at every integer, apply L'Hôpital's rule to f as x goes to infinity, then conclude with the theorem that says: if f of x tends to L, then f of n tends to L. Write all three steps, they are where the method marks are.

Q: Why is 1 to the power infinity not equal to 1?

Because the base and the exponent move at the same time. The base gets closer to 1 while the exponent grows, and the result depends on which effect wins. One plus one over n, to the power n, tends to e; one plus one over n squared, to the power n, tends to 1. Take the natural logarithm, compute its limit as a quotient, then exponentiate.

Q: How do I prove that a recursive sequence converges?

First prove by induction that the sequence is bounded, then that it is monotone, usually by factoring the difference between two consecutive terms. The Monotone Sequence Theorem then guarantees a limit. Only after that, let n go to infinity in the recursion to get an equation for the limit, solve it, and reject any solution that violates the bounds you proved.

Q: Does every bounded sequence converge?

No. The sequence that alternates between minus one and one is bounded and never settles. A bounded sequence converges when it is also monotone, which is the Monotone Sequence Theorem. The converse direction is true: every convergent sequence is bounded, so an unbounded sequence always diverges.

Q: What is the difference between a sequence and a series in MATH 141?

A sequence is a list of numbers, and its limit is where the terms themselves settle. A series is the sum of the terms of a sequence, and its convergence is decided by the sequence of its partial sums. The two questions are different: the terms one over n tend to zero, yet their sum grows without bound, which the next chapter of the course proves.

This sheet is not a summary of section 11.1 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on sequences in MATH 141 at McGill University, and which precise gesture avoids each loss.

Sequences look easy after MATH 140, and that is the danger: most limits are computed correctly and then lose half their marks for a missing justification. Every value below is exact and done by hand, as on the exam, and every trap comes with the sentence that earns the method mark.

The thread of the chapter

The limit of a sequence is decided by its TAIL, and every shortcut needs its permit first: L'Hôpital needs a function $f(x)$ of a real variable, the squeeze needs two bounds with the SAME limit, a form $1^\infty$ needs a logarithm, and the equation $L = f(L)$ of a recursion needs a proof that the limit EXISTS before it is solved.

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

Limit of a sequence
Squeeze theorem
Geometric sequence
Monotone Sequence Theorem
Bounded sequence
Recursive sequence
L'Hôpital's rule
Indeterminate form
Continuity and limits
Growth rates of sequences
Factorial
The number e

The essentials

A sequence lives on the integers: go through f(x) to use calculus

• $a_n \to L$ means that ALL the terms from some rank on are as close to $L$ as we like. The first terms, even a thousand of them, decide nothing.
• Bridge theorem: if $\lim_{x\to\infty} f(x) = L$ and $f(n) = a_n$ for every integer $n$ , then $a_n \to L$ . This is the ONLY licence for L'Hôpital: it is applied to $f$ , never to $a_n$ .
• The converse is false: $\sin(\pi n) = 0 \to 0$ while $\sin(\pi x)$ has no limit. A function that diverges proves nothing about its sequence.
• Continuity: if $a_n \to L$ and $g$ is continuous at $L$ , then $g(a_n) \to g(L)$ . This brings a limit back from $\ln a_n$ to $a_n = e^{\ln a_n}$ .
• Absolute value: $|a_n| \to 0$ if and only if $a_n \to 0$ . For a limit other than $0$ the equivalence is false: $|(-1)^n| \to 1$ .

The dots

a_n = \sin(\pi n)

are all on the axis, so the sequence converges to

0

, while the curve through them oscillates forever: the theorem only goes from

f(x)

a_n

The write-up the marker wants has three lines: define $f(x)$ , compute $\lim_{x\to\infty} f(x)$ , return to $a_n$ by naming the theorem. The first and third lines are where the method marks are.

Existence before value: bounded, monotone, and recursions

• Monotone Sequence Theorem: bounded AND monotone implies convergent. Bounded alone is not enough, $(-1)^n$ ; monotone alone is not enough, $\sqrt n$ .
• The limit of a decreasing sequence bounded below is its GREATEST lower bound, not the bound you found. $\frac{n+1}{2n}$ is bounded below by $0$ and tends to $\frac{1}{2}$ .
• Monotone by the sign of $a_{n+1} - a_n$ , by comparing $\frac{a_{n+1}}{a_n}$ with $1$ (positive terms), or by the sign of $f'(x)$ when $a_n = f(n)$ . Eventually monotone is enough.
• Recursion $a_{n+1} = g(a_n)$ : bounds by induction, then monotonicity, then existence, then $L = g(L)$ with $g$ continuous, then reject the candidates that violate the bounds.
• $L = g(L)$ only lists the possible FINITE limits: $b_{n+1} = 2b_n + 1$ gives $L = -1$ , yet $b_n = 2^n - 1 \to \infty$ .

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.

Standard limits, and the forms that decide nothing

Read a line as: a sequence of the first column, under the condition of the second, has the limit of the third. The red lines are forms, not answers: two sequences of the same form can have different limits.

Sequence	Condition	Limit
$r^n$	$\|r\| < 1$	$0$
Example: $\left(-\frac{3}{4}\right)^n \to 0$ : the sign alternates, the size shrinks.
$r^n$	$r > 1$	$\infty$
Example: $\left(\frac{5}{4}\right)^n \to \infty$ , and $\frac{2^n + 3^{n+1}}{3^n - 4} \to 3$ after dividing by $3^n$ .
$r^n$	$r \le -1$	diverges
Example: $(-1)^n$ jumps between $-1$ and $1$ ; $\left(-\frac{5}{4}\right)^n$ alternates with growing size.
$n^{1/n}$ and $a^{1/n}$	$a > 0$	$1$
Example: $\ln\left(n^{1/n}\right) = \frac{\ln n}{n} \to 0$ , so $n^{1/n} \to e^0 = 1$ .
$\left(1 + \frac{x}{n}\right)^n$	any real $x$	$e^x$
Example: $\left(1 - \frac{3}{n}\right)^n \to e^{-3}$ and $\left(\frac{n}{n+1}\right)^n \to e^{-1}$ .
$\frac{n^p}{b^n}$ and $\frac{b^n}{n!}$	$p > 0$ , $b > 1$	$0$
Example: $\frac{n^{100}}{1.01^n} \to 0$ and $\frac{10^n}{n!} \to 0$ , although both start by growing.
$(1 + a_n)^{b_n}$	$a_n \to 0$ , $b_n \to \infty$	form $1^\infty$ settles nothing
Example: $\left(1 + \frac{1}{n}\right)^n \to e$ , not $1$ . Same form, other result: $\left(1 + \frac{1}{n^2}\right)^n \to 1$ while $\left(1 + \frac{1}{n}\right)^n \to e$ : same form, two limits. What to do: Take the logarithm: $b_n \ln(1 + a_n)$ , then a quotient and L'Hôpital through $f(x)$ , then exponentiate.
$u_n - v_n$	$u_n \to \infty$ , $v_n \to \infty$	form $\infty - \infty$ settles nothing
Example: $\sqrt{n^2 + 3n} - n \to \frac{3}{2}$ , not $0$ . Same form, other result: $\sqrt{n^2 + 3n} - n \to \frac{3}{2}$ while $\sqrt{n^2 + 3} - n \to 0$ . What to do: Multiply by the conjugate, or factor out the dominant term, then divide by it.

The growth ladder behind the sixth line: $\ln n \ll n^p \ll b^n \ll n! \ll n^n$ . In a quotient, the highest rung present decides.

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. Applying L'Hôpital to the sequence itself

1 method mark, on every limit of this kind in the paper

What not to write

“ $\lim \frac{\ln n}{n} = \lim \frac{1/n}{1} = 0$ by L'Hôpital.”

What to write

“Let $f(x) = \frac{\ln x}{x}$ . By L'Hôpital, $\lim_{x\to\infty} f(x) = \lim_{x\to\infty} \frac{1/x}{1} = 0$ . Since $f(n) = a_n$ , $a_n \to 0$ .”

Why: $a_n$ is defined on the integers only and has no derivative. The rule is applied to a function of a real variable, and the bridge theorem brings the result back to the sequence: both sentences must be written.

2. Concluding from the function that the sequence diverges

the whole question

What not to write

“ $\sin(\pi x)$ has no limit at infinity, so $\sin(\pi n)$ diverges.”

What to write

“ $\sin(\pi n) = 0$ for every integer $n$ , so the sequence is constant and converges to $0$ .”

Why: If $f(x) \to L$ then $f(n) \to L$ ; the converse is false, because the integers can land on special points of $f$ . When $f(x)$ has no limit, go back to the sequence itself. The figure of the essentials shows the dots sitting on the zeros of the curve.

3. Answering 1 to a form 1 to the infinity

2 to 3 marks, the whole limit

What not to write

“ $1 + \frac{1}{n} \to 1$ and $1^n = 1$ , so $\left(1 + \frac{1}{n}\right)^n \to 1$ .”

What to write

“ $\ln a_n = n \ln\left(1 + \frac{1}{n}\right) = \frac{\ln(1 + t)}{t}$ with $t = \frac{1}{n} \to 0^+$ , which tends to $1$ by L'Hôpital, so $a_n \to e^1 = e$ .”

Three sequences of the same form

1^\infty

settle on three different lines:

1

e

and

e^2

. The form alone decides nothing.

Why: The base and the exponent move together; freezing the base at its limit while the exponent keeps growing is not a limit law. The logarithm turns the power into a product, and the product into a quotient that L'Hôpital can handle.

4. Answering 0 to infinity minus infinity

2 marks

What not to write

“ $\sqrt{n^2 + 3n} \to \infty$ and $n \to \infty$ , so $\sqrt{n^2 + 3n} - n \to 0$ .”

What to write

“ $\sqrt{n^2 + 3n} - n = \frac{3n}{\sqrt{n^2 + 3n} + n} = \frac{3}{\sqrt{1 + 3/n} + 1} \to \frac{3}{2}$ .”

Why: Two quantities that tend to infinity can differ by any amount. The conjugate turns the difference into a quotient in which the dominant terms can be compared.

5. Squeezing between bounds that do not share their limit

the whole question, and the credibility of the rest of the paper

What not to write

“ $-1 \le (-1)^n \frac{n}{n+1} \le 1$ , so by the squeeze theorem the sequence converges.”

What to write

“The bounds $-1$ and $1$ have different limits, so the squeeze theorem does not apply. The even terms tend to $1$ , the odd terms to $-1$ : the sequence diverges.”

Why: The theorem needs $b_n \to L$ AND $c_n \to L$ with the same $L$ . A bounded sequence can diverge. Choose bounds that shrink to the target: $\frac{\sin^2 n}{n}$ lies between $0$ and $\frac{1}{n}$ , both tending to $0$ .

6. Trusting the first terms instead of the tail

the whole question

What not to write

“ $\frac{10^n}{n!}$ goes $10, 50, 166, 416, \dots$ and keeps growing, so it tends to infinity.”

What to write

“ $\frac{a_{n+1}}{a_n} = \frac{10}{n+1} \le \frac{10}{11}$ for $n \ge 10$ , so $0 < a_n \le a_{10}\left(\frac{10}{11}\right)^{n-10} \to 0$ .”

Why: The sequence climbs until $a_9 = a_{10}$ , larger than $2700$ , then collapses. A limit is a statement about ALL the terms beyond some rank, and the factorial always overtakes the exponential eventually.

7. Taking the lower bound for the limit

2 marks

What not to write

“ $p_n = \frac{n+1}{2n}$ is decreasing and bounded below by $0$ , so $p_n \to 0$ .”

What to write

“ $p_n$ is decreasing and bounded below by $0$ , so it converges to some $L \ge 0$ . Since $p_n = \frac{1}{2} + \frac{1}{2n}$ , $L = \frac{1}{2}$ .”

Why: The Monotone Sequence Theorem gives EXISTENCE, and the limit is the greatest lower bound, which is rarely the one you wrote. The value comes from a separate computation.

8. Solving L = f(L) before proving that the limit exists

the whole question, and a negative limit for a sequence of positive terms

What not to write

“ $b_1 = 1$ , $b_{n+1} = 2b_n + 1$ . Let $L = \lim b_n$ : $L = 2L + 1$ , so $L = -1$ .”

What to write

“By induction $b_n = 2^n - 1$ , so $b_n \to \infty$ : the sequence has no finite limit, and the equation $L = 2L + 1$ does not apply.”

Why: Letting $n \to \infty$ in the recursion ASSUMES that the limit exists. Prove it first (bounded and monotone by induction); the equation then lists the candidates and the bounds choose.

9. Giving the limit of the logarithm as the answer

1 mark, and an answer that contradicts $n^{1/n} \ge 1$

What not to write

“ $\ln\left(n^{1/n}\right) = \frac{\ln n}{n} \to 0$ , so $n^{1/n} \to 0$ .”

What to write

“ $\frac{\ln n}{n} \to 0$ and the exponential is continuous, so $n^{1/n} = e^{(\ln n)/n} \to e^0 = 1$ .”

Why: The logarithm is a detour, not the destination: the last line must come back through $e^{(\cdot)}$ , naming the continuity of the exponential. A limit of $0$ for a sequence whose terms are all $\ge 1$ is impossible, and a two second check would have caught it.

Which method to choose

Which tool, by the FORM of the general term

Look at the shape of a_n before writing anything: the form picks the tool

If a quotient of powers of $n$ , or of exponentials → divide numerator and denominator by the dominant term

Example: $\frac{3n^2 - n + 4}{5n^2 + 2n} \to \frac{3}{5}$ ; $\frac{5^n - 4^{n+2}}{5^{n+1} + 3^n} \to \frac{1}{5}$
If a difference of two terms that both tend to $\infty$ , often with a root → multiply by the conjugate, then divide by the dominant term

Example: $\sqrt{n^2 + 3n} - n \to \frac{3}{2}$
If a factor $(-1)^n$ , $\sin n$ or $\cos n$ → squeeze, or $|a_n| \to 0$ ; never try to compute $\lim \cos n$

Example: $\frac{1}{n} \le \frac{2 + \cos n}{n} \le \frac{3}{n}$ , so the limit is $0$
If $\ln n$ , $e^n$ , a power against an exponential, form $\frac{\infty}{\infty}$ or $\frac{0}{0}$ → define $f(x)$ , apply L'Hôpital to $f$ , come back to $a_n$

Example: $\frac{n^2}{e^n} \to 0$ after two applications
If a variable exponent: $1^\infty$ , $\infty^0$ , $0^0$ , $n^{1/n}$ → take the logarithm, compute its limit, exponentiate by continuity

Example: $(n^2 + 1)^{1/n} \to e^0 = 1$
If a factorial $n!$ → write it as a product of fractions, or compare $\frac{a_{n+1}}{a_n}$ with a number below $1$ , then squeeze

Example: $\frac{n!}{n^n} \le \frac{1}{n}$ ; $\frac{10^n}{n!} \le a_{10}\left(\frac{10}{11}\right)^{n-10}$
If a recursion $a_{n+1} = g(a_n)$ → bounds and monotonicity by induction, then $L = g(L)$ , then reject candidates

Example: $a_{n+1} = \sqrt{6 + a_n}$ , $a_1 = 1$ : increasing, $< 3$ , limit $3$

Two branches can apply to one sequence: $\frac{(-1)^n \ln n}{n}$ takes the absolute value first, then L'Hôpital on $\frac{\ln x}{x}$ . Write which branch you are using on the first line.

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.

Solving a recursive sequence

When to use it: Any question that defines $a_1$ and $a_{n+1} = g(a_n)$ and asks whether the sequence converges, or for its limit

1 Compute two or three terms, and solve $L = g(L)$ on the side ONLY to guess the bound to prove. Do not present it as the answer yet.
2 Prove the bound by induction: base case, then assume it for $a_n$ and deduce it for $a_{n+1}$ .
3 Prove the monotonicity: factor $a_{n+1} - a_n$ (or $a_{n+1}^2 - a_n^2$ ) and read its sign from the bound.
4 Name the Monotone Sequence Theorem: the sequence converges to some $L$ , and $L$ satisfies the same non-strict bounds.
5 Let $n \to \infty$ in the recursion, naming the continuity of $g$ at $L$ ; solve $L = g(L)$ and reject every candidate outside the bounds.

Concluding sentence

“The sequence is increasing and bounded above by $3$ , so by the Monotone Sequence Theorem it converges to some $L$ with $1 \le L \le 3$ . Since $a_{n+1} \to L$ and $\sqrt{6 + x}$ is continuous, $L = \sqrt{6 + L}$ , so $L = 3$ ; $L = -2$ is rejected since $L \ge 1$ .”

The trap: Solving $L = g(L)$ first and stopping there: without existence it proves nothing, as $b_{n+1} = 2b_n + 1$ shows.

Marking: Typically 3 marks for the induction, 2 for the monotonicity, 2 for naming the theorem and the passage to the limit, 3 for the equation and the choice of candidate.

Check before you hand in

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

The limit respects the bounds

A limit of terms that are all positive cannot be negative; of terms all at least 1 cannot be 0. Compare your answer with the bounds you proved or can see.

$n^{1/n} \ge 1$ for every $n$ , so a limit of $0$ is impossible; $b_n = 2^n - 1 > 0$ , so $-1$ is impossible.
One large value of n

Plug in n = 100 or n = 1000 mentally, keeping only the dominant terms. It will not prove the limit, but it catches a factor or a sign.

$\sqrt{10300} - 100 \approx 1.49$ , close to $\frac{3}{2}$ and far from $0$ .
The candidate satisfies the recursion

Plug the limit you found back into $L = g(L)$ : squaring an equation creates false candidates.

$L = -2$ solves $L^2 = 6 + L$ but not $L = \sqrt{6 + L}$ , since $\sqrt 4 = 2 \ne -2$ .
The dominant term matches

In a quotient, the limit is the ratio of the coefficients of the dominant terms. Read them off the statement and compare with your answer.

$\frac{2n + 1}{\sqrt{9n^2 + n}}$ : dominant terms $2n$ and $3n$ , so $\frac{2}{3}$ .

The typical problem, taken apart

A recursion with two candidates, only one of which is the limit

Let $a_1 = 2$ and $a_{n+1} = 4 - \frac{3}{a_n}$ . Prove that $\{a_n\}$ converges and find its limit.

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

The curve

y = 4 - \frac{3}{x}

meets

y = x

(1, 1)

and

(3, 3)

; the staircase starting at

a_1 = 2

climbs to

(3, 3)

and never visits

1

Step 1

$a_2 = 4 - \frac{3}{2} = \frac{5}{2}$ , $a_3 = 4 - \frac{6}{5} = \frac{14}{5}$ , $a_4 = 4 - \frac{15}{14} = \frac{41}{14}$ . On the side: $L = 4 - \frac{3}{L}$ gives $L^2 - 4L + 3 = 0$ , candidates $1$ and $3$ . The terms climb toward $3$ .

Why

The side computation is a GUESS: it tells you that the bound to prove is $3$ . It is not yet the answer, because nothing says the limit exists.

Step 2

Induction, $P(n)$ : $2 \le a_n < 3$ . $P(1)$ : $a_1 = 2$ . If $2 \le a_n < 3$ , then $1 < \frac{3}{a_n} \le \frac{3}{2}$ , so $\frac{5}{2} \le a_{n+1} < 3$ , and $P(n + 1)$ holds.

Why

Dividing reverses the inequalities, and that is where the sign errors are. The upper bound works because $4 - \frac{3}{3} = 3$ : the fixed point is the natural ceiling.

Step 3

$a_{n+1} - a_n = 4 - \frac{3}{a_n} - a_n = \frac{-a_n^2 + 4a_n - 3}{a_n} = \frac{(a_n - 1)(3 - a_n)}{a_n}$ . With $2 \le a_n < 3$ , the three factors are positive: $a_{n+1} > a_n$ .

Why

The factors of the numerator are the two candidates. The bounds of step 2 decide their signs, which is why the bounds are proved BEFORE the monotonicity.

Step 4

The sequence is increasing and bounded above by $3$ : by the Monotone Sequence Theorem it converges to some $L$ with $2 \le L \le 3$ . As $L \ne 0$ , $x \mapsto 4 - \frac{3}{x}$ is continuous at $L$ , and $a_{n+1} \to L$ gives $L = 4 - \frac{3}{L}$ .

Why

Naming the theorem and the continuity is worth marks by itself. Continuity needs $L \ne 0$ , which the bound $L \ge 2$ guarantees.

Step 5

$L^2 - 4L + 3 = (L - 1)(L - 3) = 0$ , so $L = 1$ or $L = 3$ . Since $L \ge 2$ , $L = 1$ is rejected: $a_n \to 3$ .

Why

The candidate $1$ is a genuine fixed point, but it lies outside the bounds. The equation proposes, the bounds decide, and the cobweb shows it: the staircase moves away from $(1, 1)$ .

The conclusion, written out

“The sequence is increasing and bounded above by $3$ , hence convergent by the Monotone Sequence Theorem, with $2 \le L \le 3$ . Passing to the limit, $L = 4 - \frac{3}{L}$ , so $L = 1$ or $L = 3$ , and $L \ge 2$ gives $\lim a_n = 3$ .”

The classic mistake on this problem: Writing $L = 4 - \frac{3}{L}$ on the first line and answering $1$ or $3$ ; or proving the monotonicity without the bounds, then being unable to decide the sign of $(a_n - 1)(3 - a_n)$ .

Learn by heart

• A limit is decided by the tail: no finite number of terms decides it.
• $f(x) \to L$ implies $f(n) \to L$ ; never the converse. L'Hôpital on $f$ , never on $a_n$ .
• Squeeze: two bounds, SAME limit. $|a_n| \to 0 \iff a_n \to 0$ , for the limit $0$ only.
• $r^n$ : $0$ if $|r| < 1$ , $1$ if $r = 1$ , divergent otherwise. $n^{1/n} \to 1$ , $\left(1 + \frac{x}{n}\right)^n \to e^x$ .
• $\ln n \ll n^p \ll b^n \ll n! \ll n^n$ .
• $1^\infty$ , $\infty - \infty$ , $0 \cdot \infty$ , $\infty^0$ are forms, not answers: logarithm, conjugate, quotient.
• Bounded and monotone: convergent. The limit is the best bound, not your bound.
• Recursion: induction, monotonicity, existence, THEN $L = g(L)$ , then reject.

Frequently asked questions

Can I use L'Hôpital's rule on a sequence?

Not directly, because a sequence is only defined at the integers and has no derivative. Define the function f of a real variable x that matches the sequence at every integer, apply L'Hôpital's rule to f as x goes to infinity, then conclude with the theorem that says: if f of x tends to L, then f of n tends to L. Write all three steps, they are where the method marks are.

Why is 1 to the power infinity not equal to 1?

Because the base and the exponent move at the same time. The base gets closer to 1 while the exponent grows, and the result depends on which effect wins. One plus one over n, to the power n, tends to e; one plus one over n squared, to the power n, tends to 1. Take the natural logarithm, compute its limit as a quotient, then exponentiate.

How do I prove that a recursive sequence converges?

First prove by induction that the sequence is bounded, then that it is monotone, usually by factoring the difference between two consecutive terms. The Monotone Sequence Theorem then guarantees a limit. Only after that, let n go to infinity in the recursion to get an equation for the limit, solve it, and reject any solution that violates the bounds you proved.

Does every bounded sequence converge?

No. The sequence that alternates between minus one and one is bounded and never settles. A bounded sequence converges when it is also monotone, which is the Monotone Sequence Theorem. The converse direction is true: every convergent sequence is bounded, so an unbounded sequence always diverges.

What is the difference between a sequence and a series in MATH 141?

A sequence is a list of numbers, and its limit is where the terms themselves settle. A series is the sum of the terms of a sequence, and its convergence is decided by the sequence of its partial sums. The two questions are different: the terms one over n tend to zero, yet their sum grows without bound, which the next chapter of the course proves.

Practise it

Corrected exercises: Sequences, 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

Previous sheet Probability density functions Next sheet Series: geometric, telescoping, divergence test

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Get in touch for a first session. Sequences are where the series chapters are won or lost: every convergence test of the end of the course is a limit of a sequence, justified.

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Revision sheet: sequences (MATH 141)

The essentials

A sequence lives on the integers: go through f(x) to use calculus

Existence before value: bounded, monotone, and recursions

The rules in table form

Standard limits, and the forms that decide nothing

The mistakes that cost marks

1. Applying L'Hôpital to the sequence itself

2. Concluding from the function that the sequence diverges

3. Answering 1 to a form 1 to the infinity

4. Answering 0 to infinity minus infinity

5. Squeezing between bounds that do not share their limit

6. Trusting the first terms instead of the tail

7. Taking the lower bound for the limit

8. Solving L = f(L) before proving that the limit exists

9. Giving the limit of the logarithm as the answer

Which method to choose

Which tool, by the FORM of the general term

How the answer is expected to be written

Solving a recursive sequence

Check before you hand in

The typical problem, taken apart

A recursion with two candidates, only one of which is the limit

Learn by heart

Frequently asked questions

Can I use L'Hôpital's rule on a sequence?

Why is 1 to the power infinity not equal to 1?

How do I prove that a recursive sequence converges?

Does every bounded sequence converge?

What is the difference between a sequence and a series in MATH 141?

Corrected exercises: Sequences, MATH 141 at McGill

See also

Looking for a MATH 141 tutor in Montreal?