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

Revision sheet: sequences, 201-SN3-RE (201-NYB)

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

A sequence is the list of terms, a series is the sequence of their running sums. The two can behave in opposite ways on the same expression: the terms one over n tend to zero, while their running sums grow without limit. It is the most costly confusion of the end of the semester.

Q: How do you show that a recursively defined sequence converges?

In three distinct steps. First prove a bound by induction, then monotonicity, which gives the existence of the limit by the Monotone Convergence Theorem. Only then do you take the limit in the recurrence relation to find its value.

Q: Can you find the limit by solving the fixed-point equation?

Only after proving that the limit exists. The fixed-point equation assumes existence: it is obtained by taking the limit in the recurrence relation. On a sequence that doubles at each step, this shortcut would give zero while the sequence diverges to infinity.

Q: Does a bounded sequence always converge?

No. The sequence that alternates between plus one and minus one is bounded and does not converge. The theorem requires two conditions: monotone and bounded. Monotonicity forces the sequence to go in one direction only, the bound prevents it from running off to infinity, and together they force it to settle.

Q: Can L'Hôpital's rule be applied to a sequence?

Not directly, because a sequence is not differentiable. Study the associated real function, compute its limit at plus infinity, then conclude that the sequence has the same limit. This implication only works in that direction: a sequence can converge while the function has no limit.

This sheet does not define a sequence again: you know what it is. Written for Integral Calculus, 201-SN3-RE (formerly 201-NYB), it deals with what decides the mark on this chapter and the next two, namely the distinction between a sequence and a series, and the complete method for recursive sequences, where the existence of the limit must be proved BEFORE it is computed.

It is written for CEGEP students in the Science program (Sciences de la nature). The corrected exercise set on the same chapter then puts each reflex to the test.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

A sequence converges when its terms get close to a FINITE number, and that number is called its limit. Remembering this definition avoids the most costly confusion of the end of the semester: the convergence of a SEQUENCE says nothing about the convergence of the SERIES it generates.

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

The essentials

Converging, diverging, and what it means

• $\left(u_{n}\right)$ CONVERGES to $L$ if $u_{n}$ gets as close as we like to $L$ for $n$ large enough. $L$ must be FINITE.
• A sequence that tends to $\pm\infty$ DIVERGES, even if its behaviour is perfectly regular.
• A sequence that oscillates without settling, such as $(-1)^{n}$ , diverges too.
• A convergent sequence is necessarily BOUNDED; the converse is false.
• If $u_{n}=f(n)$ and if $f$ has a limit $L$ at $+\infty$ , then $u_{n}\to L$ . The CONVERSE is false.

Two sequences, two fates:

\frac{n+1}{n}

closes in on the line

y=1

, while

(-1)^{n}

jumps forever between

-1

and

1

without ever settling.

Going through the function is what allows L'Hôpital's rule on a sequence: a sequence cannot be differentiated, a function can.

The three tools that prove convergence

• MONOTONE AND BOUNDED: an increasing sequence that is bounded above converges; so does a decreasing sequence that is bounded below. BOTH conditions are needed.
• SQUEEZE: if $a_{n}\le u_{n}\le b_{n}$ with $a_{n}$ and $b_{n}$ having the same limit $L$ , then $u_{n}\to L$ .
• ASSOCIATED FUNCTION: compute $\lim_{x\to+\infty}f(x)$ , possibly with L'Hôpital's rule, then conclude about $u_{n}=f(n)$ .
• Geometric $u_{n}=r^{n}$ : converges if $-1<r\le 1$ , diverges otherwise. To $0$ if $\left|r\right|<1$ , to $1$ if $r=1$ .

On the left,

2-\frac{1}{n}

increases below the upper bound

2

: it converges. On the right,

\frac{\cos n}{n}

is squeezed between

-\frac{1}{n}

and

\frac{1}{n}

, which tend to

0

: it converges too.

The Monotone Convergence Theorem gives the EXISTENCE of the limit without computing it: that is exactly what is needed before solving the fixed-point equation.

Sequence and series: two objects, two questions

• The SEQUENCE is the list of terms $u_{1},u_{2},u_{3},\ldots$
• The SERIES is the sequence of partial SUMS $S_{n}=u_{1}+u_{2}+\cdots+u_{n}$ .
• $u_{n}\to 0$ is NECESSARY for the series to converge, but not at all sufficient.
• Counterexample to remember: $u_{n}=\frac{1}{n}$ tends to $0$ and yet $\sum\frac{1}{n}$ diverges.

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 reference sequences: converge or diverge

Four families cover almost all exam questions. Each one comes with a computed term, because a term that is still $117$ at index $50$ does not converge to $0$ .

Sequence $u_{n}$	Condition	Behaviour
$r^{n}$	$\left\|r\right\|<1$	CONVERGES to $0$
Example: $0.9^{50}\approx 0.0052$ , and $0.9^{200}\approx 7.1\times 10^{-10}$ : the decrease is slow at first, then abrupt.
$r^{n}$	$r=1$	CONVERGES to $1$
Example: The constant sequence $u_{n}=1$ converges to $1$ : a sequence does not need to move to have a limit.
$r^{n}$	$r>1$	DIVERGES to $+\infty$
Example: $1.1^{50}\approx 117.4$ : a rate of $10$ percent per period multiplies by more than a hundred in fifty periods.
$r^{n}$	$r\leq-1$	DIVERGES by oscillating
Example: $(-2)^{10}=1024$ and $(-2)^{11}=-2048$ : the terms change sign and grow, there is no limit, not even an infinite one.
$\frac{1}{n^{a}}$	$a>0$	CONVERGES to $0$
Example: $\frac{1}{\sqrt{100}}=0.1$ : even the slowest of these sequences eventually tends to $0$ .
$n^{a}r^{n}$	$\left\|r\right\|<1$	CONVERGES to $0$
Example: At $n=200$ with $a=2$ and $r=0.9$ , the term is about $2.8\times 10^{-5}$ : the exponential wins over any power.
$\left(1+\frac{1}{n}\right)^{n}$	none	CONVERGES to $\mathrm{e}$
Example: At $n=1000$ , the term is about $2.717$ , to compare with $\mathrm{e}\approx 2.718$ . The convergence is slow, like $\frac{1}{n}$ .

A sequence that tends to $\pm\infty$ DIVERGES, even if its behaviour is perfectly regular. It is the vocabulary confusion that costs the most in this chapter.

Sequence or series: the same expression, two answers

The general term and the sum of its terms do not ask the same question, and that is the thread of the whole chapter. The right-hand column is the one for the SERIES.

Term $u_{n}$	The SEQUENCE $u_{n}$	The SERIES $\sum u_{n}$
$\frac{1}{n}$	converges to $0$	DIVERGES
Example: The counterexample to remember. The sum of the first $100$ terms is about $5.19$ , that of the first million about $14.39$ : it keeps rising, without bound.
$\frac{1}{n^{2}}$	converges to $0$	CONVERGES to $\frac{\pi^{2}}{6}$
Example: That is about $1.645$ . The sum of the first $100$ terms is already $1.635$ : here the accumulation stops.
$\frac{1}{\sqrt{n}}$	converges to $0$	DIVERGES
Example: The sum of the first $10{,}000$ terms is about $198.5$ . The term tends to $0$ too slowly for the sum to stop.
$\left(\frac{1}{2}\right)^{n}$	converges to $0$	CONVERGES to $1$
Example: Summing from $n=1$ : $\frac{1/2}{1-1/2}=1$ . The sum of the first $20$ terms is already $0.99999905$ .
$(-1)^{n}$	DIVERGES by oscillating	DIVERGES
Example: The partial sums alternate between $-1$ and $0$ : they have no limit either.

$u_{n}\to 0$ is NECESSARY for the series to converge, never sufficient: rows 1 and 3 prove it. The contrapositive, on the other hand, is true and serves as a quick test: if $u_{n}$ does not tend to $0$ , the series diverges, and that is written in one 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. Confusing the sequence and the series

the whole question, and the same trap comes back over three chapters

What not to write

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

What to write

“ $\frac{1}{n}\to 0$ is necessary but not sufficient: the harmonic series DIVERGES.”

The terms

\frac{1}{n}

come down toward

0

while their partial sums climb without ever settling: the sequence converges, the series diverges.

Why: The partial sums of the harmonic series grow like $\ln n$ : they increase more and more slowly, but without bound. A term that tends to zero guarantees nothing, because infinitely many of them are being added.

2. Differentiating a sequence

1 mark for rigour, every time

What not to write

“ $u_{n}=\frac{\ln n}{n}$ , I differentiate with respect to $n$ .”

What to write

“A sequence is not differentiable: I study $f(x)=\frac{\ln x}{x}$ over the reals, whose limit is $0$ , and deduce $u_{n}\to 0$ .”

Why: The index $n$ is an integer, and differentiation needs a continuous variable. Going through the associated function is allowed and takes one sentence, which is precisely the point the marking scheme reserves. The converse, however, is false: $\sin(2\pi n)=0$ converges while $\sin(2\pi x)$ has no limit.

3. Concluding convergence from monotonicity alone

the whole question

What not to write

“ $u_{n}=n$ is increasing, so it converges.”

What to write

“Increasing AND BOUNDED ABOVE gives convergence. $u_{n}=n$ is increasing but not bounded above: it diverges to $+\infty$ .”

u_{n}=n

climbs at every index but no horizontal line holds it back: increasing with no upper bound, it diverges to

+\infty

Why: The theorem has two hypotheses, and the one that gets forgotten is the bound. An increasing sequence always does one of two things: either it runs into a ceiling and converges, or it climbs forever. It is the bound, and the bound alone, that decides.

4. Computing the limit of a recursive sequence without proving that it exists

half the question, often three marks

What not to write

“ $u_{n+1}=\sqrt{2+u_{n}}$ , so in the limit $L=\sqrt{2+L}$ , so $L=2$ .”

What to write

“I first show that the sequence is increasing and bounded above by $2$ , SO it converges; the fixed-point equation then gives $L=2$ .”

Why: The fixed-point equation assumes that the limit exists: it is obtained by taking the limit in the recurrence relation. On $u_{n+1}=2u_{n}$ with $u_{0}=1$ , the same method would give $L=2L$ so $L=0$ , while the sequence diverges to infinity.

5. Taking a fixed point as a proof of convergence

the whole question

What not to write

“The equation $L=f(L)$ has a solution, so the sequence converges to it.”

What to write

“A fixed point is a CANDIDATE limit. Convergence is proved separately, by monotonicity and a bound, or by the squeeze theorem.”

b_{n+1}=2b_{n}

has the fixed point

0

, but the staircase starting from

b_{1}=1

moves away from it: a fixed point is only a candidate.

Why: A fixed-point equation can have a solution without the sequence getting close to it: $u_{n+1}=-u_{n}$ with $u_{0}=1$ has the fixed point $0$ and oscillates between $1$ and $-1$ . The existence of the limit and its value are two separate questions, and the order matters.

6. Believing that bounded implies convergent

the whole question

What not to write

“ $u_{n}=(-1)^{n}$ is bounded by $1$ , so it converges.”

What to write

“It is bounded but not monotone: it oscillates between $-1$ and $1$ and has no limit.”

(-1)^{n}

stays between

-1

and

1

without ever settling: bounded is not enough, monotonicity is needed too.

Why: The theorem requires monotone AND bounded, in that logical order: monotonicity forces the sequence to go in one direction only, the bound prevents it from running off to infinity. Without monotonicity, a bounded sequence can jump forever between two values.

7. Using the converse of the step through the function

1 to 2 marks for rigour

What not to write

“ $u_{n}=\sin(2\pi n)$ converges to $0$ , so $\sin(2\pi x)$ tends to $0$ .”

What to write

“The implication only goes one way: the function has no limit, while the sequence, which only sees its integers, equals $0$ everywhere.”

Why: The sequence only looks at the integers, the function looks at all the reals: the function therefore knows MORE. What is true for all the reals is true for the integers, never the other way around. It is also why L'Hôpital's rule on the function concludes about the sequence, and not the reverse.

8. Including $r=-1$ in the convergence condition of a geometric sequence

1 mark, and the whole discussion of a geometric series

What not to write

“ $u_{n}=r^{n}$ converges as soon as $\left|r\right|\le 1$ .”

What to write

“ $r^{n}$ converges for $-1<r\le 1$ : the value $r=-1$ is EXCLUDED, the sequence oscillates there between $-1$ and $1$ .”

Why: The interval of convergence is not symmetric: it is closed on the right at $1$ , where the sequence is constant, and open on the left at $-1$ , where it oscillates. This asymmetry comes back exactly in the study of power series in a later chapter.

Which method to choose

Which method for the limit of a sequence

Look at how the sequence is GIVEN: by an explicit formula or by a recurrence relation. The two lead to completely different methods.

If explicit formula, quotient of polynomials or of powers → factor out the dominant term, or go to the function and apply L'Hôpital's rule

Example: $\frac{3n^{2}-1}{5n^{2}+n}\to\frac{3}{5}$
If explicit formula with a natural bracket → the SQUEEZE theorem, bounding above and below by two sequences with the same limit

Example: $-\frac{1}{n}\le\frac{\sin n}{n}\le\frac{1}{n}$ gives $0$
If explicit formula of the form $r^{n}$ or $n^{a}r^{n}$ → comparison of growth rates: the exponential wins over any power

Example: $\frac{n^{10}}{2^{n}}\to 0$
If sequence defined RECURSIVELY → prove monotonicity and a bound by induction, conclude that the limit exists, THEN solve $L=f(L)$

Example: $u_{n+1}=\sqrt{2+u_{n}}$ gives $L=2$
If the sequence alternates in sign → study $\left|u_{n}\right|$ : if it tends to $0$ , so does the sequence

Example: $\frac{(-1)^{n}}{n}\to 0$ because $\frac{1}{n}\to 0$

A recursive sequence requires THREE distinct steps, each one graded: monotonicity, bound, then fixed point. Skipping one of the first two costs more than getting the third wrong.

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.

A recursive sequence, written up for the grader

When to use it: Any sequence given by $u_{0}$ and a relation $u_{n+1}=f(u_{n})$ .

1 Compute the first three or four terms, to guess the direction of variation and the bound.
2 Prove the BOUND by induction: “Suppose $u_{n}<2$ ; then $u_{n+1}=\sqrt{2+u_{n}}<\sqrt{4}=2$ .”
3 Prove MONOTONICITY, often by studying the sign of $u_{n+1}-u_{n}$ or by using the bound.
4 Conclude that the limit EXISTS by the Monotone Convergence Theorem.
5 Take the limit in the recurrence relation, solve, and reject the impossible solutions.

Concluding sentence

Since the sequence is increasing and bounded above by $2$ , it converges; taking the limit in $u_{n+1}=\sqrt{2+u_{n}}$ , we get $L=\sqrt{2+L}$ , so $L^{2}-L-2=0$ , so $L=2$ , the solution $L=-1$ being rejected because the sequence is positive.

The trap: Rejecting the negative root without saying so. The sentence “ $L=-1$ is impossible because all the terms are positive” is worth half a mark and proves that you know the equation has two solutions.

Marking: Typically 2 marks for the bound proved by induction, 2 marks for monotonicity, 1 mark for the existence conclusion, 1 mark for the fixed point and the rejection.

Check before you hand in

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

Compute the first terms

Three or four terms on the calculator immediately show whether the sequence goes up, goes down or oscillates, and toward which value it is heading.

$u_{0}=1$ gives $1.732$ , then $1.932$ , then $1.983$ : increasing and visibly capped by $2$ ✓.
Check that the limit found is a fixed point

Substitute the limit into the recurrence relation: it must reproduce itself exactly.

$\sqrt{2+2}=2$ ✓. A value that does not reproduce itself signals a mistake in solving the equation.
Check that the limit and the bound are consistent

The limit of an increasing sequence that is bounded above is always less than or equal to the bound, and greater than the first term.

$1\le L\le 2$ , and $L=2$ does lie in this bracket ✓. A limit of $3$ would contradict the proven upper bound.
Identify the question asked: sequence or series

Reread the statement and spot whether it is about the general term or the sum. The two questions have different answers on the same expression.

For $\frac{1}{n}$ : the SEQUENCE converges to $0$ , the SERIES diverges. Answering the wrong question cancels the answer.

The typical problem, taken apart

A recursive sequence, from the three steps to the limit

Let the sequence be defined by $u_{0}=1$ and $u_{n+1}=\sqrt{2+u_{n}}$ for all $n\ge 0$ .

Show that it converges, then find its limit.

The curve

y=\sqrt{2+x}

meets the line

y=x

at the single point

(2,2)

: it is the candidate limit, and the curve lies above the line to the left of this point, which makes the sequence go up.

Step 1

First terms: $u_{0}=1$ , $u_{1}=\sqrt{3}\approx 1.732$ , $u_{2}=\sqrt{3.732}\approx 1.932$ , $u_{3}\approx 1.983$ .

Why

Three values are enough to see the sequence go up and level off toward $2$ . They guide the proof: we are going to prove increasing and bounded above by $2$ , not the opposite.

Step 2

UPPER BOUND by induction. Base case: $u_{0}=1<2$ . Inductive step: if $u_{n}<2$ , then $2+u_{n}<4$ , so $u_{n+1}=\sqrt{2+u_{n}}<\sqrt{4}=2$ . Hence $u_{n}<2$ for every $n$ .

Why

We prove the bound BEFORE monotonicity, because the bound will be used to establish that the sequence increases. The square root being increasing, it preserves the inequality: that is the argument that makes the inductive step work.

Step 3

INCREASING. $u_{n+1}^{2}-u_{n}^{2}=2+u_{n}-u_{n}^{2}=-\left(u_{n}-2\right)\left(u_{n}+1\right)$ . Since $0<u_{n}<2$ , the first factor is negative and the second positive, so the difference is POSITIVE. The terms being positive, $u_{n+1}>u_{n}$ .

Why

We compare the SQUARES to avoid the square root, which is allowed between positive numbers. Factoring the quadratic makes the sign obvious, and the bracket proved in the previous step is exactly what lets us conclude.

Step 4

The sequence is increasing and bounded above by $2$ : by the Monotone Convergence Theorem, it CONVERGES to a finite limit $L$ .

Why

This sentence is the pivot of the write-up. It gives existence without giving the value, and it is what allows the next step. Without it, taking the limit would be a calculation on an object that nothing proves exists.

Step 5

Taking the limit: $L=\sqrt{2+L}$ , so $L^{2}=2+L$ , that is $L^{2}-L-2=0$ and $(L-2)(L+1)=0$ . The candidates are $L=2$ and $L=-1$ .

Why

We square, which can create extraneous solutions: that is precisely why the next step exists. Two roots for a single limit: we must decide.

Step 6

Rejection: all the terms are positive, so $L\ge 0$ and $L=-1$ is impossible. The limit is $L=2$ . Check: $\sqrt{2+2}=2$ ✓, and $1\le 2\le 2$ respects the proven bracket ✓.

Why

The rejection sentence is graded, just as rejecting a root outside the domain was in optimization. It proves that you know where the second solution comes from and why it does not fit.

The conclusion, written out

The sequence is increasing and bounded above by $2$ , so it converges; its limit satisfies $L=\sqrt{2+L}$ , which gives $L=2$ , the root $L=-1$ being rejected because the terms are positive.

The classic mistake on this problem: Writing directly $L=\sqrt{2+L}$ so $L=2$ , without proving convergence. On $u_{n+1}=2u_{n}$ with $u_{0}=1$ , this shortcut would give $L=2L$ , so $L=0$ , while the sequence doubles at each step and diverges to infinity.

Learn by heart

• Converging means tending to a FINITE number. Tending to infinity is diverging.
• Monotone AND bounded gives convergence. Both conditions, not just one.
• A sequence cannot be differentiated: go through the associated function, and say so.
• The implication from function to sequence is true; the converse is false.
• Recursive sequence: monotonicity, bound, EXISTENCE, then fixed point. In that order.
• A fixed point is a candidate limit, never a proof of convergence.
• $r^{n}$ converges for $-1<r\le 1$ : the interval is not symmetric.
• $u_{n}\to 0$ is necessary for $\sum u_{n}$ to converge, never sufficient.

Frequently asked questions

What is the difference between a sequence and a series?

A sequence is the list of terms, a series is the sequence of their running sums. The two can behave in opposite ways on the same expression: the terms one over n tend to zero, while their running sums grow without limit. It is the most costly confusion of the end of the semester.

How do you show that a recursively defined sequence converges?

In three distinct steps. First prove a bound by induction, then monotonicity, which gives the existence of the limit by the Monotone Convergence Theorem. Only then do you take the limit in the recurrence relation to find its value.

Can you find the limit by solving the fixed-point equation?

Only after proving that the limit exists. The fixed-point equation assumes existence: it is obtained by taking the limit in the recurrence relation. On a sequence that doubles at each step, this shortcut would give zero while the sequence diverges to infinity.

Does a bounded sequence always converge?

No. The sequence that alternates between plus one and minus one is bounded and does not converge. The theorem requires two conditions: monotone and bounded. Monotonicity forces the sequence to go in one direction only, the bound prevents it from running off to infinity, and together they force it to settle.

Can L'Hôpital's rule be applied to a sequence?

Not directly, because a sequence is not differentiable. Study the associated real function, compute its limit at plus infinity, then conclude that the sequence has the same limit. This implication only works in that direction: a sequence can converge while the function has no limit.

Practise it

Corrected exercises: Sequences, 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.

16 corrected exercises
160 points
225 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

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Revision sheet: sequences, 201-SN3-RE (201-NYB)

The essentials

Converging, diverging, and what it means

The three tools that prove convergence

Sequence and series: two objects, two questions

The rules in table form

The reference sequences: converge or diverge

Sequence or series: the same expression, two answers

The mistakes that cost marks

1. Confusing the sequence and the series

2. Differentiating a sequence

3. Concluding convergence from monotonicity alone

4. Computing the limit of a recursive sequence without proving that it exists

5. Taking a fixed point as a proof of convergence

6. Believing that bounded implies convergent

7. Using the converse of the step through the function

8. Including r=−1r=-1r=−1 in the convergence condition of a geometric sequence

Which method to choose

Which method for the limit of a sequence

How the answer is expected to be written

A recursive sequence, written up for the grader

Check before you hand in

The typical problem, taken apart

A recursive sequence, from the three steps to the limit

Learn by heart

Frequently asked questions

What is the difference between a sequence and a series?

How do you show that a recursively defined sequence converges?

Can you find the limit by solving the fixed-point equation?

Does a bounded sequence always converge?

Can L'Hôpital's rule be applied to a sequence?

Corrected exercises: Sequences, 201-SN3-RE (201-NYB)

See also

Looking for a 201-SN3-RE (201-NYB) tutor in Montreal?

8. Including $r=-1$ in the convergence condition of a geometric sequence