MATH 38

Mathematical Analysis III

I. F. Evidente
IMSP (UPLB)

Previously...

Tests for Infinite Series of Nonnegative Terms:

1
2
3

Integral Test
Limit Comparison Test
Comparison Test

Alternating Series Test

Special Series: geometric, harmonic, constant, p-series, alternating

harmonic series

Outline

Four Convergence Theorems

Absolute and Conditional Convergence

Absolute Convergence Tests

Root Test
Ratio Test

Theorem (Alternating Series Test for Convergence)

Let

a n be an alternating series.

Theorem (Alternating Series Test for Convergence)

Let an be an alternating series. The alternating series is convergent if
both of the following conditions hold:
P

Theorem (Alternating Series Test for Convergence)

Let an be an alternating series. The alternating series is convergent if
both of the following conditions hold:
P

lim |a n | = 0

Theorem (Alternating Series Test for Convergence)

Let an be an alternating series. The alternating series is convergent if
both of the following conditions hold:
P

lim |a n | = 0

n
2

{|a n |} is a decreasing sequence.

Remark
1

If an alternating series does not satisfy the above conditions, you

cannot conclude that the alternating series is divergent.

Remark
1

If an alternating series does not satisfy the above conditions, you

cannot conclude that the alternating series is divergent.

If the first condition is not satisfied, use the divergence test to prove
that the sequence is divergent.

Example
(1)n n
X
n=1 n + 1

Example
(1)n n
X
n=1 n + 1

|a n |

Example
(1)n n
X
n=1 n + 1

(1)n n

|a n | =
n +1

Example
(1)n n
X
n=1 n + 1

(1)n n
= n

|a n | =
n +1 n +1

Example
(1)n n
X
n=1 n + 1

(1)n n
= n

|a n | =
n +1 n +1

lim

n n + 1

Example
(1)n n
X
n=1 n + 1

(1)n n
= n

|a n | =
n +1 n +1

lim

n n + 1

=1

Example
(1)n n
X
n=1 n + 1

(1)n n
= n

|a n | =
n +1 n +1

lim

n n + 1

=1

lim

n
n +1

Example
(1)n n
X
n=1 n + 1

(1)n n
= n

|a n | =
n +1 n +1

lim

n n + 1

=1

lim

n
= 1
n +1

Example
(1)n n
X
n=1 n + 1

(1)n n
= n

|a n | =
n +1 n +1

lim

n n + 1

(1)n n
n n + 1

This implies: lim

=1

lim

n
= 1
n +1

Example
(1)n n
X
n=1 n + 1

(1)n n
= n

|a n | =
n +1 n +1

lim

n n + 1

=1

lim

n
= 1
n +1

(1)n n
does not exist (see remark after Thm 1.1.3).
n n + 1

This implies: lim

Example
(1)n n
X
n=1 n + 1

(1)n n
= n

|a n | =
n +1 n +1

lim

n n + 1

=1

lim

n
= 1
n +1

(1)n n
does not exist (see remark after Thm 1.1.3).
n n + 1

This implies: lim

The series is divergent by the Divergence Test.

Outline

Four Convergence Theorems

Absolute and Conditional Convergence

Absolute Convergence Tests

Root Test
Ratio Test

Theorem (Two of the Four Convergence Theorems)

Suppose

a n and

b n are infinite series and c is a nonzero constant.

Theorem (Two of the Four Convergence Theorems)

Suppose
1

If

a n and

b n are infinite series and c is a nonzero constant.

a n converges, then

Theorem (Two of the Four Convergence Theorems)

Suppose
1

If

b n are infinite series and c is a nonzero constant.

P
P
P
a n converges, then ca n converges and ca n = c a n .

a n and

Theorem (Two of the Four Convergence Theorems)

Suppose
1

b n are infinite series and c is a nonzero constant.

P
P
P
P
a n converges, then ca n converges and ca n = c a n . If a n

a n and

If
diverges, then
P

Theorem (Two of the Four Convergence Theorems)

Suppose
1

b n are infinite series and c is a nonzero constant.

P
P
P
P
If an converges, then ca n converges and ca n = c an . If an
P
diverges, then c a n diverges.
P

a n and

Theorem (Two of the Four Convergence Theorems)

Suppose
1

b n are infinite series and c is a nonzero constant.

P
P
P
P
If an converges, then ca n converges and ca n = c an . If an
P
diverges, then c a n diverges.
P
P
If an and bn differ only by a finite number of terms, then either
P

a n and

both converge or both diverge.

Theorem (Two of the Four Convergence Theorems)

Suppose
1

b n are infinite series and c is a nonzero constant.

P
P
P
P
If an converges, then ca n converges and ca n = c an . If an
P
diverges, then c a n diverges.
P
P
If an and bn differ only by a finite number of terms, then either
P

a n and

both converge or both diverge.

Simply put:

Theorem (Two of the Four Convergence Theorems)

Suppose
1

b n are infinite series and c is a nonzero constant.

P
P
P
P
If an converges, then ca n converges and ca n = c an . If an
P
diverges, then c a n diverges.
P
P
If an and bn differ only by a finite number of terms, then either
P

a n and

both converge or both diverge.

Simply put:
1

Convergence or divergence of a series is unaffected by multiplying the

terms of the series by a constant.

Theorem (Two of the Four Convergence Theorems)

Suppose
1

b n are infinite series and c is a nonzero constant.

P
P
P
P
If an converges, then ca n converges and ca n = c an . If an
P
diverges, then c a n diverges.
P
P
If an and bn differ only by a finite number of terms, then either
P

a n and

both converge or both diverge.

Simply put:
1

Convergence or divergence of a series is unaffected by multiplying the

terms of the series by a constant.

Convergence or divergence of a series is unaffected by deleting or

adding a finite number of terms.

Theorem (Additional Two of the Four Convergence Theorems)

1

Suppose

a n and

b n both converge. Then

Theorem (Additional Two of the Four Convergence Theorems)

1

Suppose
P

a n and

b n both converge. Then

(a n b n ) both converge and

P
(a n + b n ) and

Theorem (Additional Two of the Four Convergence Theorems)

1

P
P
P
Suppose an and bn both converge. Then (an + bn ) and
P
P
P
P
(a b n ) both converge and (a n + b n ) = a n + b n and
P n
P
P
(a n b n ) = a n b n

Theorem (Additional Two of the Four Convergence Theorems)

1

P
P
P
Suppose an and bn both converge. Then (an + bn ) and
P
P
P
P
(a b n ) both converge and (a n + b n ) = a n + b n and
P n
P
P
(a n b n ) = a n b n

If one series diverges and the other converges, then

Theorem (Additional Two of the Four Convergence Theorems)

1

P
P
P
Suppose an and bn both converge. Then (an + bn ) and
P
P
P
P
(a b n ) both converge and (a n + b n ) = a n + b n and
P n
P
P
(a n b n ) = a n b n
P
If one series diverges and the other converges, then (an + bn )

diverges.

Example
Determine whether or not the following series converge:
1

X
n=1

1
2n+1

2n
3n

Example
Determine whether or not the following series converge:
1

X
n=1

1
2n+1
n (convergent)
2n
3

Example
Determine whether or not the following series converge:
1

1
2n+1
n (convergent)
n
3
n=1 2

5
X
(1)n
+
n
n=1 n

Example
Determine whether or not the following series converge:
1

1
2n+1
n (convergent)
n
3
n=1 2

5
X
(1)n
+
(divergent)
n
n=1 n

Example
Determine whether or not the following series converge:
1

1
2n+1
n (convergent)
n
3
n=1 2

5
X
(1)n
+
(divergent)
n
n=1 n

What is the sum of the series in (1)?

Example
Determine whether or not the following series converge:
1

1
2n+1
n (convergent)
n
3
n=1 2

5
X
(1)n
+
(divergent)
n
n=1 n

What is the sum of the series in (1)?

Remark
If an and bn both diverge, then nothing can be said about the
P
convergence of (an + bn ).
P

Outline

Four Convergence Theorems

Absolute and Conditional Convergence

Absolute Convergence Tests

Root Test
Ratio Test

Motivation:
Given

a n , we consider the convergence of

|a n |.

Motivation:
Given

a n , we consider the convergence of

Definition
Let

a n be an infinite series.

|a n |.

Motivation:
Given

a n , we consider the convergence of

|a n |.

Definition
Let
1

a n be an infinite series.
P
a n is said to be absolutely convergent if

Motivation:
Given

a n , we consider the convergence of

|a n |.

Definition
Let
1

a n be an infinite series.
P
P
a n is said to be absolutely convergent if |a n | is convergent.

Motivation:
Given

a n , we consider the convergence of

|a n |.

Definition
Let
1
2

a n be an infinite series.
P
P
a n is said to be absolutely convergent if |a n | is convergent.
P
a n is said to be conditionally convergent if

Motivation:
Given

a n , we consider the convergence of

|a n |.

Definition
Let
1
2

a n be an infinite series.
P
P
a n is said to be absolutely convergent if |a n | is convergent.
P
P
a n is said to be conditionally convergent if a n is convergent,
P
but |an | is divergent.

Motivation:
Given

a n , we consider the convergence of

|a n |.

Definition
Let
1
2

a n be an infinite series.
P
P
a n is said to be absolutely convergent if |a n | is convergent.
P
P
a n is said to be conditionally convergent if a n is convergent,
P
but |an | is divergent.

Note
Walang conditionally divergent :)

Example
1

(1)n+1
X
2n
n=1

Example
1

(1)n+1
X
2n
n=1
(1)n+1
X

2n

n=1

Example
1

(1)n+1
X
2n
n=1
(1)n+1
1
X
X

=
2n
n
n=1
n=1 2

Example
1

(1)n+1
X
2n
n=1
(1)n+1
1
X
X

=
2n
n
n=1
n=1 2

(convergent)

Example
1

(1)n+1
X
2n
n=1
(1)n+1
1
X
X

=
2n
n
n=1
n=1 2

Thus,

(convergent)

(1)n+1
X
is absolutely convergent.
2n
n=1

Example
1

(1)n+1
X
2n
n=1
(1)n+1
1
X
X

=
2n
n
n=1
n=1 2

Thus,
2

(convergent)

(1)n+1
X
is absolutely convergent.
2n
n=1

(1)n
X
n
n=1

Example
1

(1)n+1
X
2n
n=1
(1)n+1
1
X
X

=
2n
n
n=1
n=1 2

Thus,
2

(convergent)

(1)n+1
X
is absolutely convergent.
2n
n=1

(1)n
X
n
n=1

(1)n
X

n=1

Example
1

(1)n+1
X
2n
n=1
(1)n+1
1
X
X

=
2n
n
n=1
n=1 2

Thus,
2

(convergent)

(1)n+1
X
is absolutely convergent.
2n
n=1

(1)n
X
n
n=1

(1)n
1
X
X

n =
n=1
n=1 n

Example
1

(1)n+1
X
2n
n=1
(1)n+1
1
X
X

=
2n
n
n=1
n=1 2

Thus,
2

(convergent)

(1)n+1
X
is absolutely convergent.
2n
n=1

(1)n
X
n
n=1

(1)n
1
X
X

n =
n=1
n=1 n

(divergent)

Example
1

(1)n+1
X
2n
n=1
(1)n+1
1
X
X

=
2n
n
n=1
n=1 2

Thus,
2

(1)n+1
X
is absolutely convergent.
2n
n=1

(1)n
X
n
n=1

Thus,

(convergent)

(1)n
1
X
X

n =
n=1
n=1 n

(divergent)

(1)n
X
is conditionally convergent.
n
n=1

Definition
Let
1
2

a n be an infinite series.
P
P
a n is said to be absolutely convergent if |a n | is convergent.
P
P
a n is said to be conditionally convergent if a n is convergent,
P
but |an | is divergent.

Definition
Let
1
2

a n be an infinite series.
P
P
a n is said to be absolutely convergent if |a n | is convergent.
P
P
a n is said to be conditionally convergent if a n is convergent,
P
but |an | is divergent.

Question
1

Is a conditionally convergent series convergent?

Definition
Let
1
2

a n be an infinite series.
P
P
a n is said to be absolutely convergent if |a n | is convergent.
P
P
a n is said to be conditionally convergent if a n is convergent,
P
but |an | is divergent.

Question
1

Is a conditionally convergent series convergent?

Is an absolutely convergent series convergent?

Theorem
If

a n is absolutely convergent, then

a n is convergent.

Theorem
If

a n is absolutely convergent, then

Let {an }
n=1 be a sequence.

a n is convergent.

Theorem
If

a n is absolutely convergent, then

Let {an }
n=1 be a sequence.
a 1 + a 2 + ... + a k

a n is convergent.

Theorem
If

a n is absolutely convergent, then

a n is convergent.

Let {an }
n=1 be a sequence.
a 1 + a 2 + ... + a k |a 1 + a 2 + ... + a k |

Theorem
If

a n is absolutely convergent, then

a n is convergent.

Let {an }
n=1 be a sequence.
a 1 + a 2 + ... + a k |a 1 + a 2 + ... + a k | |a 1 | + |a 2 | + ...|a k |

Theorem
If

a n is absolutely convergent, then

a n is convergent.

Let {an }
n=1 be a sequence.
a 1 + a 2 + ... + a k |a 1 + a 2 + ... + a k | |a 1 | + |a 2 | + ...|a k |

k
X
n=1

an

k
X
n=1

|a n |

Theorem
If

a n is absolutely convergent, then

a n is convergent.

Let {an }
n=1 be a sequence.
a 1 + a 2 + ... + a k |a 1 + a 2 + ... + a k | |a 1 | + |a 2 | + ...|a k |

k
X

an

n=1

lim

k
X

k n=1

k
X

|a n |

n=1

ak

lim

k
X

k n=1

|a k |

Theorem
If

a n is absolutely convergent, then

a n is convergent.

Let {an }
n=1 be a sequence.
a 1 + a 2 + ... + a k |a 1 + a 2 + ... + a k | |a 1 | + |a 2 | + ...|a k |

k
X

an

n=1

lim

k
X

k n=1

X
n=1

k
X

|a n |

n=1

ak

an

lim

k
X

k n=1

X
n=1

|a n |

|a k |

Theorem
If

a n is absolutely convergent, then

a n is convergent.

Let {an }
n=1 be a sequence.
a 1 + a 2 + ... + a k |a 1 + a 2 + ... + a k | |a 1 | + |a 2 | + ...|a k |

k
X

an

n=1

lim

k
X

k n=1

X
n=1

k
X

|a n |

n=1

ak

an

lim

k
X

k n=1

X
n=1

|a n |

|a k |
<

Problem
Draw a Venn diagram illustrating the relationship between the set of
convergent series, the set of absolutely convergent series, the set of
conditionally convergent series and the set of divergent series.

Problem
Draw a Venn diagram illustrating the relationship between the set of
convergent series, the set of absolutely convergent series, the set of
conditionally convergent series and the set of divergent series.

Problem
Draw a Venn diagram illustrating the relationship between the set of
convergent series, the set of absolutely convergent series, the set of
conditionally convergent series and the set of divergent series.
CONVERGENT DIVERGENT

Problem
Draw a Venn diagram illustrating the relationship between the set of
convergent series, the set of absolutely convergent series, the set of
conditionally convergent series and the set of divergent series.
CONVERGENT DIVERGENT
absolutely
convergent

conditionally
convergent

Outline

Four Convergence Theorems

Absolute and Conditional Convergence

Absolute Convergence Tests

Root Test
Ratio Test

Theorem (Root Test)

Let
1

a n be an infinite series.
p
If lim n |an | = L < 1, then

Theorem (Root Test)

Let
1

a n be an infinite series.
p
If lim n |an | = L < 1, then the series is absolutely convergent.

Theorem (Root Test)

Let
1

a n be an infinite series.
p
If lim n |an | = L < 1, then the series is absolutely convergent.
n
p
p
If lim n |an | = L > 1 or lim n |an | = , then

Theorem (Root Test)

Let
1

a n be an infinite series.
p
If lim n |an | = L < 1, then the series is absolutely convergent.
n
p
p
If lim n |an | = L > 1 or lim n |an | = , then the series is divergent.

Theorem (Root Test)

Let
1

a n be an infinite series.
p
If lim n |an | = L < 1, then the series is absolutely convergent.
n
p
p
If lim n |an | = L > 1 or lim n |an | = , then the series is divergent.
n
n
p
n
If lim |an | = 1, then

Theorem (Root Test)

Let
1

a n be an infinite series.
p
If lim n |an | = L < 1, then the series is absolutely convergent.
n
p
p
If lim n |an | = L > 1 or lim n |an | = , then the series is divergent.
n
n
p
n
If lim |an | = 1, then no conclusion can be made.

Theorem (Root Test)

Let
1

a n be an infinite series.
p
If lim n |an | = L < 1, then the series is absolutely convergent.
n
p
p
If lim n |an | = L > 1 or lim n |an | = , then the series is divergent.
n
n
p
n
If lim |an | = 1, then no conclusion can be made.

(Commit to memory)

Theorem (Root Test)

Let
1

a n be an infinite series.
p
If lim n |an | = L < 1, then the series is absolutely convergent.
n
p
p
If lim n |an | = L > 1 or lim n |an | = , then the series is divergent.
n
n
p
n
If lim |an | = 1, then no conclusion can be made.

(Commit to memory)

Theorem (Root Test)

Let
1

a n be an infinite series.
p
If lim n |an | = L < 1, then the series is absolutely convergent.
n
p
p
If lim n |an | = L > 1 or lim n |an | = , then the series is divergent.
n
n
p
n
If lim |an | = 1, then no conclusion can be made.

(Commit to memory)

Note
1

p
1
n
|a n | = |a n | /n

Theorem (Root Test)

Let
1

a n be an infinite series.
p
If lim n |an | = L < 1, then the series is absolutely convergent.
n
p
p
If lim n |an | = L > 1 or lim n |an | = , then the series is divergent.
n
n
p
n
If lim |an | = 1, then no conclusion can be made.

(Commit to memory)

Note
1

p
1
n
|a n | = |a n | /n

(1) Apply absolute value, (2) Get the n th root, (3) Get the limit.

TIP:
The Root Test is useful when the index of summation is a power that can
be "extracted".

Theorem (Root Test)

Let
1

a n be an infinite series.
p
If lim n |an | = L < 1, then the series is absolutely convergent.
n
p
p
If lim n |an | = L > 1 or lim n |an | = , then the series is divergent.
n
n
p
n
If lim |an | = 1, then no conclusion can be made.

Example
1

1
X
n
n=1 n

Theorem (Root Test)

Let
1

a n be an infinite series.
p
If lim n |an | = L < 1, then the series is absolutely convergent.
n
p
p
If lim n |an | = L > 1 or lim n |an | = , then the series is divergent.
n
n
p
n
If lim |an | = 1, then no conclusion can be made.

Example
1

1
X
(absolutely convergent)
n
n=1 n

Theorem (Root Test)

Let
1

a n be an infinite series.
p
If lim n |an | = L < 1, then the series is absolutely convergent.
n
p
p
If lim n |an | = L > 1 or lim n |an | = , then the series is divergent.
n
n
p
n
If lim |an | = 1, then no conclusion can be made.

Example
1

1
X
(absolutely convergent)
n
n=1 n

X
2n n
n
(1)
n +1
n=2

Theorem (Root Test)

Let
1

a n be an infinite series.
p
If lim n |an | = L < 1, then the series is absolutely convergent.
n
p
p
If lim n |an | = L > 1 or lim n |an | = , then the series is divergent.
n
n
p
n
If lim |an | = 1, then no conclusion can be made.

Example
1

1
X
(absolutely convergent)
n
n=1 n

X
2n n
n
(divergent)
(1)
n +1
n=2

Theorem (Root Test)

Let
1

a n be an infinite series.
p
If lim n |an | = L < 1, then the series is absolutely convergent.
n
p
p
If lim n |an | = L > 1 or lim n |an | = , then the series is divergent.
n
n
p
n
If lim |an | = 1, then no conclusion can be made.

Example
1

1
X
(absolutely convergent)
n
n=1 n

X
2n n
n
(divergent)
(1)
n +1
n=2
2n+1
X
2n
n=1 n

Theorem (Root Test)

Let
1

a n be an infinite series.
p
If lim n |an | = L < 1, then the series is absolutely convergent.
n
p
p
If lim n |an | = L > 1 or lim n |an | = , then the series is divergent.
n
n
p
n
If lim |an | = 1, then no conclusion can be made.

Example
1

1
X
(absolutely convergent)
n
n=1 n

X
2n n
n
(divergent)
(1)
n +1
n=2
2n+1
X
(absolutely convergent)
2n
n=1 n

Theorem (Ratio Test)

Let

a n be an infinite series for which each term is nonzero.

Theorem (Ratio Test)

Let
1

a n be an infinite series for which each term is nonzero.

a n+1

= L < 1, then
If lim

n a

Theorem (Ratio Test)

Let
1

a n be an infinite series for which each term is nonzero.

a n+1

= L < 1, then the series is absolutely convergent.

If lim

n a

Theorem (Ratio Test)

Let
1

a n be an infinite series for which each term is nonzero.

a n+1

= L < 1, then the series is absolutely convergent.

If lim
n a n

a n+1
a n+1

= , then
= L > 1 or lim
If lim
n a n
n a n

Theorem (Ratio Test)

Let
1

a n be an infinite series for which each term is nonzero.

a n+1

= L < 1, then the series is absolutely convergent.

If lim
n a n

a n+1
a n+1

= , then the series is divergent.

= L > 1 or lim
If lim
n a n
n a n

Theorem (Ratio Test)

Let
1

a n be an infinite series for which each term is nonzero.

a n+1

= L < 1, then the series is absolutely convergent.

If lim
n a n

a n+1
a n+1

= , then the series is divergent.

= L > 1 or lim
If lim
n a n
n a n

a n+1

= 1, then
If lim
n a n

Theorem (Ratio Test)

Let
1

a n be an infinite series for which each term is nonzero.

a n+1

= L < 1, then the series is absolutely convergent.

If lim
n a n

a n+1
a n+1

= , then the series is divergent.

= L > 1 or lim
If lim
n a n
n a n

a n+1

= 1, then no conclusion can be made.

If lim
n a n

Theorem (Ratio Test)

Let
1

a n be an infinite series for which each term is nonzero.

a n+1

= L < 1, then the series is absolutely convergent.

If lim
n a n

a n+1
a n+1

= , then the series is divergent.

= L > 1 or lim
If lim
n a n
n a n

a n+1

= 1, then no conclusion can be made.

If lim
n a n

(Commit to memory)

Theorem (Ratio Test)

Let
1

a n be an infinite series for which each term is nonzero.

a n+1

= L < 1, then the series is absolutely convergent.

If lim
n a n

a n+1
a n+1

= , then the series is divergent.

= L > 1 or lim
If lim
n a n
n a n

a n+1

= 1, then no conclusion can be made.

If lim
n a n

(Commit to memory)

Note
1

a n+1
1

a = |a n+1 ||a n |
n

(1) Apply the absolute value, (2) Simplify, (3) Get the limit.

TIP:
The Ratio Test is useful when the series involves factorials, or the index of
summation is a power that cannot be "extracted".

Example
1

1
X
n=1 n!

Example
1

1
X
(absolutely convergent)
n=1 n!

Example
1

1
X
(absolutely convergent)
n=1 n!

(1)n+1 3n
X
2
n=1 2n + 3

Example
1

1
X
(absolutely convergent)
n=1 n!

(1)n+1 3n
X
(divergent)
2
n=1 2n + 3

Example
1

1
X
(absolutely convergent)
n=1 n!

(1)n+1 3n
X
(divergent)
2
n=1 2n + 3

n5
X
n=1 (2n)!

Example
1

1
X
(absolutely convergent)
n=1 n!

(1)n+1 3n
X
(divergent)
2
n=1 2n + 3

n5
X
(absolutely convergent)
n=1 (2n)!

Remark
Two additional special series:
1

1
X
is convergent.
n=1 n!
1
X
n=1 n

is convergent.

These may come in handy...

Remark
To change the lower index of an infinite series:

X
n=k

f (n) =

These may come in handy...

Remark
To change the lower index of an infinite series:

X
n=k

f (n) =

X
n=k+1

These may come in handy...

Remark
To change the lower index of an infinite series:

X
n=k

f (n) =

X
n=k+1

f (n 1) =

These may come in handy...

Remark
To change the lower index of an infinite series:

X
n=k

f (n) =

X
n=k+1

f (n 1) =

X
n=k1

These may come in handy...

Remark
To change the lower index of an infinite series:

X
n=k

f (n) =

X
n=k+1

f (n 1) =

X
n=k1

f (n + 1)

These may come in handy...

Remark
To change the lower index of an infinite series:

X
n=k

Example
1

1
=
n=0 n + 1

f (n) =

X
n=k+1

f (n 1) =

X
n=k1

f (n + 1)

These may come in handy...

Remark
To change the lower index of an infinite series:

X
n=k

Example
1

X
1
=
n=0 n + 1
n=1

f (n) =

X
n=k+1

f (n 1) =

X
n=k1

f (n + 1)

These may come in handy...

Remark
To change the lower index of an infinite series:

f (n) =

n=k

Example
1

1
X
1
=
n=0 n + 1
n=1

X
n=k+1

f (n 1) =

X
n=k1

f (n + 1)

These may come in handy...

Remark
To change the lower index of an infinite series:

f (n) =

n=k

Example
1

1
X
1
=
n=0 n + 1
n=1 n

X
n=k+1

f (n 1) =

X
n=k1

f (n + 1)

These may come in handy...

Remark
To change the lower index of an infinite series:

f (n) =

n=k

Example
1

1
X
1
=
n=0 n + 1
n=1 n

X
ar n1 =
n=1

X
n=k+1

f (n 1) =

X
n=k1

f (n + 1)

These may come in handy...

Remark
To change the lower index of an infinite series:

f (n) =

n=k

Example
1

1
X
1
=
n=0 n + 1
n=1 n

X
X
ar n1 =
n=1

n=0

X
n=k+1

f (n 1) =

X
n=k1

f (n + 1)

These may come in handy...

Remark
To change the lower index of an infinite series:

f (n) =

n=k

Example
1

1
X
1
=
n=0 n + 1
n=1 n

X
X
ar n1 =
ar
n=1

n=0

X
n=k+1

f (n 1) =

X
n=k1

f (n + 1)

These may come in handy...

Remark
To change the lower index of an infinite series:

f (n) =

n=k

Example
1

1
X
1
=
n=0 n + 1
n=1 n

X
X
ar n1 =
ar n
n=1

n=0

X
n=k+1

f (n 1) =

X
n=k1

f (n + 1)

These may come in handy...

Remark
To find the common ratio and first term of a geometric series, put it in the
form

n=0

ar n

These may come in handy...

Remark
To find the common ratio and first term of a geometric series, put it in the
form

n=0

Example
1

2n+1
X
n
n=0 3
1
X
n
n=1 2
(1)n+1 5n1
X
n=1

3n 2n+1

ar n

Proof: Convergence of Geometric Series

Case
r > 0 and r 6= 1

Note: If 0 < r < 1

lim r n+1 =

Proof: Convergence of Geometric Series

Case
r > 0 and r 6= 1

Note: If 0 < r < 1

lim r n+1 = 0

Proof: Convergence of Geometric Series

Case
r > 0 and r 6= 1

Note: If 0 < r < 1

lim r n+1 = 0

If r > 1

lim r n+1 =

Proof: Convergence of Geometric Series

Case
r > 0 and r 6= 1

Note: If 0 < r < 1

lim r n+1 = 0

If r > 1

lim r n+1 =

Proof: Convergence of Geometric Series

Given

X
n=0

ar n such that r > 0 and r 6= 1.

Proof: Convergence of Geometric Series

Given

X
n=0

ar n such that r > 0 and r 6= 1. The n th partial sum of is

Proof: Convergence of Geometric Series

Given

ar n such that r > 0 and r 6= 1. The n th partial sum of is

n=0

sn =

a(1 r n )
1r

Proof: Convergence of Geometric Series

Given

ar n such that r > 0 and r 6= 1. The n th partial sum of is

n=0

sn =

When does the SPS converge?

a(1 r n )
1r

Proof: Convergence of Geometric Series

Given

ar n such that r > 0 and r 6= 1. The n th partial sum of is

n=0

sn =

When does the SPS converge?

lim s n =

a(1 r n )
1r

Proof: Convergence of Geometric Series

Given

ar n such that r > 0 and r 6= 1. The n th partial sum of is

n=0

sn =

a(1 r n )
1r

When does the SPS converge?

a(1 r n )
=
n
1r

lim s n = lim

Proof: Convergence of Geometric Series

Given

ar n such that r > 0 and r 6= 1. The n th partial sum of is

n=0

sn =

a(1 r n )
1r

When does the SPS converge?

a(1 r n )
a
=
lim (1 r n )
n
1r
1 r n

lim s n = lim

Proof: Convergence of Geometric Series

Given

ar n such that r > 0 and r 6= 1. The n th partial sum of is

n=0

sn =

a(1 r n )
1r

When does the SPS converge?

a(1 r n )
a
=
lim (1 r n )
n
1r
1 r n

lim s n = lim

If 0 < r < 1:

Proof: Convergence of Geometric Series

Given

ar n such that r > 0 and r 6= 1. The n th partial sum of is

n=0

sn =

a(1 r n )
1r

When does the SPS converge?

a(1 r n )
a
=
lim (1 r n )
n
1r
1 r n

lim s n = lim

If 0 < r < 1:
lim s n =

Proof: Convergence of Geometric Series

Given

ar n such that r > 0 and r 6= 1. The n th partial sum of is

n=0

sn =

a(1 r n )
1r

When does the SPS converge?

a(1 r n )
a
=
lim (1 r n )
n
1r
1 r n

lim s n = lim

If 0 < r < 1:
lim s n =

a
lim (1 r n ) =
1 r n

Proof: Convergence of Geometric Series

Given

ar n such that r > 0 and r 6= 1. The n th partial sum of is

n=0

sn =

a(1 r n )
1r

When does the SPS converge?

a(1 r n )
a
=
lim (1 r n )
n
1r
1 r n

lim s n = lim

If 0 < r < 1:
lim s n =

a
a
lim (1 r n ) =
1 r n
1r

Proof: Convergence of Geometric Series

Given

ar n such that r > 0 and r 6= 1. The n th partial sum of is

n=0

sn =

a(1 r n )
1r

When does the SPS converge?

a(1 r n )
a
=
lim (1 r n )
n
1r
1 r n

lim s n = lim

If 0 < r < 1:
lim s n =

a
a
lim (1 r n ) =
1 r n
1r

The series converges in this case.

Proof: Convergence of Geometric Series

Given

ar n such that r > 0 and r 6= 1. The n th partial sum of is

n=0

sn =

a(1 r n )
1r

When does the SPS converge?

a(1 r n )
a
=
lim (1 r n )
n
1r
1 r n

lim s n = lim

If 0 < r < 1:
lim s n =

a
a
lim (1 r n ) =
1 r n
1r

The series converges in this case.

If r > 1:

Proof: Convergence of Geometric Series

Given

ar n such that r > 0 and r 6= 1. The n th partial sum of is

n=0

sn =

a(1 r n )
1r

When does the SPS converge?

a(1 r n )
a
=
lim (1 r n )
n
1r
1 r n

lim s n = lim

If 0 < r < 1:
lim s n =

a
a
lim (1 r n ) =
1 r n
1r

The series converges in this case.

If r > 1:
lim s n =

Proof: Convergence of Geometric Series

Given

ar n such that r > 0 and r 6= 1. The n th partial sum of is

n=0

sn =

a(1 r n )
1r

When does the SPS converge?

a(1 r n )
a
=
lim (1 r n )
n
1r
1 r n

lim s n = lim

If 0 < r < 1:
lim s n =

a
a
lim (1 r n ) =
1 r n
1r

The series converges in this case.

If r > 1:
lim s n =

a
lim (1 r n ) =
1 r n

Proof: Convergence of Geometric Series

Given

ar n such that r > 0 and r 6= 1. The n th partial sum of is

n=0

sn =

a(1 r n )
1r

When does the SPS converge?

a(1 r n )
a
=
lim (1 r n )
n
1r
1 r n

lim s n = lim

If 0 < r < 1:
lim s n =

a
a
lim (1 r n ) =
1 r n
1r

The series converges in this case.

If r > 1:
lim s n =

a
lim (1 r n ) =
1 r n

Proof: Convergence of Geometric Series

Given

ar n such that r > 0 and r 6= 1. The n th partial sum of is

n=0

sn =

a(1 r n )
1r

When does the SPS converge?

a(1 r n )
a
=
lim (1 r n )
n
1r
1 r n

lim s n = lim

If 0 < r < 1:
lim s n =

a
a
lim (1 r n ) =
1 r n
1r

The series converges in this case.

If r > 1:
lim s n =

a
lim (1 r n ) =
1 r n

The series diverges in this case.

Summary

What are the four convergence theorems?

Summary

What are the four convergence theorems?

What is the difference between general convergence, absolute

convergence and conditional convergence?

Summary

What are the four convergence theorems?

What is the difference between general convergence, absolute

convergence and conditional convergence?
What the two absolute convergence tests were taken up today?

Summary

What are the four convergence theorems?

What is the difference between general convergence, absolute

convergence and conditional convergence?
What the two absolute convergence tests were taken up today?

When is it useful to use the root test?

Summary

What are the four convergence theorems?

What is the difference between general convergence, absolute

convergence and conditional convergence?
What the two absolute convergence tests were taken up today?

1
2

When is it useful to use the root test?

When is it useful to use the ratio test?

END OF CHAPTER 1 :)

END OF CHAPTER 1 :)
Announcement:
Chapter 1 Quiz will be on December 12, Thursday (Lecture Class)