INFINITE SEQUENCES AND SERIES

E LECTRONIC VERSION OF LECTURE

Hoang Hai Ha
HoChiMinh City University of Technology
Faculty of Applied Science, Department of Applied Mathematics
Email: hoanghaiha@hcmut.edu.v

September 10, 2020

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O UTLINE

1 S ERIES

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O UTLINE

1 S ERIES

2 N ON - NEGATIVE SERIES

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O UTLINE

1 S ERIES

2 N ON - NEGATIVE SERIES

3 D’ ALAMBERT AND C AUCHY TEST FOR CONVERGENCE

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O UTLINE

1 S ERIES

2 N ON - NEGATIVE SERIES

3 D’ ALAMBERT AND C AUCHY TEST FOR CONVERGENCE

4 A LTERNATING S ERIES

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O UTLINE

1 S ERIES

2 N ON - NEGATIVE SERIES

3 D’ ALAMBERT AND C AUCHY TEST FOR CONVERGENCE

4 A LTERNATING S ERIES

5 P OWER SERIES

Hoang Hai Ha (HCMUT-OISP) INFINITE SEQUENCES AND SERIES September 10, 2020 2 / 32
O UTLINE

1 S ERIES

2 N ON - NEGATIVE SERIES

3 D’ ALAMBERT AND C AUCHY TEST FOR CONVERGENCE

4 A LTERNATING S ERIES

5 P OWER SERIES

6 S UM OF POWER SERIES

Hoang Hai Ha (HCMUT-OISP) INFINITE SEQUENCES AND SERIES September 10, 2020 2 / 32
 Series

S ERIES

D EFINITION 1.1
Series is an infinite sequence:

a 1 + a 2 + ... + a n + ....
∞
P
is denoted by : an
n=1

Question: How do we find this sum?

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 Series

E XAMPLE 1.1
1 + 2 + 3 + .... + n + .... ⇒ it would be impossible.

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 Series

E XAMPLE 1.1
1 + 2 + 3 + .... + n + .... ⇒ it would be impossible.
1
2
+ 14 + 18 + 16
1
+ ... + 21n + ...

∞
P 1
It is reasonable to write that 2n
=1
n=1

Hoang Hai Ha (HCMUT-OISP) INFINITE SEQUENCES AND SERIES September 10, 2020 4 / 32
 Series

PARTIAL SUM

D EFINITION 1.2
We define the partial sum as below

S 1 =a 1
S 2 =a 1 + a 2
S 3 =a 1 + a 2 + a 3
..
.
S n =a 1 + a 2 + ... + a n

Partial sums generate sequence S n .

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 Series

C ONVERGENCE

D EFINITION 1.3
∞
P
If lim S n = S is finite, then the series a n is called
n→∞ n=1
∞
P
convergent and denoted: a n = S. The number S is
n=1
called SUM of the series. If lim S n = ∞ or does not exist,
n→∞
∞
P
then a n is called divergent.
n=1

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 Series

E XAMPLE 1.2
∞
P 1
Find the sum of
n=1 n(n + 1)

S OLUTION

1 1 1
Sn = + +···
1·2 2·3 n(n + 1)
1 1 1 1 1 1
= − + − + ... + −
1 2 2 3 n n +1
1
=1 − =1
n +1
∞
P 1
⇒ =1
n=1 n(n + 1)

Hoang Hai Ha (HCMUT-OISP) INFINITE SEQUENCES AND SERIES September 10, 2020 7 / 32
 Series

E XAMPLE 1.3
∞
P
Prove that n is divergent.
n=1

S OLUTION
n(n + 1)
S n = 1 + 2 + ... + n =
2

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 Series

E XAMPLE 1.3
∞
P
Prove that n is divergent.
n=1

S OLUTION
n(n + 1)
S n = 1 + 2 + ... + n =
2

lim S n = ∞
n→∞
∞
P
Then n is divergent.
n=1

Hoang Hai Ha (HCMUT-OISP) INFINITE SEQUENCES AND SERIES September 10, 2020 8 / 32
 Series

G EOMETRIC SERIES
∞
qn
P
C ONSIDER THE SERIES
n=1
2 n
S n = q + q + ... + q

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 Series

G EOMETRIC SERIES
∞
qn
P
C ONSIDER THE SERIES
n=1
2 n
S n = q + q + ... + q
q(1 − q n )
Sn =
1−q

0 if |q| < 1


lim q n = ∞ if q ≥ 1
n→∞ 

not exits if q ≤ −1

Hoang Hai Ha (HCMUT-OISP) INFINITE SEQUENCES AND SERIES September 10, 2020 9 / 32
 Series

G EOMETRIC SERIES
∞
qn
P
C ONSIDER THE SERIES
n=1
2 n
S n = q + q + ... + q
q(1 − q n )
Sn =
1−q

0 if |q| < 1


lim q n = ∞ if q ≥ 1
n→∞ 

not exits if q ≤ −1
∞
q
qn =
X
if |q| < 1
n=1 1−q
Hoang Hai Ha (HCMUT-OISP) INFINITE SEQUENCES AND SERIES September 10, 2020 9 / 32
 Series

E XAMPLE 1.4
∞ 2n + 3n
P
Find the sum of n+1
n=1 5

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 Series

N ECESSARY CONDITION FOR CONVERGENCE

T HEOREM 1.1
∞
P
If a n is convergent then lim a n = 0.
n=1 n→∞
∞
P
Theorem means: for any series a n , if lim a n 6= 0 then
n=1 n→∞
series is divergent.
E XAMPLE 1.5
∞
P n2
Show that 2
is divergent.
n=1 5n + 4

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 Series

P ROPERTIES

+∞
P
1 The series a n is conv/div together with the series
n=1
+∞
P
a n , (n 0 > 1) is convergent.
n=n 0

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 Series

P ROPERTIES

+∞
P
1 The series a n is conv/div together with the series
n=1
+∞
P
a n , (n 0 > 1) is convergent.
n=n 0
+∞
P
2 The series a n is conv/div together with
n=1
+∞
α.a n (α 6= 0 ∈ R)
P
n=1
+∞ +∞
α.a n = α.
P P
an .
n=1 n=1

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 Series

+∞
P +∞
P
3 If the series a n and the series b n are convergent
n=1 n=1
and have the sums S 1 , S 2 respectively then the series
+∞
P
(a n + b n ) is also convergent and has the sum
n=1
S 1 + S 2.
+∞
X +∞
X +∞
X
(a n + b n ) = an + bn .
n=1 n=1 n=1

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 Non-negative series Definition

N ON - NEGATIVE SERIES

D EFINITION 2.1
+∞
P
The series a n is called the non-negative series if
n=1
a n Ê 0, n ∈ N.
∞
P
Note: Non-negative series a n is divergent if and only if
n=1
lim S n = +∞.
n→∞

Hoang Hai Ha (HCMUT-OISP) INFINITE SEQUENCES AND SERIES September 10, 2020 14 / 32
 Non-negative series Some Basic Series

S OME B ASIC S ERIES

+∞
q n is convergent if |q| < 1 and is divergent if
P
1

n=1
|q| Ê 1.(Geometric series)
P 1
+∞
2
α
is convergent if α > 1 and is divergent if α É 1.
n=1 n

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 Non-negative series The Comparision Test

T HEOREM 2.1 ( T HE C OMPARISION T EST )

+∞
P +∞
P
Suppose that a n and b n are series with
n=1 n=1
non-negative terms:

0 É a n É b n , ∀n Ê n 0
+∞
P +∞
P
1 If b n is convergent, then a n is also convergent.
n=1 n=1
+∞
P +∞
P
2 If a n is divergent, then b n is also divergent.
n=1 n=1

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 Non-negative series The Comparision Test

E XAMPLE 2.1
n2 + 1 +∞
P
Determine whether the series 4
converges or
n=1 n + n + 1
diverges.

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 Non-negative series The Comparision Test

T HEOREM 2.2
+∞
P +∞
P
Suppose that a n and
b n are series with
n=1 n=1
an
non-negative terms, K = lim
n→+∞ b n
+∞
P +∞
P
1 K > 0. Either both series a n and b n converge or
n=1 n=1
both diverge.

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 Non-negative series The Comparision Test

T HEOREM 2.2
+∞
P +∞
P
Suppose that a n and
b n are series with
n=1 n=1
an
non-negative terms, K = lim
n→+∞ b n
+∞
P +∞
P
1 K > 0. Either both series a n and b n converge or
n=1 n=1
both diverge.
+∞
P +∞
P
2 K = 0. If b n is convergent then a n is convergent.
n=1 n=1

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 Non-negative series The Comparision Test

T HEOREM 2.2
+∞
P +∞
P
Suppose that a n and
b n are series with
n=1 n=1
an
non-negative terms, K = lim
n→+∞ b n
+∞
P +∞
P
1 K > 0. Either both series a n and b n converge or
n=1 n=1
both diverge.
+∞
P +∞
P
2 K = 0. If b n is convergent then a n is convergent.
n=1 n=1
+∞
P +∞
P
3 K = +∞. If b n is divergent then a n is divergent.
n=1 n=1

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 Non-negative series The Comparision Test

D EFINITION 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)

x→a
We denote the equivalent functions by f (x) ∼ g (x).

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 Non-negative series The Comparision Test

T HE BASIC EQUIVALENCE

lnα x << x β << a x , (α, β > 0, a > 1)

T HEOREM 2.3
Suppose the functions f 1 , f 2 , ..., f n approach to ∞, and f k is
the one that approaches fastest( biggest), then the sum
f 1 + f 2 + ... + f n ∼ f k

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 Non-negative series The Comparision Test

E XAMPLE 2.2
Determine whether the following series converges or
diverges.
X e n + n3
+∞

n=1 2
n + ln3 n
+∞
X 2n 2 + 5n + 1
p
6 2
n=1 n + 3n + 2

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 D’alambert and Cauchy test for convergence

D’ ALAMBERT AND C AUCHY TEST FOR CONVERGENCE

D’ ALAMBERT AND C AUCHY TEST

¯ ¯
∞
P ¯ a n+1 ¯
Given series a n , define D := lim ¯¯ ¯ or
p n=1 n→∞ an ¯
D := lim n |a n |
n→∞
1 D < 1 ⇒ series is convergent.

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 D’alambert and Cauchy test for convergence

D’ ALAMBERT AND C AUCHY TEST FOR CONVERGENCE

D’ ALAMBERT AND C AUCHY TEST

¯ ¯
∞
P ¯ a n+1 ¯
Given series a n , define D := lim ¯¯ ¯ or
p n=1 n→∞ an ¯
D := lim n |a n |
n→∞
1 D < 1 ⇒ series is convergent.
2 D > 1 ⇒ series is divergent.

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 D’alambert and Cauchy test for convergence

D’ ALAMBERT AND C AUCHY TEST FOR CONVERGENCE

D’ ALAMBERT AND C AUCHY TEST

¯ ¯
∞
P ¯ a n+1 ¯
Given series a n , define D := lim ¯¯ ¯ or
p n=1 n→∞ an ¯
D := lim n |a n |
n→∞
1 D < 1 ⇒ series is convergent.
2 D > 1 ⇒ series is divergent.
3 D = 1 No conclusion.

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 D’alambert and Cauchy test for convergence

E XAMPLE 3.1
Determine whether the following series conv or div:
∞ n
X 2
n=1 n!

∞ n2
n (n + 1)
X
(−1) 2
n=1 n n 2n

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 Alternating Series

D EFINITION 4.1
+∞
(−1)n a n , (a n Ê 0, ∀n Ê n 0 ) is called the
P
The series
n=1
alternating series.

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 Alternating Series

D EFINITION 4.1
+∞
(−1)n a n , (a n Ê 0, ∀n Ê n 0 ) is called the
P
The series
n=1
alternating series.

T HEOREM 4.1 (A LTERNATING S ERIES T EST )

+∞
(−1)n a n satisfies
P
If the alternating series
n=1
1 lim a n = 0
n→+∞
2 a n is decreasing sequence.
then the series is convergent.

Hoang Hai Ha (HCMUT-OISP) INFINITE SEQUENCES AND SERIES September 10, 2020 24 / 32
 Alternating Series

E XAMPLE 4.1
+∞ 1
(−1)n+1 p for convergence or
P
Test the series
n=1 n
divergence.

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 Alternating Series

E XAMPLE 4.1
+∞ 1
(−1)n+1 p for convergence or
P
Test the series
n=1 n
divergence.

E XAMPLE 4.2
+∞ ln n
(−1)n+1
P
Test the series for convergence or
n=1 n
divergence.

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 Power series

P OWER SERIES

D EFINITION 5.1
∞
a n (x − x 0 )n is called power
P
The series in the form
n=1
series.

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 Power series

P OWER SERIES

D EFINITION 5.1
∞
a n (x − x 0 )n is called power
P
The series in the form
n=1
series.

T HEOREM 5.1 (R ADIUS OF CONVERGENCE )

∞
a n (x − x 0 )n . A non negative number
P
Given power series
n=1
R is called radius of convergence of power series if the
series convergent if |x − x 0 | < R and divergent if
|x − x 0 | > R.

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 Power series

F INDING RADIUS OF CONVERGENCE

D’ ALAMBERT OR C AUCHY FORMULA

¯ ¯
¯ an ¯
R = lim ¯¯ ¯
n→∞ a n+1 ¯

1
R = lim p
n→∞ n |a |
n

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 Power series

E XAMPLE 5.1
Determine radius of convergence for the following series
∞ n(x + 2)n
P
1 .
n=1 3n+1
∞
P n x 2n
2 (−1) 2n
n=1 2 (n!)2
∞
n!x n .
P
3

n=1

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 Sum of power series

S UM OF THE SERIES
∞
a n (x − x 0 )n has radius
P
Consider that the power series
n=1
convergence R, then for x ∈ (x 0 − R, x 0 + R), if
∞
a n (x − x 0 )n .
X
S(x) =
n=1

then S(x) is differentiable on interval (x 0 − R, x 0 + R) then

∞
na n (x − x 0 )n−1 .
X
S 0 (x) =
n=1

∞
an
Z
(x − x 0 )n+1 .
X
S(x)d x =
n=1 n + 1

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 Sum of power series

E XAMPLE 6.1
∞
P 1
Find the sum of n
.
n=1 n6

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 Sum of power series

E XAMPLE 6.1
∞
P 1
Find the sum of n
.
n=1 n6

E XAMPLE 6.2
∞ n +2
(−1)n
P
Find the sum of .
n=1 3n

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 Sum of power series

S OLUTION FOR E XAMPLE (6.1)

1 ∞
P
Denote: A = n
and consider power series:
n=1 n6
∞ xn
P 1
S(x) = , then A = S( ).
n=1 n 6
We have:
∞ xn
µ ¶
P 1
S(x) = ⇒ A=S .
n=1 n 6
∞ 1
Taking derivative both sides: S 0 (x) = x n−1 =
P
.
n=1 1−x
Then µS(x)
¶ = − ln |1 − x| +C , since S(0) = 0 then C = 0.
1 5
A=S = − ln .
6 6
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 Sum of power series

S OLUTION FOR EXAMPLE (6.2)

∞ n +2
(−1)n
P
Denote: A = and take power series
n=1 3n µ ¶
∞
P n 1
S(x) = (n + 2)x , then A = S − . We have:
n=1 3
∞ ∞
(n + 1)x n + x n = S 1 (x) + S 2 (x).
P P
S(x) =
n=1 n=1
¶0 ³
x
µ ´0
∞ ∞
P n P n+1
S 1 (x) = (n + 1)x = x = −x =
n=1 n=1 1−x
1
− 1.
(1 − x)2
1 x
Then S(x) = − 1 +
(1 − x)2 1−x
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