INFINITE SERIES

ELECTRONIC VERSION OF LECTURE

Dr. Lê Xuân Đại

HoChiMinh City University of Technology
Faculty of Applied Science, Department of Applied Mathematics
Email: ytkadai@hcmut.edu.vn

HCMC— 2016.
Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 1 / 42
OUTLINE

1 SERIES

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 2 / 42

OUTLINE

1 SERIES

2 NON-NEGATIVE SERIES

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 2 / 42

OUTLINE

1 SERIES

2 NON-NEGATIVE SERIES

3 ALTERNATING SERIES

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 2 / 42

OUTLINE

1 SERIES

2 NON-NEGATIVE SERIES

3 ALTERNATING SERIES

4 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 2 / 42

 Series Definition

DEFINITION 1.1
The expression of the form
a0 + a1 + a2 + . . . + an + . . . ,

where ai is a real number, i = 0, 1, 2, . . . , n, . . . is

called an infinite series. We denote it by
∞
P
an .
n=0

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 3 / 42

 Series Definition

DEFINITION 1.2
n
The sum Sn = ak = a0 + a1 + a2 + . . . + an is
P
k=0
called the n-th partial sums of the series
∞
P
an .
n=0

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 4 / 42

 Series Definition

DEFINITION 1.2
n
The sum Sn = ak = a0 + a1 + a2 + . . . + an is
P
k=0
called the n-th partial sums of the series
∞
P
an .
n=0

DEFINITION 1.3
∞
The series an is called convergent, if
P
n=0
lim Sn = S exists as a real number. The
n→∞
number S is called the sum of the series.
Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 4 / 42
 Series Definition

EXAMPLE 1.1
Consider
∞
1 1 1 X 1
1+ + +...+ n +... = n
.
2 4 2 n=0 2

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 5 / 42

 Series Definition

EXAMPLE 1.1
Consider
∞
1 1 1 X 1
1+ + +...+ n +... = n
.
2 4 2 n=0 2

1
1 1 − 2n+1
µ ¶
1 1 1
Sn = 1 + + + . . . + n = = 2 1 − n+1
2 4 2 1 − 21 2

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 5 / 42

 Series Definition

EXAMPLE 1.1
Consider
∞
1 1 1 X 1
1+ + +...+ n +... = n
.
2 4 2 n=0 2

1
1 1 − 2n+1
µ ¶
1 1 1
Sn = 1 + + + . . . + n = = 2 1 − n+1
2 4 2 1 − 21 2
µ ¶
1
lim Sn = lim 2 1 − n+1 = 2.
n→∞ n→∞ 2
Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 5 / 42
 Series Definition

DEFINITION 1.4
∞
The series an is called divergent, if
P
n=0
lim Sn = ∞ or does not exist.
n→∞

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 6 / 42

 Series Definition

DEFINITION 1.4
∞
The series an is called divergent, if
P
n=0
lim Sn = ∞ or does not exist.
n→∞

EXAMPLE 1.2
∞
Determine whether the series qn , q ∈ R is
P
n=0
convergent or divergent. If it is convergent,
find its sum.

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 6 / 42

 Series Definition

n+1
 1−q

, q 6= 1
Sn = 1 + q + q2 + . . . + qn = 1−q
n + 1, q = 1


Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 7 / 42

 Series Definition

n+1
 1−q

, q 6= 1
Sn = 1 + q + q2 + . . . + qn = 1−q
n + 1, q = 1


1 − qn+1
1
If |q| 6= 1 then lim Sn = lim =
n→∞ n→∞ 1 − q

 1 , |q| < 1

n+1 ¶
q
µ
1
lim − = 1−q
n→∞ 1 − q 1−q ∞, |q| > 1

2
If q = 1 then lim Sn = lim n + 1 = ∞
n→∞ n→∞
3
If q = −1 then lim S2k+1 = 0, lim S2k = 1.
k→∞ k→∞

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 7 / 42

 Series Definition

n+1
 1−q

, q 6= 1
Sn = 1 + q + q2 + . . . + qn = 1−q
n + 1, q = 1


1 − qn+1
1
If |q| 6= 1 then lim Sn = lim =
n→∞ n→∞ 1 − q

 1 , |q| < 1

n+1 ¶
q
µ
1
lim − = 1−q
n→∞ 1 − q 1−q ∞, |q| > 1

2
If q = 1 then lim Sn = lim n + 1 = ∞
n→∞ n→∞
3
If q = −1 then lim S2k+1 = 0, lim S2k = 1.
k→∞ k→∞
∞ 1
If |q| < 1 then the sum of series is qn =
P
·
n=0 1−q
Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 7 / 42
 Series Definition

EXAMPLE 1.3
∞ 1
Find the sum of the series
P
n=1 n(n + 1)

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 8 / 42

 Series Definition

EXAMPLE 1.3
∞ 1
Find the sum of the series
P
n=1 n(n + 1)

1 1 1
Sn = + +...+ ·
1.2 2.3 n(n + 1)

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 8 / 42

 Series Definition

EXAMPLE 1.3
∞ 1
Find the sum of the series
P
n=1 n(n + 1)

1 1 1
Sn = + +...+ ·
1.2 2.3 n(n + 1)
1 1 1
= − , n ∈ N·
n(n + 1) n n + 1

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 8 / 42

 Series Definition

EXAMPLE 1.3
∞ 1
Find the sum of the series
P
n=1 n(n + 1)

1 1 1
Sn = + +...+ ·
1.2 2.3 n(n + 1)
1 1 1
= − , n ∈ N·
n(n + 1) n n + 1
1 1 1 1 1 1 1
Sn = − + − +...+ − = 1− .
1 2 2 3 n n+1 n+1

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 8 / 42

 Series Definition

EXAMPLE 1.3
∞ 1
Find the sum of the series
P
n=1 n(n + 1)

1 1 1
Sn = + +...+ ·
1.2 2.3 n(n + 1)
1 1 1
= − , n ∈ N·
n(n + 1) n n + 1
1 1 1 1 1 1 1
Sn = − + − +...+ − = 1− .
1 2 2 3 n n+1 n+1
µ ¶
1
lim Sn = lim 1 − = 1.
n→∞ n→∞ n+1
Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 8 / 42
 Series Necessary Condition

THEOREM 1.1 (NECESSARY CONDITION)

+∞
If the series an is covergent then
P
n=1
lim an = 0.
n→+∞

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 9 / 42

 Series Necessary Condition

THEOREM 1.1 (NECESSARY CONDITION)

+∞
If the series an is covergent then
P
n=1
lim an = 0.
n→+∞
+∞
PROOF. an converges ⇔ lim Sn = S.
P
n=1 n→∞

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 9 / 42

 Series Necessary Condition

THEOREM 1.1 (NECESSARY CONDITION)

+∞
If the series an is covergent then
P
n=1
lim an = 0.
n→+∞
+∞
PROOF. an converges ⇔ lim Sn = S.
P
n=1 n→∞

lim an = lim (Sn − Sn−1 ) =

n→+∞ n→+∞
= lim Sn − lim Sn−1 = S − S = 0
n→+∞ n→+∞

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 9 / 42

 Series Necessary Condition

THEOREM 1.1 (NECESSARY CONDITION)

+∞
If the series an is covergent then
P
n=1
lim an = 0.
n→+∞
+∞
PROOF. an converges ⇔ lim Sn = S.
P
n=1 n→∞

lim an = lim (Sn − Sn−1 ) =

n→+∞ n→+∞
= lim Sn − lim Sn−1 = S − S = 0
n→+∞ n→+∞
+∞
Note. If lim an = 0, we can not conclude that an is
P
n→+∞ n=1
convergent.
Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 9 / 42
 Series Necessary Condition

EXAMPLE 1.4
+∞ 1
Determine whether the series p is convergent or
P
n=1 n
divergent.

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 10 / 42

 Series Necessary Condition

EXAMPLE 1.4
+∞ 1
Determine whether the series p is convergent or
P
n=1 n
divergent.
Necessary Condition is satisfied:
1
lim an = lim p = 0. But the series is divergent.
n→+∞ n→+∞ n

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 10 / 42

 Series Necessary Condition

EXAMPLE 1.4
+∞ 1
Determine whether the series p is convergent or
P
n=1 n
divergent.
Necessary Condition is satisfied:
1
lim an = lim p = 0. But the series is divergent.
n→+∞ n→+∞ n

1 1 1 p
Sn = 1 + p + . . . + p Ê n · p = n, n ∈ N
2 n n

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 10 / 42

 Series Necessary Condition

EXAMPLE 1.4
+∞ 1
Determine whether the series p is convergent or
P
n=1 n
divergent.
Necessary Condition is satisfied:
1
lim an = lim p = 0. But the series is divergent.
n→+∞ n→+∞ n

1 1 1 p
Sn = 1 + p + . . . + p Ê n · p = n, n ∈ N
2 n n
p
lim Sn Ê lim n = +∞ ⇒ lim Sn = +∞
n→+∞ n→+∞ n→+∞
1 +∞
Therefore, the series p is divergent.
P
n=1 n
Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 10 / 42
 Series Necessary Condition

COROLLARY 1.1 (TEST FOR DIVERGENCE)

If lim an does not exist or if lim an 6= 0 then
n→∞ n→∞
+∞
the series an is divergent.
P
n=1

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 11 / 42

 Series Necessary Condition

COROLLARY 1.1 (TEST FOR DIVERGENCE)

If lim an does not exist or if lim an 6= 0 then
n→∞ n→∞
+∞
the series an is divergent.
P
n=1

EXAMPLE 1.5
µ ¶3n+2
+∞ 2n + 3
Show that the series n5
P
n=1 2n + 1
diverges.

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 11 / 42

 Series Necessary Condition

µ ¶3n+2 µ ¶ 2n+1 . 2(3n+2)

5 2n + 3 2 2 2n+1
an = n = n5 . 1 +
2n + 1 2n + 1
n→∞
⇒ an −−−→ ∞
So the series diverges by the Test for
Divergence.

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 12 / 42

 Series Some properties of convergent series
+∞
10 The series an is convergent if and only
P
n=1
+∞
if the series an , (n0 > 1) is convergent.
P
n=n0

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 13 / 42

 Series Some properties of convergent series
+∞
10 The series an is convergent if and only
P
n=1
+∞
if the series an , (n0 > 1) is convergent.
P
n=n0
+∞
P nP
0 −1 +∞
P
an = an + an
n=1 n=1 n=n0
+∞
20 If the series an is convergent then the
P
n=1
+∞
series α.an (α ∈ R) is also convergent.
P
n=1
+∞ +∞
α.an = α.
P P
an .
n=1 n=1
Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 13 / 42
 Series Some properties of convergent series

+∞ +∞
0
3 If the series an and the series bn are
P P
n=1 n=1
convergent and have the sums S1, S2
+∞
respectively then the series (an + bn ) is
P
n=1
also convergent and has the sum S1 + S2.
+∞
X +∞
X +∞
X
(an + bn ) = an + bn .
n=1 n=1 n=1

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 14 / 42

 Non-negative series Definition

DEFINITION 2.1
+∞
The series an is called the non-negative
P
n=1
series if an Ê 0, n ∈ N.
Remarks.
1. For the non-positive series
+∞ +∞
an , (an Ê 0, n ∈ N).
X X
(−an ) = −
n=1 n=1

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 15 / 42

 Non-negative series Definition

2. The sequence of partial sums of

+∞
non-negative series an is not decreasing,
P
n=1
since Sn+1 − Sn = an+1 Ê 0. The sequence
n o∞ +∞
has a finite limit (i.e. the series
P
Sn an
n=1 n=1
n o∞
is convergent) if and only if Sn is
n=1
bounded above.

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 16 / 42

 Non-negative series Definition

2. The sequence of partial sums of

+∞
non-negative series an is not decreasing,
P
n=1
since Sn+1 − Sn = an+1 Ê 0. The sequence
n o∞ +∞
has a finite limit (i.e. the series
P
Sn an
n=1 n=1
n o∞
is convergent) if and only if Sn is
n=1
bounded above.
+∞
3. The series an is divergent if and only if
P
n=1
n o∞
the sequence Sn is not bounded above,
n=1
i.e. lim Sn = +∞.
n→∞
Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 16 / 42
 Non-negative series The Integral Test

THE INTEGRAL TEST

THEOREM 2.1
Suppose f (x) is a continuous, positive, decreasing
function on [1, +∞) and let an = f (n). Then the series
∞
f (n) is convergent if and only if the improper
P
n=1 Z ∞
integral f (x)dx is convergent.
1

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 17 / 42

 Non-negative series The Integral Test

THE INTEGRAL TEST

THEOREM 2.1
Suppose f (x) is a continuous, positive, decreasing
function on [1, +∞) and let an = f (n). Then the series
∞
f (n) is convergent if and only if the improper
P
n=1 Z ∞
integral f (x)dx is convergent.
Z ∞ 1
∞
If f (x)dx is convergent, then f (n) is
1
P
1 n=1
convergent.

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 17 / 42

 Non-negative series The Integral Test

THE INTEGRAL TEST

THEOREM 2.1
Suppose f (x) is a continuous, positive, decreasing
function on [1, +∞) and let an = f (n). Then the series
∞
f (n) is convergent if and only if the improper
P
n=1 Z ∞
integral f (x)dx is convergent.
Z ∞ 1
∞
If f (x)dx is convergent, then f (n) is
1
P
1 n=1
convergent.
Z ∞ ∞
If f (x)dx is divergent, then f (n) is divergent.
2
P
1 n=1

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 17 / 42

 Non-negative series The Integral Test

EXAMPLE 2.1
∞ 1 ∞ 1
Test the series , and the series for
P P
2
n=2 n ln n n=2 n ln n
convergence or divergence.

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 18 / 42

 Non-negative series The Integral Test

EXAMPLE 2.1
∞ 1 ∞ 1
Test the series , and the series for
P P
2
n=2 n ln n n=2 n ln n
convergence or divergence.

Z∞ Z∞
dx d(ln x) h i∞
= = ln ln x = +∞
x ln x ln x 2
2 2

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 18 / 42

 Non-negative series The Integral Test

EXAMPLE 2.1
∞ 1 ∞ 1
Test the series , and the series for
P P
2
n=2 n ln n n=2 n ln n
convergence or divergence.

Z∞ Z∞
dx d(ln x) h i∞
= = ln ln x = +∞
x ln x ln x 2
2 2
Z∞ Z∞
dx d(ln x) 1 ∞
· ¸
1
= = − =
x ln2 x 2
ln x ln x 2 ln 2
2 2

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 18 / 42

 Non-negative series The Integral Test

EXAMPLE 2.1
∞ 1 ∞ 1
Test the series , and the series for
P P
2
n=2 n ln n n=2 n ln n
convergence or divergence.

Z∞ Z∞
dx d(ln x) h i∞
= = ln ln x = +∞
x ln x ln x 2
2 2
Z∞ Z∞
dx d(ln x) 1 ∞
· ¸
1
= = − =
x ln2 x 2
ln x ln x 2 ln 2
2 2
∞ ∞
X 1 X 1
is divergent, 2
is convergent
n=2 n ln n n=2 n ln n

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 18 / 42

 Non-negative series Some Basic Series

SOME BASIC SERIES

+∞
qn is convergent if |q| < 1 and is
1
P
n=1
divergent if |q| Ê 1.
+∞ 1
is convergent if α > 1 and is
2
P
α
n=1 n
divergent if α É 1.
+∞ 1
is convergent if α > 1 or if
3
P
α β
n=2 n ln n
α = 1, β > 1 and is divergent if α < 1 or if
α = 1, β É 1.
Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 19 / 42
 Non-negative series The Comparision Test

THEOREM 2.2 (THE COMPARISION TEST )

+∞ +∞
Suppose that an and bn are series with
P P
n=1 n=1
non-negative terms:
0 É an É bn , ∀n Ê n0
+∞ +∞
If bn is convergent, then an is also
1
P P
n=1 n=1
convergent.
+∞ +∞
If an is divergent, then bn is also
2
P P
n=1 n=1
divergent.
Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 20 / 42
 Non-negative series The Comparision Test

EXAMPLE 2.2
5 + (−1)n .3 +∞
Determine whether the series
P
n=1 2n+3
converges or diverges.

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 21 / 42

 Non-negative series The Comparision Test

EXAMPLE 2.2
5 + (−1)n .3 +∞
Determine whether the series
P
n=1 2n+3
converges or diverges.

5 + (−1)n .3 8 1
0 É an = É = = bn , n Ê 1.
2n+3 2n+3 2n
P 1
+∞ 1/2 +∞
We know that so bn is
P
n
= = 1
n=1 2 1 − 1/2 n=1
+∞
convergent. Therefore, an is convergent.
P
n=1
Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 21 / 42
 Non-negative series The Limit Comparison Test

THEOREM 2.3
+∞ +∞
Suppose that an and
bn are series with
P P
n=1 n=1
an
non-negative terms, K = lim
n→+∞ bn
+∞ +∞
K = 0. If bn is convergent then an is
1
P P
n=1 n=1
convergent.
+∞ +∞
K > 0. Either both series an and bn converge
2
P P
n=1 n=1
or both diverge.
+∞ +∞
K = +∞. If bn is divergent then an is
3
P P
n=1 n=1
divergent.
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 Non-negative series The Limit Comparison Test

DEFINITION 2.2
The functions f (x) and g(x) are called
equivalent as x → a if

f (x)
lim =1 (1)
x→a g(x)

We denote the equivalent functions by

x→a
f (x) ∼ g(x).

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 23 / 42

 Non-negative series The Limit Comparison Test

SOME BASIC EQUIVALENT FUNCTIONS I

As x → 0 the following functions are

equivalent:
1
1
sin x ∼ x, tan x ∼ x, 1 − cos x ∼ x2
2
2
arctan x ∼ x, arcsin x ∼ x
3
ax − 1 ∼ x. ln a, (a > 0, a 6= 1), ex − 1 ∼ x
x
4
loga (1 + x) ∼ x loga e = , (a > 0, a 6= 1),
ln a
ln(1 + x) ∼ x

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 24 / 42

 Non-negative series The Limit Comparison Test

SOME BASIC EQUIVALENT FUNCTIONS II

p x
5
(1 + x)µ − 1 ∼ µ.x, (µ ∈ R), 1 + x − 1 ∼ ,
p 2
x
1 + x − 1 ∼ , (n ∈ N)
n

n
x2
6
sinh x ∼ x, cosh x − 1 ∼
2

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 25 / 42

 Non-negative series The Limit Comparison Test

THEOREM 2.4
If (
u(x) → 0 as x → a
f (x) ∼ g(x) as x → 0
then f (u(x)) ∼ g(u(x)) as x → a.

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 26 / 42

 Non-negative series The Limit Comparison Test

THEOREM 2.4
If (
u(x) → 0 as x → a
f (x) ∼ g(x) as x → 0
then f (u(x)) ∼ g(u(x)) as x → a.
THEOREM 2.5
If g(x) → A 6= 0 and f (x) ∼ f (x) as x → a, then
f (x).g(x) ∼ A.f (x) as x → a

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 26 / 42

 Non-negative series The Limit Comparison Test

THEOREM 2.6
If f (x) ∼ f (x) and g(x) ∼ g(x) as x → a, then

f (x) x→a f (x)

∼ (2)
g(x) g(x)

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 Non-negative series The Limit Comparison Test

EXAMPLE 2.3
+∞ e n + n3
Determine whether the series
P
3
n=1 2n + ln n
converges or diverges.

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 Non-negative series The Limit Comparison Test

EXAMPLE 2.3
+∞ e n + n3
Determine whether the series
P
3
n=1 2n + ln n
converges or diverges.
We have
en ³ e ´n
e n + n3 n→∞
an = ∼ n= = bn .
2n + ln3 n 2 2
+∞ +∞
We know that bn is divergent, thus
P P
an
n=1 n=1
is divergent.
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 Non-negative series The Limit Comparison Test

EXAMPLE 2.4
ln 1 + sin n1
¡ ¢
+∞
Determine whether the series
P
n=1 n + ln2 n
converges or diverges.

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 29 / 42

 Non-negative series The Limit Comparison Test

EXAMPLE 2.4
ln 1 + sin n1
¡ ¢
+∞
Determine whether the series
P
n=1 n + ln2 n
converges or diverges.
We have
ln(1 + sin n1 ) n→∞ sin n1 n→∞ 1
an = ∼ ∼ 2 = bn .
n + ln2 n n n
+∞ +∞
We know that bn is convergent, thus
P P
an
n=1 n=1
is convergent.
Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 29 / 42
 Non-negative series The Limit Comparison Test

EXAMPLE 2.5
Determine whether
´ the series
P p 3³
+∞ π
n cosh − 1 converges or diverges.
n=1 n

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 30 / 42

 Non-negative series The Limit Comparison Test

EXAMPLE 2.5
Determine whether
´ the series
P p 3³
+∞ π
n cosh − 1 converges or diverges.
n=1 n

We have
p
3
π n→∞ 3/2 1 π
³ ´2 1 π2
an = n (cosh −1) ∼ n . = = bn .
n 2 n 2 n1/2
+∞ +∞
We know that bn is divergent, thus
P P
an
n=1 n=1
is divergent.
Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 30 / 42
 Alternating Series

DEFINITION 3.1
+∞
The series (−1)n an , (an Ê 0, ∀n Ê n0 or
P
n=1
an É 0, ∀n Ê n0 ) is called the alternating series.

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 Alternating Series

DEFINITION 3.1
+∞
The series (−1)n an , (an Ê 0, ∀n Ê n0 or
P
n=1
an É 0, ∀n Ê n0 ) is called the alternating series.

THEOREM 3.1 (ALTERNATING SERIES TEST )

+∞
If the alternating series (−1)n an satisfies
P
n=1
1
lim an = 0
n→+∞
2
an+1 É an , ∀n Ê n0 ,
then the series is convergent.
Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 31 / 42
 Alternating Series

EXAMPLE 3.1
+∞ ln n
Test the series (−1)n+1 p for convergence
P
n=1 n
or divergence.

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 Alternating Series

EXAMPLE 3.1
+∞ ln n
Test the series (−1)n+1 p for convergence
P
n=1 n
or divergence.
ln n ln x 2 − ln x
an = p , f (x) = p , f 0 (x) = p < 0, ∀x > e2 .
n x 2x x
1
lim an = 0
n→+∞
n o+∞
2
the sequence an is decreasing.
n=8
+∞
(−1)n+1 an is convergent by the Alternating
P
n=1
Series Test.
Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 32 / 42
 Absolute Convergence and the Ratio and Root Tests Absolute Convergence

DEFINITION 4.1
+∞
A series an is called absolutely convergent
P
n=1
+∞
if the series of absolute values |an | is
P
n=1
convergent.

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 Absolute Convergence and the Ratio and Root Tests Absolute Convergence

DEFINITION 4.1
+∞
A series an is called absolutely convergent
P
n=1
+∞
if the series of absolute values |an | is
P
n=1
convergent.

THEOREM 4.1
+∞
If a series |an | is convergent then the series
P
n=1
+∞
an is convergent.
P
n=1

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 Absolute Convergence and the Ratio and Root Tests Absolute Convergence

DEFINITION 4.2
+∞
A series an is called conditionally
P
n=1
+∞
convergent if the series an is convergent,
P
n=1
+∞
but the series |an | is divergent.
P
n=1

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 Absolute Convergence and the Ratio and Root Tests Absolute Convergence

EXAMPLE 4.1
arctan(−n)n +∞
Test the series for the
P
p
4
n=1 2n6 + 3n + 1
convergence or divergence.

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 35 / 42

 Absolute Convergence and the Ratio and Root Tests Absolute Convergence

EXAMPLE 4.1
arctan(−n)n +∞
Test the series for the
P
p
4
n=1 2n6 + 3n + 1
convergence or divergence.
arctan(−n)n
We have an = p 4
. Consider |an | =
6
2n + 3n + 1
| arctan(−n)n | π/2 n→∞ π/2
p4
É p
4
∼ p4
= bn .
6 6 2n 6/4
2n + 3n + 1 2n + 3n + 1

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 Absolute Convergence and the Ratio and Root Tests Absolute Convergence

EXAMPLE 4.1
arctan(−n)n +∞
Test the series for the
P
p
4
n=1 2n6 + 3n + 1
convergence or divergence.
arctan(−n)n
We have an = p 4
. Consider |an | =
6
2n + 3n + 1
| arctan(−n)n | π/2 n→∞ π/2
p4
É p4
∼ p4
= bn .
6 6 2n 6/4
2n + 3n + 1 2n + 3n + 1
+∞
We know that bn is convergent, thus
P
n=1
+∞ +∞
|an | is convergent, an is convergent.
P P
n=1 n=1
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 Absolute Convergence and the Ratio and Root Tests The Ratio Test

THEOREM 4.2 ¯ ¯
+∞ ¯ an+1 ¯
For
P
an , D = lim ¯¯ ¯.
n=1 n→+∞ an ¯
+∞
If D < 1, then the series an is absolutely
1
P
n=1
convergent and therefore convergent.
+∞
If D > 1 or D = +∞, then the series an is
2
P
n=1
divergent.
3
If D = 1, the Ratio Test is inconclusive.

Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 36 / 42

 Absolute Convergence and the Ratio and Root Tests The Ratio Test

Note. If D = 1, the test gives no information.

+∞ 1
For instance, for the divergent series
P
¯ ¯ n=1 n
¯ an+1 ¯
we have D = lim ¯¯ ¯ = lim n = 1
n→+∞ an ¯ n→+∞ n + 1
P 1
+∞
Whereas for the convergent series 2
we
n=1 n
n2
¯ ¯
¯ an+1 ¯
have D = lim ¯¯ ¯ = lim = 1.
n→+∞ an ¯ n→+∞ (n + 1)2

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 Absolute Convergence and the Ratio and Root Tests The Ratio Test

EXAMPLE 4.2
3n .n! +∞
Test the series for the convergence or
P
n
n=1 n
divergence.

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 Absolute Convergence and the Ratio and Root Tests The Ratio Test

EXAMPLE 4.2
3n .n! +∞
Test the series for the convergence or
P
n
n=1 n
divergence.
3n .n!
We have an = n · Consider
n
¯ an+1 ¯ 3n+1 .(n + 1)! nn
¯ ¯ ³ n ´n
n→∞ 3
¯ ¯= · = 3. −−−→ > 1
¯ a ¯
n (n + 1)n+1 3n .n! n+1 e
+∞
Therefore an is divergent by the Ratio
P
n=1
Test.
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 Absolute Convergence and the Ratio and Root Tests The Ratio Test

EXAMPLE 4.3
Test the convergence of the series
+∞
P 2.5.8 . . . (3n − 1)
·
n=1 1.6.11 . . . (5n − 4)

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 Absolute Convergence and the Ratio and Root Tests The Ratio Test

EXAMPLE 4.3
Test the convergence of the series
+∞
P 2.5.8 . . . (3n − 1)
·
n=1 1.6.11 . . . (5n − 4)
¯ ¯
2.5.8 . . . (3n − 1) ¯ an+1 ¯
an = . Consider ¯¯ ¯=
1.6.11 . . . (5n − 4) an ¯
2.5.8 . . . (3n − 1)(3n + 2) 1.6.11 . . . (5n − 4)
= =
1.6.11 . . . (5n − 4)(5n + 1) 2.5.8 . . . (3n − 1)
3n + 2 n→∞ 3
= −−−→ < 1.
5n + 1 5
+∞
Therefore an is convergent by the Ratio
P
n=1
Test.
Dr. Lê Xuân Đại (HCMUT-OISP) INFINITE SERIES HCMC— 2016. 39 / 42
 Absolute Convergence and the Ratio and Root Tests The Root Test

THEOREM 4.3
+∞ p
n
For
P
an , C = lim |an |.
n=1 n→+∞
+∞
If C < 1, then the series an is absolutely
1
P
n=1
convergent and therefore convergent.
+∞
If C > 1 or D = +∞, then the series an is
2
P
n=1
divergent.
3
If C = 1, the Ratio Test is inconclusive.

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 Absolute Convergence and the Ratio and Root Tests The Root Test

EXAMPLE 4.4
Test the
µ convergence for the series
¶ n
+∞
P 5 3n + 2
n .
n=1 4n + 3

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 Absolute Convergence and the Ratio and Root Tests The Root Test

EXAMPLE 4.4
Test the
µ convergence for the series
3n + 2 n
¶
+∞
P 5
n .
n=1 4n + 3
µ ¶n
3n + 2
an = n5 . Consider
4n + 3
pn
p
n 3n + 2 n→∞ 3
|an | = n5 · −−−→ < 1.
4n + 3 4
+∞
Thus an converges by the Root Test.
P
n=1

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 Absolute Convergence and the Ratio and Root Tests The Root Test

THANK YOU FOR YOUR ATTENTION

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