University of Namibia
S3511MC: CALCULUS I
Calculus I
Author:
Dr J. Mushanyu
Department:
Computing, Mathematical
& Statistical Science
March 2, 2025
Abstract
The true sign of intelligence is not knowledge but imagination. Knowledge is so
important but imagination is more than that.
—Albert Einstein
Calculus represents a milestone in the history of mathematics and Western thought. It builds
upon the foundational ideas of influential figures such as Archimedes, Fermat, Newton, Leibniz,
and Cauchy, forming a cornerstone of modern science and technology. A basic understanding of
calculus is essential for any well-educated individual, while a deep knowledge is indispensable for
scientific literacy.
Calculus consists of two main branches: differential calculus and integral calculus. Differential
calculus focuses on rates of change, with a central concept being the derivative, which quantifies
various types of change in mathematical and physical contexts. On the other hand, integral calculus
deals with the summation and amalgamation of quantities, encapsulated by the concept of the
integral. Together, these branches provide powerful tools for analyzing and solving problems in
diverse fields, from physics and engineering to economics and biology.
CHAPTER 1
THE BASICS
The true sign of intelligence is not knowledge but imagination
—Albert Einstein.
1.1
Number Systems
Mathematics has its own language, with numbers serving as its alphabet. This language gains
structure through the use of connective symbols, rules of operation, and rigorous logical reasoning.
In calculus, the primary number systems we utilize are the natural numbers, integers, rational
numbers, and real numbers. Let us explore each of these in detail:
1. The natural numbers are the system of positive counting numbers 1, 2, 3 . . . . We denote the
set of all natural numbers by N.
N = {1, 2, 3, 4, 5, 6, 7, 8, . . . }.
2. The integers are the positive and negative whole numbers and zero, . . . , −3, −2, −1, 0, 1, 2, 3, . . . .
We denote the set of all integers by Z.
Z = {. . . , −4, −3, −2, −1, 0, 1, 2, 3, 4, . . . }.
3. The rational numbers are quotients of integers or fractions, such as 32 , − 54 . Any number
p
of the form , with p, q ∈ Z and q ̸= 0, is a rational number. We denote the set of all rational
q
numbers by Q.
p
Q=
p, q ∈ Z, q ̸= 0 .
q
1
4. The real numbers are the set of all decimals, both terminating and non-terminating. We
denote the set of all real numbers by R. A decimal number of the form x = 3.16792 is actually
a rational number, for it represents
x = 3.16792 =
316792
.
100000
A decimal number of the form
m = 4.27519191919 . . . ,
with a group of digits that repeats itself interminably, is also a rational number. To see this,
notice that
100 · m = 427.519191919 . . .
and therefore we may subtract
100m = 427.519191919 . . .
m = 4.27519191919 . . .
Subtracting, we see that
99m = 423.244
or
423244
.
99000
So, as we asserted, m is a rational number or quotient of integers. To indicate recurring
decimals we sometimes place dots over the repeating cycle of digits, e.g., m = 4.2751̇9̇,
19
= 3.16̇.
6
m=
Another kind of decimal number is one which has a non-terminating decimal expansion that
does not keep repeating. An example is π = 3.14159265 . . . . Such a number is irrational, that
is, it cannot be expressed as the quotient of two integers.
In summary : There are three types of real numbers : (i) terminating decimals, (ii) nonterminating decimals that repeat, (iii) non-terminating decimals that do not repeat. Types
(i) and (ii) are rational numbers. Type (iii) are irrational numbers.
The geometric representation of real numbers as points on a line is called the real axis. Between
any two rational numbers on the line there are infinitely many rational numbers. This leads
us to call the set of rational numbers an everywhere dense set.
Real numbers are characterised by three fundamental properties :
(a) algebraic means formalisations of the rules of calculation (addition, subtraction, multiplication, division). Example : 2(3 + 5) = 2 · 3 + 2 · 5 = 6 + 10 = 16.
1
3
(b) order denote inequalities. Example : − < .
4
3
(c) completeness implies that there are “no gaps” on the real line.
Algebraic properties of the reals for addition (a, b, c ∈ R) are :
(A1) a + (b + c) = (a + b) + c. associativity
2
(A2) a + b = b + a. commutativity
(A3) There is a 0 such that a + 0 = a. identity
(A4) There is an x such that a + x = 0.
inverse
Why these rules? They define an algebraic structure (commutative group). Now define analogous algebraic properties for multiplication :
(M1) a(bc) = (ab)c.
(M2) ab = ba.
(M3) There is a 1 such that a · 1 = a.
(M4) There is an x such that ax = 1 for a ̸= 0.
Finally, connect multiplication and addition :
(D) a(b + c) = ab + ac. distributivity
These 9 rules define an algebraic structure called a field.
Order properties of the reals are :
(O1) for any a, b ∈ R, a ≤ b or b ≤ a. totality of ordering I
(O2) if a ≤ b and b ≤ a, then a = b. totality of ordering II
(O3) if a ≤ b and b ≤ c, then a ≤ c. transitivity
(O4) if a ≤ b, then a + c ≤ b + c. order under addition
(O5) if a ≤ b and c ≥ 0, then ac ≤ bc. order under multiplication
Some useful rules for calculations with inequalities are : If a, b, c are real numbers, then :
(a) if a < b and c < 0 ⇒ bc < ac.
(b) if a < b ⇒ −b < −a.
1
(c) if a > 0 ⇒ > 0.
a
(d) if a and b are both positive or negative, then a < b ⇒
1
1
< .
b
a
The completeness property can be understood by the following construction of the real
numbers : Start with the counting numbers 1, 2, 3, . . . .
• N = {1, 2, 3, 4, . . . } natural numbers ⇒ Can we solve a + x = b for x?
• Z = {. . . , −2, −1, 0, 1, 2, . . . } integers ⇒ Can we solve ax = b for x?
• Q = { pq |p, q ∈ Z, q ̸= 0} rational numbers ⇒ Can we solve x2 = 2 for x?
√
• R real numbers, for example : The positive solution to the equation x2 = 2 is 2. This
is an irrational number whose decimal representation is not eventually repeating.
3
⇒ N⊂Z⊂Q⊂R
In summary, the real numbers R are complete in the sense that they correspond to all points on
the real line, i.e., there are no “holes” or “gaps”, whereas the rationals have “holes” (namely
the irrationals).
You Try It : What type of real number is 3.41287548754875 . . . ? Can you express this
number in more compact form?
1.2
Intervals
Definition 1.2.1. A subset of the real line is called an interval if it contains at least two numbers
and all the real numbers between any of its elements.
Examples :
1. x > −2 defines an infinite interval. Geometrically, it corresponds to a ray on the real line.
2. 3 ≤ x ≤ 6 defines a finite interval. Geometrically, it corresponds to a line segment on the real
line.
Finite Intervals. Let a and b be two points such that a < b. By the open interval (a, b) we mean
the set of all points between a and b, that is, the set of all x such that a < x < b. By the closed
interval [a, b] we mean the set of all points between a and b or equal to a or b, that is, the set of all
x such that a ≤ x ≤ b. The points a and b are called the endpoints of the intervals (a, b) and [a, b].
By a half-open interval we mean an open interval (a, b) together with one of its endpoints. There
are two such intervals : [a, b) is the set of all x such that a ≤ x < b and (a, b] is the set of all x such
that a < x ≤ b.
Infinite Intervals. Let a be any number. The set of all points x such that a < x is denoted by
(a, ∞), the set of all points x such that a ≤ x is denoted by [a, ∞). Similarly, (−∞, b) denotes the
set of all points x such that x < b and (−∞, b] denotes the set of all x such that x ≤ b.
4
1.3
Solving Inequalities
Solve inequalities to find intervals of x ∈ R. Set of all solutions is the solution set of the inequality.
Example 1.3.1.
1.
2x − 1 < x + 3
2x < x + 4
x < 4.
2. For what values of x is x + 3(2 − x) ≥ 4 − x?
x + 3(2 − x)
x + 6 − 3x
6 − 2x
2
≥
≥
≥
≥
4 − x when
4−x
4−x
x ⇒ x ≤ 2.
3. For what values of x is (x − 4)(x + 3) < 0?
Case 1: (x − 4) > 0 and (x + 3) < 0, =⇒ x > 4 and x < −3.
Impossible since x cannot be both greater than 4 and less than −3.
Case 2: (x − 4) < 0 and (x + 3) > 0, =⇒ x < 4 and x > −3 =⇒ −3 < x < 4.
Exercise 1.3.1. You Try It: Solve the inequality
1.4
2
3
<
.
x−1
2x + 1
The Absolute Value
It is a quantity that gives the magnitude or size of a real number. The absolute value or modulus
of a real number x, denoted by |x|, is given by
x,
if x ≥ 0
|x| =
−x,
if x < 0.
Geometrically, |x| is the distance between x and 0. For example, | − 6| = 6, |5| = 5, |0| = 0.
5
1.4.1
Properties of the Absolute Value
1. The absolute value of a real number x is non-negative, that is, |x| ≥ 0.
2. The absolute value of a real number x is zero if and only if x = 0, that is, |x| = 0 ⇐⇒ x = 0.
3. In general, if x and y are any two numbers, then
(a) −|x| ≤ x ≤ |x|.
(b) | − x| = |x| and |x − y| = |y − x|.
(c) |x| = |y| implies x = ±y.
(d) |xy| = |x| · |y| and
x
|x|
=
if y ̸= 0.
y
|y|
(e) |x + y| ≤ |x| + |y|. (Triangle inequality)
4. If a is any positive number, then
(a) |x| = a if and only if x = ±a.
(b) |x| < a if and only if −a < x < a.
(c) |x| > a if and only if x > a or x < −a.
(d) |x| ≤ a if and only if −a ≤ x ≤ a.
(e) |x| ≥ a if and only if x ≥ a or x ≤ −a.
Example 1.4.1. Show that for all real numbers x, | − x| = |x|.
Solution 1.4.1.1.
If x ∈ R, then either x > 0, x = 0 or x < 0. If x > 0, then −x < 0. Thus, |−x| = −(−x) = x = |x|,
that is, | − x| = |x|.
If x = 0, then | − x| = | − 0| = |0| = 0, that is, | − x| = |x|.
If x < 0, then −x > 0. Now |x| = −x = | − x| since −x > 0.
Therefore in all cases | − x| = |x|.
Solving an Equation with Absolute Values
Example 1.4.2. Solve the equation |2x − 3| = 7.
Solution 1.4.2.1.
Since |2x − 3| = 7, we have that 2x − 3 = ±7. Hence there are two possibilities,
2x − 3 = 7
2x = 10
x = 5
2x − 3 = −7
2x = −4
x = −2
The solutions of |2x − 3| = 7 are x = 5 and x = −2.
6
Solving Inequalities Involving Absolute values
Example 1.4.3. Solve the inequality 5 −
2
< 1.
x
Solution 1.4.3.1.
We have
5−
2
2
< 1 ⇐⇒ −1 < 5 − < 1
x
x
2
⇐⇒ −6 < − < −4
x
1
⇐⇒ 3 > > 2
x
1
1
⇐⇒
<x< .
3
2
Example 1.4.4.
Solve the inequalities and show the solution set on the real line. (a) |2x − 3| ≤ 1
(b) |2x − 3| ≥ 1.
Solution 1.4.4.1.
(a)
|2x − 3| ≤ 1 ⇐⇒ −1 ≤ 2x − 3 ≤ 1
⇐⇒ 2 ≤ 2x ≤ 4
⇐⇒ 1 ≤ x ≤ 2.
The solution set is the closed interval [1, 2].
(b)
|2x − 3| ≥ 1 ⇐⇒ 2x − 3 ≥ 1 or 2x − 3 ≤ −1
⇐⇒ x ≥ 2 or x ≤ 1.
The solution set is (−∞, 1] ∪ [2, ∞).
Exercise 1.4.1. You Try It: Solve the inequality 4|x| < 7x − 6.
1.5
The Principle of Mathematical Induction
It is an important property of the positive integers (natural numbers) and is used in proving statements involving all positive integers when it is known for, for example, that the statements are valid
for n = 1, 2, 3, . . . but it is suspected or conjectured that they hold for all positive integers.
7
1.5.1
Steps
1. Initial Step: Prove the statement for n = 1 or some other positive integer.
2. Inductive Assumption: Assume the statement true for n = k, where k ∈ Z+ .
3. Inductive Step: From the assumption in 2 prove the statement must be true for n = k + 1.
4. Conclusion: Since the statement is true for n = 1 (from 1) it must (from 3) be true for
n = 1 + 1 = 2 and from this for n = 2 + 1 = 3, and so on, so must be true for all positive
integers.
Example 1.5.1. For any positive integer n,
1 + 2 + ··· + n =
n(n + 1)
.
2
Solution 1.5.1.1.
1. Initial Step.
Prove for n = 1, 1 =
2
1(1 + 1)
= = 1, which is clearly true.
2
2
2. Inductive Assumption.
Assume that the statement holds for n = k, that is,
1 + 2 + ··· + k =
k(k + 1)
.
2
3. Inductive Step.
Prove for n = k + 1. So
k(k + 1)
+ (k + 1) (by inductive hypothesis)
2
k(k + 1) + 2(k + 1)
=
2
k 2 + 3k + 2
=
2
(k + 1)(k + 2)
=
2
1 + 2 + · · · + k + (k + 1) =
so holds for n = k + 1.
4. Conclusion.
Hence by induction, 1 + 2 + · · · + n =
n(n + 1)
is true for any positive integer n.
2
Example 1.5.2. Prove that for any natural number
1 + 3 + 5 + · · · + 2n − 1 = n2 .
8
Solution 1.5.2.1.
1. Initial Step.
Prove for n = 1, 1 = 12 = 1, so it is true.
2. Inductive Assumption.
Assume that the statement holds for n = k, that is,
1 + 3 + 5 + · · · + 2k − 1 = k 2 .
3. Inductive Step.
Prove for n = k + 1. We have
1 + 3 + 5 + · · · + (2k − 1) + 2(k + 1) − 1 = k 2 + 2k + 1
= (k + 1)2 .
(by inductive hypothesis)
So it is true for n = k + 1.
4. Conclusion.
Hence by induction 1 + 3 + 5 + · · · + 2n − 1 = n2 is true for all natural numbers n.
Example 1.5.3. Prove that 3n > 2n for all natural numbers n.
Solution 1.5.3.1.
1. Initial Step.
Prove for n = 1 =⇒ 31 = 3 > 21 = 2, which is true.
2. Inductive Assumption.
Assume the statement holds for n = k, that is, 3k > 2k .
3. Inductive Step.
Prove for n = k + 1.
3k+1 =
>
>
>
3k · 3
2k · 3 by inductive hypothesis
2k · 2 since 3 > 2
2k+1 ,
which is true.
4. Conclusion.
Hence, by induction 3n > 2n for all natural numbers n.
Example 1.5.4. Prove that for any integer n ≥ 1, 22n − 1 is divisible by 3.
9
Solution 1.5.4.1.
1. Initial Step.
Prove for n = 1 =⇒ 22 − 1 = 3 and is divisible by 3, hence its true.
2. Inductive Assumption.
Assume that the statement holds for n = k, that is, for k ≥ 1, 22k − 1 is divisible by 3, i.e.,
22k − 1 = 3l, for some l ∈ Z.
3. Inductive Step.
Prove for n = k + 1.
22(k+1) − 1 =
=
=
=
=
4 · 22k − 1 but 22k = 3l + 1 by the inductive hypothesis
4(3l + 1) − 1
12l + 4 − 1
12l + 3
3(4l + 1),
which is true.
4. Conclusion.
Hence, by induction 22n − 1 is divisible by 3 for all n ≥ 1.
10
CHAPTER 2
SEQUENCES AND LIMITS OF SEQUENCES
Mathematically, a sequence is a function whose domain is the set of positive integers. Although a
sequence is a function, it is common to represent sequences by subscript notation rather than by
the standard function notation.
Definition 2.0.1. A sequence is a set of numbers u1 , u2 , u3 , . . . in a definite order of arrangement
and formed according to a definite rule.
Each number in the sequence is called a term and un is called the nth term. The sequence
u1 , u2 , u3 , . . . is written briefly as {un }.
Example 2.0.1. Listing the terms of a sequence.
1. The terms of the sequence {un } = 2n are u1 = 2, u2 = 4, u3 = 6, u4 = 8, and so on.
2. The terms of the sequence {un } = 3 + (−1)n are u1 = 3 + (−1)1 = 2, u2 = 3 + (−1)2 = 4,
u3 = 3 + (−1)3 = 2, u4 = 3 + (−1)4 = 4, and so on.
1
1
1
1
1
3. The terms of the sequence {un } =
are u1 = , u2 =
= , u3 =
=
3 · 6 · 9 · · · 3n
3
3·6
18
3·6·9
1
1
1
, u4 =
=
, and so on.
162
3 · 6 · 9 · 12
1944
4. The terms of the sequence {un } = 2 + 4 + 8 + · · · + 2n are u1 = 2, u2 = 2 + 4 = 6,
u3 = 2 + 4 + 8 = 14, u4 = 2 + 4 + 8 + 16 = 30, and so on.
The sequence is called finite or infinite according as there are or are not a finite number of terms.
Finding Sequence Pattern
11
Sometimes you may be required to discover a pattern in the sequence and to determine the nth
term of the sequence. For the specified nth term to be correct, it must be able to generate all the
terms of the sequence. The process of determining the nth term from the observed pattern of the
first several terms of a sequence is an example of inductive reasoning.
Example 2.0.2. Find a sequence {an } whose first five terms are given as follows:
1.
2 4 8 16 32
, , , , ,···
1 3 5 7 9
an =
2n
.
2n − 1
2.
1 1 1 1 1
, , , , ,···
2 4 8 16 32
an =
1
.
2n
3.
1 1 1 1 1
, , , , ,···
2 4 8 15 26
an =
4.
−2 8 −26 80 −242
, ,
, ,
,···
1 2 6 24 120
an
6
.
(n + 1)(n2 − n + 6)
n
3 −1
n
= (−1)
.
n!
Recursion Formula or Recurrence Relations
So far we have seen that a sequence {Un } may be defined by giving a formula for {Un } in terms of
n. For example
2n2 − 5n + 4
Un = √
.
n2 + 1
We can also define sequences by giving a relation or formula that connect successive terms of a
sequence and specifying the value or values of the first term or the first and second terms etc. The
formula or relation linking the terms is called a recursion formula or recurrence relation.
Example 2.0.3.
Find the values of the first four terms of the sequence defined by
un+1 =
2
,
un
u1 = 1,
n ∈ N.
Solution 2.0.3.1.
2
2
= =2
u1
1
2
2
= u2+1 =
= =1
u2
2
2
2
= u3+1 =
= = 2.
u3
1
u2 = u1+1 =
u3
u4
12
Exercise 2.0.1. You Try It: Define recursively
a0 = a1 = 1,
and
an = an−1 + 2an−2 , n ≥ 2.
Find a6 recursively.
Fibonacci Sequence
Leonardo of Pisa who was nicknamed Fibonacci, set a modelling exercise in his arithmetic book of
1202 that studied a hypothetical growth of rabbit population. The model considers at the beginning (say at the beginning of the year) of the breeding season, an initial population of a pair of
immature rabbits, male and female, which after one reproductive season produce a pair of male and
female immature rabbits. The model assumes that rabbits reproduce once every month, and they
can reproduce when they are one month old. Lastly, the model considers that no rabbits will die
during the modelling time. Their offspring pairs are assumed to follow exactly the same breeding
traits in the next breeding season and so on. The question is then to determine the number of
pairs of rabbits at the end of the reproductive period. For instance, for simplicity, one can start by
asking the following question, how many rabbits would we have after 1 year? This question can be
illustrated using Figure 2.1 below.
Figure 2.1: The Fibonacci rabbit problem.
This can be modelled by denoting the number of pairs (male and female) of rabbits by Nt . Normalizing the reproductive period to 1, we have at the tth reproductive stage
Nt+1 = Nt + Nt−1 ,
t = 2, 3, . . . .
13
(2.1)
This gives, with N1 = N2 = 1, the Fibonacci sequence, namely
1, 1, 2, 3, 5, 8, 13, . . . .
Each term in the sequence is simply the sum of the previous two.
2.1
Limits of Sequences
1
Let us consider the sequence un = . This sequence has the terms 1, 12 , 13 , 14 , . . . . We see that the
n
terms of the sequence tend to or approach 0.
Definition 2.1.1. A number L is called the limit of an infinite sequence a1 , a2 , a3 , . . . or {an }, if
for any positive number ε, we can find a positive number N depending on ε such that |an − L| < ε
for all integers n > N . We write lim an = L.
n→∞
If {an } is a convergent sequence, it means that the terms an can be made arbitrarily close to L for
n sufficiently large.
Example 2.1.1.
If un = 3 +
1
3n + 1
7 10
=
, the sequence is 4, , , . . . and we can show that
n
n
2 3
lim un = 3.
n→∞
14
If the limit of a sequence exists, the sequence is called convergent, otherwise, it is called divergent.
Example 2.1.2. Prove that lim
1
n→∞ n
= 0.
Proof.
1
1
1
1
1
−0 =
= < ε. But n > . So N = . Taking N
n
n
n
ε
ε
1
1
to be the smallest integer greater than , we have, lim = 0.
n→∞ n
ε
Let ε > 0, we can find N (ε) such that
1
= 0 if p ∈ N.
n→∞ np
Exercise 2.1.1. You Try It: Prove that lim
Example 2.1.3. Use the definition of a limit to prove that lim
2n − 1
n→∞ 3n + 2
2
= .
3
Proof.
Let ε > 0, we can find N (ε) such that
3(2n − 1) − 2(3n + 2)
6n − 3 − 6n − 4
−7
7
2n − 1 2
−
=
=
=
=
<ε
3n + 2 3
3(3n + 2)
3(3n + 2)
3(3n + 2)
3(3n + 2)
7
<ε
3(3n + 2)
7 − 6ε
⇒ n>
.
9ε
⇒
7 − 6ε
7 − 6ε
. So taking N to be the smallest integer greater than
, we have
9ε
9ε
2n − 1
2n − 1 2
2
−
= .
< ε , i.e., lim
n→∞ 3n + 2
3n + 2 3
3
Take N =
2.2
Theorems on Limits
If lim an = A and lim bn = B, then
n→∞
n→∞
1. lim (an + bn ) = lim an + lim bn = A + B.
n→∞
n→∞
n→∞
2. lim (an − bn ) = lim an − lim bn = A − B.
n→∞
n→∞
n→∞
3. lim (an · bn ) = ( lim an )( lim bn ) = AB.
n→∞
n→∞
n→∞
15
lim an
an
A
= n→∞
=
if lim bn = B ̸= 0.
n→∞ bn
lim bn
B n→∞
4. lim
n→∞
5. The limit of a convergent sequence {un } of real numbers is unique.
Proof.
We prove the above Theorem part 5. We must show that if lim un = l1 and lim un = l2 , then
n→∞
n→∞
ε
l1 = l2 . By hypothesis, given any ε > 0, we can find N such that |un − l1 | < when n > N and
2
ε
|un − l2 | < when n > N . Then
2
|l1 − l2 | = |l1 − un + un − l2 | ≤ |l1 − un | + |un − l2 | <
ε ε
+ = ε,
2 2
i.e., |l1 − l2 | is less than any positive ε (however small) and so must be zero, i.e., l1 − l2 = 0 =⇒
l1 = l2 .
Example 2.2.1.
If lim an = A and lim bn = B, prove that lim (an + bn ) = A + B.
n→∞
n→∞
n→∞
Proof.
We must show that for any ε > 0, we can find N > 0, such that |(an + bn ) − (A + B)| < ε for all
n > N . We have
|(an + bn ) − (A + B)| = |(an − A) + (bn − B)| ≤ |an − A| + |bn − B|.
ε
for all n > N1 and
By hypothesis, given ε > 0 we can find N1 and N2 such that |an − A| <
2
ε
|bn − B| < for all n > N2 . Then
2
|(an + bn ) − (A + B)| <
ε ε
+ =ε
2 2
for all n > N where N = max(N1 , N2 ). Hence lim (an + bn ) = A + B.
n→∞
2.3
Limits of Combination of Sequences
We want to be able to evaluate limits, for example, of the form lim
n→∞
16
1
3
2− + 2
n n
5 − 2n2
or lim
.
n→∞ 4 + 3n + 2n2
Example 2.3.1.
lim
n→∞
3
1
2− + 2
n n
1
1
+ 3 lim 2 = 2 − 0 + 0 = 2.
n→∞ n
n→∞ n
= lim 2 − lim
n→∞
Example 2.3.2.
3 − n5
3n2 − 5n
3
3+0
lim
= lim
= .
=
2
6
2
n→∞ 5n + 2n − 6
n→∞ 5 +
5+0+0
5
− n2
n
Example 2.3.3.
√
√
√
√
lim ( n + 1 − n) = lim ( n + 1 − n) ·
n→∞
2.4
n→∞
√
√ n+1+ n
1
√
= lim √
√
√ = 0.
n→∞
n+1+ n
n+1+ n
Sequences Tending to Infinity
n tends to infinity, n → ∞ (n grows or increases beyond any limit ). Infinity is not a number and
the sequences that tend to infinity are not convergent.
We write lim an = ∞, if for each positive number M , we can find a positive number N (depending
n→∞
on M ) such that an > M for all n > N .
Similarly, we write lim an = −∞, if for each positive number M , we can find a positive number N
n→∞
such that an < −M for all n > N .
Example 2.4.1.
Prove that (a) lim 32n−1 = ∞
n→∞
(b) lim (1 − 2n) = −∞.
n→∞
Proof.
(a) If for each positive number M we can find a positive number N such that an > M for all n > N ,
1 ln M
then 32n−1 > M when (2n − 1) ln 3 > ln M , i.e., n >
+ 1 . Taking N to be the smallest
2 ln 3
1 ln M
greater than
+ 1 , then lim 32n−1 = ∞.
n→∞
2 ln 3
(b) If for each positive number M , we can find a positive number N such that an < −M for all
n > N , i.e., 1 − 2n < −M when 2n − 1 > M or n > 12 (M + 1). Taking N to be the smallest integer
greater than 12 (M + 1), we have lim (1 − 2n) = −∞.
n→∞
17
Theorem 2.4.1. Squeeze Theorem.
If lim an = l = lim bn and there exists an N such that an ≤ cn ≤ bn , for all n > N , then
n→∞
n→∞
lim cn = l.
n→∞
cos n
.
n→∞
n
Example 2.4.2. Find lim
Solution 2.4.2.1.
We know that −1 ≤ cos n ≤ 1
cos n
1
1
cos n
1
cos n
1
≤ =⇒ − lim ≤ lim
≤ lim =⇒ 0 ≤ lim
≤0
=⇒ − ≤
n→∞ n
n→∞
n→∞ n
n→∞
n
n
n
n
n
cos n
=⇒ lim
= 0.
n→∞
n
Theorem 2.4.2. Absolute Value Theorem.
For the sequence {an }, if lim |an | = 0, then lim an = 0.
n→∞
n→∞
Proof.
Consider the two sequences {|an |} and {−|an |}. Applying the Squeeze Theorem, we have
−|an | ≤ an ≤ |an |.
Since
lim −|an | = lim |an | = 0,
n→∞
we conclude that
2.5
n→∞
lim an = 0.
n→∞
Bounded and Monotonic Sequences
A sequence that tends to a limit l is said to be convergent and the sequence converges to l. A
sequence may tend to +∞ or −∞, and is said to be divergent and it diverges to +∞ or −∞.
If un ≤ M for n = 1, 2, 3, . . . , where M is a constant, we say that the sequence {un } is bounded
above and M is called an upper bound. The smallest upper bound is called the least upper bound
(l.u.b).
If un ≥ m, the sequence is bounded below and m is called a lower bound. The largest lower bound
is called the greatest lower bound (g.l.b).
If m ≤ un ≤ M , the sequence is called bounded, indicated by |un | ≤ P . (Every convergent sequence
is bounded, but the converse is not necessarily true)
18
If un+1 ≥ un , the sequence is called monotonic increasing and if un+1 > un it is called strictly
increasing. If un+1 ≤ un , the sequence is called monotonic decreasing, while if un+1 < un it is
strictly decreasing.
Example 2.5.1.
1. The sequence 1, 1.1, 1.11, 1.111, . . . is bounded and monotonic increasing.
2. The sequence 1, −1, 1, −1, 1, . . . is bounded but not monotonic increasing or decreasing.
Definition 2.5.1. A null sequence is a sequence that converges to 0, e.g., un =
1
, n ≥ 11.
n − 10
If {un } does not tend to a limit or +∞ or −∞, we say that {un } oscillates (or is an oscillating
sequence). It can oscillate finitely (bounded) or infinitely (unbounded).
Example 2.5.2.
un = (−1)n ,
un = (−1)n n.
Theorem 2.5.1. If a sequence {un } is bounded and monotonic, then it converges.
2.6
The Cauchy Convergence Criterion
In the previous section, we have observed that if a sequence is bounded and monotonic, then it
converges. In this section, we shall observe that bounded monotonic sequences are not the only
convergent sequences. There are sequences that are not monotonic that are convergent. An example
of such a sequence is
(−1)n
.
un = 3 +
n
1
This sequence converges to 3 but it is not monotonic. We see that if n is even, then un = 3 + so
n
1
1
1
1
that un+1 = 3 +
. Since >
, we have that un > un+1 . But if n is odd, then un = 3 −
n+1
n
n+1
n
1
1
1
. Since − < −
, we have that un < un+1 . Thus, the sequence is not
so that un+1 = 3 −
n+1
n
n+1
monotonic yet it converges to 3.
At this point, we now establish a condition that characterises all convergent sequences of real
numbers. This condition is embodied in the Cauchy Convergence Criterion.
Theorem 2.6.1. Cauchy Convergence Criterion
A sequence {an } converges if and only if given ε > 0, there exists a positive integer N such that,
|am − an | < ε for all m, n > N .
19
Remark
Note that this condition entails that the terms of the sequence {an } tend to or approach each
other as n becomes large. Thus, converging sequences are precisely those sequences whose terms
tend to or approach one another.
Example 2.6.1. Let sn = 1 +
1
1 1
+ + · · · + . Show that the sequence {sn } does not converge.
2 3
n
Solution 2.6.1.1.
1 1
1
1
1
1
1 1
1
|s2n − sn | = 1 + + + · · · + +
+
+ ··· +
− 1 + + + ··· +
2 3
n n+1 n+2
n+n
2 3
n
=
1
1
1
1
1
1
1
1
1
+
+ ··· +
=
+
+ ··· +
>
+
+ ··· +
n+1 n+2
n+n
n+1 n+2
n+n
n+n n+n
n+n
=
n
n
1
=
= .
n+n
2n
2
1
1
1
. Hence, if we pick ε < , say ε = , we cannot find a positive
2
2
4
1
integer N such that for all n > N , |s2n − sn | < . Note that here, m = 2n. We see then that the
4
sequence {sn } cannot converge.
Thus, for any n,
|s2n − sn | ≥
20
CHAPTER 3
THEORY OF INFINITE SERIES
One important application of infinite sequences is in representing infinite summations. If {an } is
an infinite sequence, then
∞
X
an = a1 + a2 + a3 + · · ·
n=1
is called an infinite series (or simply a series). The numbers a1 , a2 , a3 , . . . are called the terms of
the series. To find the sum of an infinite series, consider the following sequence of partial sums.
S1
S2
S3
..
.
= a1
= a1 + a2
= a1 + a2 + a3
.
..
..
= ..
.
.
Sn = a1 + a2 + a3 + · · · + an .
If this sequence of partial sums converges, then the series is said to converge and has the sum
indicated in the following definition.
3.1
Definition of Convergent and Divergent Series
For the infinite series
X
an , the nth partial sum is given by
Sn = a1 + a2 + a3 + · · · + an .
X
If the sequence of partial sums {Sn } converges to S, then the series
an converges. The limit S
is called the sum of the series. If {Sn } diverges, then the series diverges.
21
Example 3.1.1. The series
∞
X
1
1
1 1 1
+
+
+
+ · · · has the following partial sums.
=
n
2
2
4
8
16
n=1
1
2
1 1
3
+ =
2 4
4
1 1 1
7
+ + =
2 4 8
8
..
..
.. ..
.
.
. .
1
2n − 1
1 1 1
+ + + ··· + n =
.
2 4 8
2
2n
S1 =
S2 =
s3 =
..
. =
sn =
2n − 1
= 1, it follows that the series converges and its sum is 1.
n→∞
2n
Because lim
Example 3.1.2. The nth partial sum of the series
∞ X
1
1
1
1 1
1 1
−
= 1−
+
−
+
−
+ ···
n n+1
2
2 3
3 4
n=1
is given by Sn = 1 −
1
. Because the limit of Sn is 1, the series converges and its sum is 1.
n+1
Example 3.1.3. The series
∞
X
1 = 1 + 1 + 1 + · · · diverges, because Sn = n and the sequence of
n=1
partial sums diverges.
The series in Example (3.1.2) is a telescoping series. That is, it is of the form
(b1 − b2 ) + (b2 − b3 ) + (b3 − b4 ) + (b4 − b5 ) + · · ·
note that b2 is canceled by the second term, b3 is canceled by the third term and so on. Because the
nth partial sum of the series is Sn = b1 − bn+1 , it follows that a telescoping series will only converge
if and only if bn approaches a finite number as n → ∞. Moreover, if the series converges, then its
sum is
S = b1 − lim bn+1 .
n→∞
Example 3.1.4.
Find the sum of the series
∞
X
n=1
2
4n2 − 1
.
Solution 3.1.4.1.
Using partial fractions, we can write
an =
2
4n2 − 1
=
2
1
1
=
−
.
(2n − 1)(2n + 1)
2n − 1 2n + 1
22
From the telescoping form, we can see that the nth partial sum is
1 1
1
1
1 1
1
−
+
−
+ ··· +
−
=1−
.
Sn =
1 3
3 5
2n − 1 2n + 1
2n + 1
Thus, the series converges and its sum is 1. That is,
∞
X
2
1
= lim Sn = lim 1 −
= 1.
n→∞
4n2 − 1 n→∞
2n + 1
n=1
3.2
Geometric Series
A geometric series with ratio r is given by
∞
X
arn = a + ar + ar2 + · · · + arn + · · · , a ̸= 0.
n=0
Theorem 3.2.1. A geometric series with ratio r diverges if |r| ≥ 1. If 0 < |r| < 1, then the series
∞
X
a
, 0 < |r| < 1.
converges to the sum
arn =
1
−
r
n=0
Example 3.2.1. The geometric series
n
2
∞
∞
X
X
3
1
1
1
3
=
= 3(1) + 3
+3
+ ···
n
2
2
2
2
n=0
n=0
has a ratio of r = 21 with a = 3. Because 0 < |r| < 1, the series converges and its sum is
a
3
= 6.
S=
=
1−r
1 − 12
3.2.1
If
X
Properties of Infinite Series
an = A and
X
bn = B and c is a real number, then the following series converge to the
X
X
X
X
indicated sums. (i)
can = cA (ii)
(an ± bn ) =
an ±
bn = A ± B.
n-th Term Test for Divergence
Limit of n−th Term of a Convergent Series
If the series
X
an converges, then the sequence {an } converges to 0.
If the sequence {an } does not converge to 0, then the series
23
X
an diverges.
3.3
Test for Convergence or Divergence of Series
In this and the following section, we will study several convergence tests that apply to series with
positive terms.
3.3.1
p− Series and Harmonic Series
A series of the form
∞
X
1
n=1
n
=
p
1
1
1
+
+
+ ···
1p 2p 3p
is a p−series, where p is a positive constant. For p = 1, the series
∞
X
1
n=1
n
=1+
1 1
+ + ···
2 3
is the harmonic series.
Theorem 3.3.1. The p−series
∞
X
1
n=1
np
=
1
1
1
+ p + p + ···
p
1
2
3
(i) converges if p > 1 and (ii) diverges if 0 < p ≤ 1.
Example 3.3.1. From the Theorem it follows that the harmonic series
∞
X
1
n=1
n
=1+
1 1
+ + ···
2 3
diverges.
3.4
Comparisons of Series
3.4.1
Direct Comparison Test
This is a test for positive-term series. It allows you to compare a series having complicated terms
with a simpler series whose convergence or divergence is known.
24
Direct Comparison Test
Theorem 3.4.1. Let 0 ≤ an ≤ bn for all n.
1. If
∞
X
bn converges, then
n=1
2. If
∞
X
n=1
∞
X
an converges.
n=1
an diverges, then
∞
X
bn diverges.
n=1
Example 3.4.1. Determine the convergence or divergence of
Solution: This series resembles
∞
X
1
n=1
3n
∞
X
1
.
n
2
+
3
n=1
(Convergent geometric series). Term-by-term comparison
yields
1
1
<
= bn , n ≥ 1.
2 + 3n
3n
Thus, by the Direct Comparison Test, the series converges.
an =
Example 3.4.2. Determine the convergence or divergence of
Solution: The series resembles
∞
X
1
n=1 n
1
2
∞
X
1
√ .
2
+
n
n=1
(Divergent p−series). Term-by-term comparison yields
1
1
√ ≤√ ,
2+ n
n
n≥1
which does not meet the requirements for divergence. Still expecting the series to diverge, we can
∞
X
1
compare the given series with
(Divergent Harmonic series). In this case, term-by-term comn
n=1
parison yields
1
1
√ = bn , n ≥ 4
an = ≤
n
2+ n
and, by the Direct Comparison Test, the given series diverges.
3.4.2
Limit Comparison Test or Quotient Test
Often a given series closely resembles a p−series or a geometric series, yet we cannot establish the
term-by-term comparison necessary to apply the Direct Comparison Test. We can apply a second
comparison test, called the Limit Comparison Test.
25
Limit Comparison Test
Suppose that an > 0 and bn > 0 and lim
n→∞
an
bn
= L.
X
X
1. If L is finite and positive, then the two series
an and
bn , either both converge or both
diverge.
X
X
2. If L = 0 and
bn converges, then
an converges.
3. If L = ∞ and
X
bn diverges, then
X
an diverges.
Example 3.4.3. Show that the following harmonic series diverges.
∞
X
1
,
an
+
b
n=1
∞
X
1
a > 0, b > 0.
1
an+b
we have lim 1
n→∞
n
n
1
=
> 0. Because this
n→∞ an + b
n
a
n=1
limit is finite and positive, we can conclude from the Limit Comparison Test that the given series
diverges.
Solution: By comparison with
= lim
The limit Comparison Test works well for comparing a messy algebraic series with a p−series. In
choosing an appropriate p−series, we must choose one with an nth term of the same magnitude as
the nth term of the given series.
Given series
∞
X
1
n=1
∞
X
3n2 − 4n + 5
1
3n − 2
n=1
∞
2
X
n − 10
4n5 + n3
n=1
3.5
√
Comparison series
∞
X
1
n2
n=1
∞
X
1
√
n
n=1
∞
∞
2
X
X
n
1
=
5
n
n3
n=1
n=1
Conclusion
Both series converge.
Both series diverge.
Both series converge.
Alternating Series
So far, most series we have dealt with have had positive terms. In this section, we will study series
that contain both positive and negative terms. The simplest such series is an alternating series,
26
whose terms alternate in sign. For example, the geometric series
n X
∞ ∞
X
1
1
1
1 1 1
−
− ···
=
(−1)n n = 1 − + − +
2
2
2 4 8 16
n=0
n=0
is an alternating geometric series with r = − 12 . Alternating series occur in two ways, either the odd
terms are negative or the even terms are negative.
Alternating Series Test
Let an > 0. The alternating series
∞
X
n
(−1) an and
n=1
∞
X
(−1)n+1 an converge, if the following two
n=1
conditions are met.
1. an+1 ≤ an for all n.
2. lim an = 0.
n→∞
Example 3.5.1. Determine the convergence or divergence of
∞
X
1
(−1)n+1 .
n
n=1
1
1
1
≤
for all n and the limit (as n → ∞) of
is 0, we can apply the
n+1
n
n
Alternating Series Test to conclude that the series converges. (This series is called the alternating
harmonic series)
Solution: Because
Example 3.5.2. Determine the convergence or divergence of
∞
X
n
.
n−1
(−2)
n=1
Solution: To apply the Alternating Series Test, note that, for n ≥ 1,
n
1
≤
2
n+1
n−1
2
n
≤
n
2
n+1
n−1
(n + 1)2
≤ n2n
n+1
n
≤ n−1 .
n
2
2
Hence, an+1 =
n+1
n
≤ n−1 = an for all n. Furthermore, by L’Hospital’s rule,
n
2
2
lim
x
x→∞ 2x−1
= lim
1
x→∞ 2x−1 (ln 2)
= 0 =⇒ lim
n
n→∞ 2n−1
Therefore, by the Alternating Series Test, the given series converges.
27
= 0.
Cases for which the Alternating Series Test Fails
Example 3.5.3. The alternating series
∞
X
(−1)n+1 (n + 1)
n=1
n
=
2 3 4 5 6
− + − + − ···
1 2 3 4 5
passes the first condition in the alternating series test because an+1 ≤ an for all n. We cannot apply
the Alternating Series Test, because the series does not pass the second condition.
The alternating series
2 1 2 1 2 1 1 1
− + − + − + − + ···
1 1 2 2 3 3 2 4
passes the second condition because an approaches 0 as n → ∞. We cannot apply the Alternating
Series Test, however, because the series does not pass the first condition.
3.6
Absolute and Conditional Convergence
Occasionally, a series may have both positive and negative terms and not be an alternating series,
for example, the series
∞
X
sin 1 sin 2 sin 3
sin n
=
+
+
+ ···
2
n
1
4
9
n=1
has both positive and negative terms, yet it is not an alternating series. One way to obtain some
information about the convergence of this series is to investigate the convergence of the series
∞
X
sin n
. By direct comparison, we have | sin n| ≤ 1, for all n, so
2
n
n=1
sin n
1
≤ 2 , n ≥ 1.
2
n
n
∞
X
sin n
Thus, by the Direct Comparison Test, the series
converges. But the question still is
2
n
n=1
∞
X
sin n
“Does the original series
converge?”
2
n
n=1
X
X
Theorem 3.6.1 (Absolute Convergence). If the series
|an | converges, then the series
an
also converges.
The converse of the Theorem is not true. For example, the alternating harmonic series
∞
X
(−1)n+1
1 1 1
= 1 − + − + ···
n
2 3 4
n=1
converges by the Alternating Series Test. Yet the harmonic series diverges. This type of convergence
is called conditional.
28
Definition of Absolute and Conditional Convergence
1.
X
an is absolutely convergent if
X
2.
X
an is conditionally convergent if
|an | converges.
X
an converges but
X
|an | diverges.
3. An absolutely convergent series converges.
Example 3.6.1. Determine whether the following series are convergent or divergent. Classify any
convergent series as absolutely or conditionally convergent.
n(n+1)
∞
X
1
1
(−1) 2
1 1
−
+
+
− ···.
(a)
=
−
n
3
3
9
27
81
n=1
Solution: This in not an alternating series. However, because
n(n+1)
∞
X
(−1) 2
n=1
3n
=
∞
X
1
n=1
3n
is a convergent geometric series, so the given series is absolutely convergent, hence convergent.
(b)
∞
X
1
1
1
1
(−1)n
=−
+
−
+
− ···.
ln(n
+
1)
ln
2
ln
3
ln
4
ln
5
n=1
Solution: In this case, the alternating series test indicates that the given series converges. However,
the series
∞
X
1
(−1)n
1
1
=
+
+
+ ···
ln(n + 1)
ln 2 ln 3 ln 4
n=1
diverges by direct comparison with terms of the harmonic series. Therefore, the given series is
conditionally convergent.
∞
X
1
(−1)n
1
1
1
√
(c)
= −√ + √ − √ + √ − · · · .
n
1
2
3
4
n=1
Solution: The given series converges by the Alternating Series Test. Moreover, because the
p−series
∞
X
1
1
1
1
(−1)n
√
= √ + √ + √ + √ + ···
n
1
2
3
4
n=1
diverges, the given series is conditionally convergent.
29
3.7
The Ratio and Root Tests
3.7.1
The Ratio Test
This is a test for absolute convergence.
Ratio Test
an+1
< 1.
n→∞
an
1.
X
an converges absolutely if lim
2.
X
an diverges if lim
an+1
an+1
> 1 or lim
= ∞.
n→∞
n→∞
an
an
3. The Ratio Test is inconclusive if lim
n→∞
an+1
= 1.
an
Although the Ratio Test is not a cure for all ills related to tests for convergence, it is particularly
useful for series that converge rapidly. Series involving factorials or exponentials are frequently of
this type.
Example 3.7.1. Determine the convergence or divergence of
∞
X
2n
n=0
Solution: Because an =
n!
.
2n
, we can write the following
n!
n+1
an+1
2
2n
lim
= lim
÷
n→∞
n→∞ (n + 1)!
an
n!
n+1
2
n!
= lim
·
n→∞ (n + 1)! 2n
2
= lim
n→∞ n + 1
= 0.
Therefore, the series converges.
Example 3.7.2. Determine whether the following series converge or diverge.
∞
∞
X
X
nn
n2 2n+1
(a)
(b)
.
3n
n!
n=1
n=0
Solution:
30
an+1
is less than 1.
an
n+2 2
3n
2
= lim (n + 1)
n→∞
3n+1
n2 2n+1
2(n + 1)2
= lim
n→∞
3n2
2
=
< 1.
3
(a) This series converges because the limit of
an+1
n→∞
an
lim
an+1
is greater than 1.
an
(n + 1)n+1 n!
lim
n→∞
(n + 1)!
nn
(n + 1)n+1 1
lim
n→∞
(n + 1)
nn
n
(n + 1)n
1
lim
= lim 1 +
n→∞
n→∞
nn
n
e > 1.
(b) This series diverges because the limit of
lim
n→∞
an+1
an
=
=
=
=
3.7.2
The Root Test
This test of convergence or divergence of series works especially well for series involving nth powers.
Root Test
Let
X
an be a series with non-zero terms.
1.
X
2.
X
p
an converges absolutely if lim n |an | < 1.
n→∞
p
p
an diverges if lim n |an | > 1 or lim n |an | = ∞.
n→∞
n→∞
p
3. The Root Test is inconclusive if lim n |an | = 1.
n→∞
Example 3.7.3. Determine the convergence or divergence of
∞
X
e2n
n=1
31
nn
.
Solution: We can apply the Root Test as follows
r
p
lim n |an | =
n→∞
lim
n→∞
n
e2n
nn
2n
en
n
n→∞ n n
e2
= lim
n→∞ n
= 0 < 1.
=
lim
Because this limit is less than 1, we can conclude that the series converges absolutely.
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