MBEYA UNIVERSITY OF SCIENCE AND TECHNOLOGY
COLLEGE OF SCIENCE AND TECHNICAL EDUCATION
DEPARTMENT OF MATHEMATICS AND STATISTICS
SEMESTER II ACADEMIC YEAR 2023/2024
NSLT 07205: ADVANCED CALCULUS
UQF 8 FIRST YEAR: LABORATORY SCIENCE AND TECHNOLOGY
LECTURE 2
PREPARED BY: MOHAMED H. MOHAMED
2.0: SERIES
2.1: Basic Concepts of Series
CONCEPTS:
1) What do we mean when we express a number as an infinite decimal? For instance, what
does it mean to write
π =3+
1
1
5
9
2
6
5
1
+ 2 + 3 + 4 + 5 + 6 + 7 + 8 + ···
10 10
10
10
10
10
10
10
where the three dots (· · · ) indicate that the sum continues forever, and the more terms
we add, the closer we get to the actual value of π.
2) In general, if we try to add the terms of an infinite sequence {an }∞
n=1 we get an expression
of the form
a1 + a2 + a3 + a4 + · · · + an + · · ·
(1)
which is called an infinite series (or just a series) and is denoted, for short, by the
symbol
∞
X
an or
X
an
n=1
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3) Does it make sense to talk about the sum of infinitely many terms?
⇒ It would be impossible to find a finite sum for the series
1 + 2 + 3 + 4 + 5 + 6 + ··· + n + ···
because if we start adding the terms we get the cumulative sums 1, 3, 6, 10, 15, 21, · · · and,
n(n + 1)
after the nth term, we get
, which becomes very large as n increases.
2
⇒ However, if we start to add the terms of the series
1 1 1
1
1
1
1
+ + +
+
+
+ ··· + n + ···
2 4 8 16 32 64
2
we get
1
1 3 7 15 31 63
, , ,
,
,
,··· 1 − n,···.
2 4 8 16 32 64
2
⇒ The table below shows that as we add more and more terms, these Partial Sums
become closer and closer to 1.
⇒ In fact, by adding sufficiently many terms of the series we can make the partial sums
as close as we like to 1. So it seems reasonable to say that the sum of this infinite series
is 1 and to write
∞
X
1
n=1
2n
=
1 1 1
1
1
1
1
+ + +
+
+
+ ··· + n + ··· = 1
2 4 8 16 32 64
2
4) We use a similar idea to determine whether or not a general series (1) has a sum. We
consider the partial sums
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S1 = a1
S2 = a1 + a2
S3 = a1 + a2 + a3
S4 = a1 + a2 + a3 + a4
and, in general
Sn = a1 + a2 + a3 + a4 + · · · + an =
n
X
ai
i=1
5) These partial sums form a new sequence {Sn }, which may or may not have a limit. If
lim Sn = S exists (as a finite number), then, as in the following definition, we call it the
n→∞
P
sum of the infinite series
an .
6) Definition:
⇒ Given a series
∞
P
an = a1 + a2 + a3 + · · · , let Sn denote its nth partial sum:
n=1
Sn =
n
X
ai = a1 + a2 + a3 + a4 + · · · + an
i=1
⇒ If the sequence {Sn } is convergent and lim Sn = S exists as a real number, then the
n→∞
P
series
an is called Convergent and we write
a1 + a2 + a3 + · · · + an + · · · = S
or
∞
X
an = S
n=1
⇒ The number
P S is called the Sum of the series. If the sequence {Sn } is divergent, then
the series
an is called Divergent.
7) Thus the sum of a series is the limit of the sequence of partial sums. So when we write
∞
P
an = S, we mean that by adding sufficiently many terms of the series we can get as
n=1
close as we like to the number S. Notice that:
∞
X
an = lim
n=1
n→∞
n
X
ai
i=1
8) EXAMPLE 1: Suppose we know that the sum of the first n terms of the series
∞
P
an
n=1
is:
S n = a1 + a2 + a3 + · · · + an =
2n
3n + 5
⇒ Then the sum of the series is the limit of the sequence {Sn }:
∞
X
2n
2
2
an = lim Sn = lim
= lim
=
5
n→∞
n→∞ 3n + 5
n→∞ 3 +
3
n
n=1
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9) In Example 1 we were given an expression for the sum of the first n terms, but it is usually
not easy to find such an expression.
10) In the following Example 2, however, we look at a famous series for which we can find an
explicit formula for Sn .
11) EXAMPLE 2: An important example of an infinite series is the geometric series:
2
3
a + ar + ar + ar + · · · + ar
n−1
+ ··· =
∞
X
arn−1
a 6= 0
n=1
⇒ Each term is obtained from the preceding one by multiplying it by the Common
Ratio r.
⇒ CASE I: If r = 1, then Sn = a + a + a + · · · + a = na → ±∞.
⇒ Since lim Sn does not exist, then the geometric series diverges in this case.
n→∞
⇒ CASE II: If r 6= 1, we have;
Sn = a + ar + ar2 + ar3 + · · · + arn−1
and rSn = ar + ar2 + ar3 + · · · + arn−1 + arn
⇒ Subtracting these equations, we get;
Sn − rSn = a − arn
a(1 − rn )
⇒ Sn =
1−r
⇒ If −1 < r < 1, then rn → 0 as n → ∞ so;
a(1 − rn )
a
a
lim Sn = lim
= lim
−
lim rn
n→∞
n→∞
n→∞
1−r
1−r
1 − r n→∞
a
a
=
−
(0)
1−r
1−r
a
⇒ lim Sn =
n→∞
1−r
⇒ Thus when −1 < r < 1 or |r| < 1, the geometric series is convergent and its sum
a
is S =
.
1−r
a(1 − rn )
, if r ≤ 1 or r > 1, then rn is divergent as n → ∞, and so
1−r
lim Sn does not exist. Therefore the geometric series diverges in those cases.
⇒ From Sn =
n→∞
⇒ We summarize the results of Example 2 as follows.
12) ⇒ The geometric series
sum is
∞
P
n=1
∞
P
arn−1 = a + ar + ar2 + · · · , is Convergent if |r| < 1 and its
n=1
ar
n−1
a
=
.
1−r
⇒ If |r| ≥ 1, the geometric series is Divergent.
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EXAMPLES
1. Find the sum of the geometric series:
5−
10 20 40
+
−
+ −···
3
9
27
SOLUTION
2
2
⇒ The first term is a = 5 and the common ratio is r = − . Since |r| = < 1, the series
3
3
is convergent and its sum is:
5−
5
10 20 40
5
+
−
+ −··· =
2 = 5 = 3
3
9
27
1 − (− 3 )
3
⇒ What do we really mean when we say that the sum of the series in this Example is 3?
Of course, we can’t literally add an infinite number of terms, one by one. But, according
to Definition, the total sum is the limit of the sequence of partial sums. So, by taking the
sum of sufficiently many terms, we can get as close as we like to the number 3. The Table
below shows the first ten partial sums Sn and the Graph below shows how the sequence
of partial sums approaches 3.
2. Is the series
∞
P
22n 31−n convergent or divergent?
n=1
SOLUTION
⇒ Let’s rewrite the nth term of the series in the form arn−1 :
∞
X
n=1
2n 1−n
2 3
=
∞
X
2 n −(n−1)
(2 ) 3
n=1
n−1
∞
∞
X
X
4n
4
=
=
4
n−1
3
3
n=1
n=1
4
⇒ We recognize this series as a geometric series with a = 4 and r = . Since r > 1, then
3
the series diverges.
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¯ = 2.3171717171... as a ratio of integers.
3. Write the number 2.317
SOLUTION
¯ = 2.3171717171... = 2.3 +
2.317
17
17
17
+ 5 + 7 + ···
3
10
10
10
⇒ After the first term we have a geometric series with a =
17
1
and r = 2 . Therefore;
3
10
10
17
17
¯ = 2.3 + 103 = 2.3 + 1000
2.317
99
1
100
1− 2
10
23
17
1147
=
+
=
10 990
495
4. Find the sum of the series
∞
P
xn , where |x| < 1.
n=0
SOLUTION
⇒ Notice that this series starts with n = 0 and so the first term is x0 = 1. Thus;
∞
X
x n = 1 + x + x2 + x3 + x4 + · · ·
n=0
⇒ This is a geometric series with a = 1 and r = x. Since |r| = |x| < 1 it converges, then
we have;
∞
X
n=0
xn =
1
1−x
13) The other series with their convergence criteria are as follows:
(i) Harmonic Series
∞ 1
P
1 1 1
⇒ The series
= 1 + + + + · · · is called Harmonic Series, which is
2 3 4
n=1 n
Divergent.
(ii) P-Series
∞ 1
P
is called P-Series.
p
n=1 n
⇒ The p-series is Convergent if p > 1 and is Divergent if p < 1.
∞ 1
P
1
1
1
1
⇒ For example, the series
= 3 + 3 + 3 + 3 + · · · is convergent because
3
1
2
3
4
n=1 n
it is a p-series with p = 3 > 1, while the series
∞ 1
∞
P
P
1
1
1
1
√
= 1+ √
+√
+√
+ · · · is divergent, because it is a p-series
1 =
3
3
3
3
n
2
3
4
n=1 n 3
n=1
1
with p = < 1.
3
⇒ The series with p = 1 is called the Harmonic Series.
⇒ The series
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(iii) Alternating Series
⇒ Alternating Series is a series whose terms are alternately positive and negative.
Here are two examples of alternating series:
∞
P
1
1 1 1 1 1
(a)
(−1)n−1 = 1 − + − + − + − · · ·
n
2 3 4 5 6
n=1
∞
P
n
1 2 3 4 5 6
(b)
(−1)n
= − + − + − + − +···
n+1
2 3 4 5 6 7
n=1
⇒ We see from these Examples that the nth term of an alternating series is of the
form:
an = (−1)n−1 bn
or an = (−1)n bn
⇒ The following Test says that, if the terms of an alternating series decrease toward
0 in absolute value, then the series Converges.
ALTERNATING SERIES TEST: If the alternating series
∞
X
(−1)n−1 bn = b1 − b2 + b3 − b4 + b5 − b6 + · · ·
bn > 0
n=1
satisfies:
(i) bn+1 ≤ bn for all n
(ii) lim bn = 0
n→∞
then the series is Convergent.
∞ (−1)n−1
P
1 1 1
⇒ For Example: The alternating harmonic series 1− + − +· · · =
2 3 4
n
n=1
satisfies:
1
1
(i) bn+1 ≤ bn because
<
n+1
n
1
(ii) lim bn = lim = 0
n→∞
n→∞ n
so the series is convergent.
⇒ Also the series
∞ (−1)n 3n
P
is alternating, but
n=1 4n − 1
3
3
3n
lim bn = lim
= lim
=
1
n→∞
n→∞
n→∞
4n − 1
4
4− n
So the condition (ii) is not satisfied, and the series diverges.
14) THEOREM: If the series
∞
P
an is convergent, then lim an = 0.
n=1
n→∞
P
⇒ NOTE 1: With any series
an we associate two P
sequences. The sequence {Sn } of
its partial sums and the sequence {an } of its terms. If
an is convergent, then the limit
of the sequence {Sn } is S (the sum of the series) and, the Theorem above asserts, the
limit of the sequence {an } is 0.
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⇒ NOTE 2: The converse of the Theorem above is not
P true in general.
⇒ That is, if lim an = 0, we cannot conclude that
an is convergent. For example,
n→∞
P1
P1
1
that an = → 0 as n → ∞, but
is divergent.
observe the harmonic series
n
n
n
15) TEST FOR DIVERGENCE:
⇒ If lim an does not exist or if lim an 6= 0, then the series
n→∞
n→∞
∞
P
an is divergent.
n=1
⇒ The Test for Divergence follows from the Theorem above because, if the series is not
divergent, then it is convergent, and so lim an = 0.
n→∞
n2
16) EXAMPLE: Show that the series
diverges.
2
n=1 5n + 4
∞
P
SOLUTION
⇒ lim an = lim
n→∞
n→∞
n2
5n2 + 4
= lim
n→∞
Test for Divergence.
17) If we find that lim an 6= 0, we know that
1
5 + n42
=
1
6= 0. So the series diverges by the
5
P
an is divergent, and if we find that lim an = 0,
n→∞
P
we know nothing about the convergence or divergence of
an .
n→∞
2.2: Tests for Convergence and Divergence of Series of
Constants
CONCEPTS:
1) More often than not, exact values of infinite series cannot be obtained. Thus, the search
turns toward information about the series. In particular, its convergence or divergence
comes in question.
2) The following tests aid in discovering this information:
(i) Comparison Test
(ii) Limit-Comparison Test/Quotient Test
(iii) Integral Test
(iv) Alternating Series Test
(v) Absolute and Conditional Convergence
(vi) Raabe’s Test
(vii) Root Test
(viii) Ratio Test
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3) The Ratio Test states that;
∞
P
an+1
= L < 1, then the series
an is Convergent.
n→∞
an
n=1
∞
P
an+1
an+1
(ii) If lim
= L > 1 or if lim
= ∞, then the series
an is Divergent.
n→∞
n→∞
an
an
n=1
an+1
(iii) If lim
= 1, the Ratio Test is Inconclusive. That is, no conclusion can be
n→∞
an
P
drawn about the convergence or divergence of
an . (If L = 1, the Ratio Test fails)
(i) If lim
4) EXAMPLE: Use the Ratio Test to test the series
n3
(−1)n n for convergence.
3
n=1
∞
P
SOLUTION
(−1)n n3
(−1)n+1 (n + 1)3
and
a
=
n+1
3n
3n+1
n+1
3
(−1) (n + 1)
3n+1
=
(−1)n n3
3n
(−1)n+1 (n + 1)3
3n
=
×
3n+1
(−1)n n3
3n
(−1)n (−1)1 (n + 1)3
=
×
3n · 31
(−1)n n3
(−1)1 (n + 1)3
1
=
× 3
3
n
3
1 (n + 1)
= −
3 n3
1 (n + 1)3
=
3 n3
3
1 n+1
=
3
n
3
1
1
=
1+
3
n
" 3 #
1
1
1+
= lim
n→∞ 3
n
3
1
1
= lim 1 +
3 n→∞
n
1
= (1 + 0)3
3
1
= <1
3
1
= <1
3
an =
⇒
an+1
an
an+1
n→∞
an
⇒ lim
an+1
n→∞
an
⇒ lim
⇒ Therefore, by the Ratio Test, the given series is Convergent.
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