HUT – DEPARTMENT OF MATH.

APPLIED
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GIẢI TÍCH 2

LEC 7: INFINITE SERIES

STEWART: PAGES 687 – 723

• PhD. NGUYEÃN QUOÁC LAÂN (November, 2014)

 SUM OF SERIES
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∞
Sequence : Infinite term a1 , ... an ... . Series : Sum ∑ an = a1 + ... + an + ...
n =1

Try to add the terms of 2 following series. Get your remark:

1 1 1 1
a / + + +L+ n +L b /1 −1 + 1 − L
2 4 8 2
1 1 1 3 1
a/ S1 = , S 2 = + = (= 1 − ),
2 2 4 4 4
1 1 1 7 1
S3 = + + = (= 1 − ) ...
2 4 8 8 8
1 1 1
⇒ Sn = 1 − n → 1 ⇒ + L + n + ... = 1
2 2 2
⎧1, n = 2k + 1
b/ Sn = ⎨ ⇒ ∃ lim S n : No sum!
⎩0, n = 2k n →∞
 DEFINITION
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Summary: To determine whether or not a series Σan has a sum,

we consider the partial sums: S1 = a1, S2 = a1 + a2, Sn = a1 + a2 + a3.
Generally: Sn = a1 + a2 + … + an ⇒ Three cases can happen:
1/ ∃ lim S n = S : finite number 2/ lim S n = ∞ 3/ ∃ lim S n
n →∞ n →∞ n →∞

Series converges & sum = S Diverges (sum = ∞ or no sum)

Definition: Given a series Σan = a1 + a2 + … + an + …, let Sn

n
denote its nth partial sum: S n = a1 + a 2 + K + a n = ∑ a k
k =1
If the sequence {Sn} converges and lim Sn = S exits as a finite
number, the series Σan is called convergent and we write: Σan
= S, which is called the sum of the series. Otherwise: divergent
 GEOMETRIC SERIES
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∞
Geometric series: ∑ aq n
= a + aq + aq 2
+ L + aq n
+L
n=0
a first term
Convergent ⇔ | q | < 1 and its sum = =
1 − q 1 − common ration
∞
1
Remember: q < 1 ⇒ 1 + q + q + L + q + L = ∑ q =
2 n n

n=0 1− q

10 20 40
Example : Is this series 5 − + − + L geometric? Find its sum
3 9 27
a − 10 3 20 9 − 40 27 2
Answer : Evaluate n+1 : = = =− =q⇒
an 5 − 10 3 20 9 3
2 5
Geometric series. q = < 1 ⇒ Convergent. Sum S = =3
3 1 − (− 2 3)
∞
4
Example : Is the series ∑ 2 2 n31−n convergent or divergent? q = : Diver.
n =1 3
 THE DIVERGENCE TEST
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∞ ∞
lim a n ≠ 0 ⇒ ∑ a n diverges (But lim a n = 0 : Nothing for ∑ an )
n→∞ n =1 n→ ∞ n =1
∞
n2 n2
Example: Study convergence of ∑ Answer : lim 2 ≠0
n =1 n →∞ n + 4 n2 + 4
⎧ f ( x ) − elementary
Recall for limit : + Substitution : ⎨ ⇒ lim f ( x ) = f (a )
⎩ f (a ) : defined x →a

α ⎧ a n
→ +∞
+ n → +∞ ⇒ α > 0 : n → +∞; q < 1 ⇒ q → 0; a > 1 ⇒ ⎨
n

⎩log a n → +∞
Pn ( x ) a0 x n + ... a0 x n Highest
+ Rational : lim = lim = lim = lim
x→∞ Qm ( x ) x→∞ b x m + ... x→∞ b x m x→∞ Highest
0 0

ln(1 + x )
n
⎛ 1⎞ sin x 1 − cos x 1 e x −1
+ lim ⎜1 + ⎟ = e; lim = 1; lim = ; lim = 1; lim =1
n →∞ ⎝ n⎠ x →0 x x →0 x 2 2 x →0 x x →0 x
 EXAMPLE
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∞ ∞ ∞
n+2 n 2 + 3n + 4 n3 + 1
Study convergence : a/ ∑ b/ ∑ 2 c/ ∑ n2 + 1
n =1 2n − 1 n =1 3n − 4n + 7 n =1
∞ ∞ ∞ n ∞ n
n2 + 1 n+2 ⎛ 5⎞ ⎛ 2n − 1 ⎞
d/ ∑ e/ ∑ ln f/ ∑ ⎜1 − ⎟ g/ ∑ ⎜ ⎟
16n + 1 n =1 3n + 1 n =1 ⎝ n⎠ n =1 ⎝ 2n + 3 ⎠
4
n =1
∞ ∞ ∞
h/ ∑ n sin
1
i/ ∑ n 2 ⎛⎜1 − cos ⎞⎟
2
(
j/ ∑ n n e − 1 )
n =1 n n =1 ⎝ n⎠ n =1

1 1 1 1
Answer : Divergence test as lim an = a/ b/ c/ ∞ d/ e/ ln f/ e −5
n →∞ 2 3 4 3
−4n
⎡ 2 n +3 ⎤ 2 n +3 −4 n
n − lim
⎛ 2n − 1 ⎞ ⎢⎛ 4 ⎞ 4 ⎥ n →∞ 2 n + 3
g/ lim ⎜ ⎟ = lim ⎜1 − ⎟ =e = e −2 h/1 i/2 j/1
n →∞ ⎝ 2 n + 3 ⎠ n→∞ ⎢⎝ 2n + 3 ⎠ ⎥
⎣ ⎦
 INTEGRAL TEST
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⎧ f (x ) ≥ 0 ∀ x ≥ N
Series Σ f(n), where f(x) satisfies: ⎨ .
⎩ f ( x ) is decreasing in [N, + ∞ )
+∞
In this case, the series ∑ f (n ) and the improper integral ∫ f ( x )dx
A
will converge or diverge at the same times.

Riemann Improper Study

+∞ ∞
1
dx a/ ∑
Integral ∫ x p
is
n =1 n2
1
convergent ⇔ p > 1 ∞
1
b/ ∑ n ln n
n=2
 ANSWER
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⎧ f (x ) ≥ 0 ∀ x ≥ N
Series Σ f(n), where f(x) satisfies: ⎨ .
⎩ f ( x ) is decreasing in [N, + ∞ )
+∞
⇒ Series ∑ f (n ) & Integral ∫ f ( x )dx : same converge or diverge
A
+∞
∞ 1 dx
⎧
⎪ a/ ∑
1 a/ f ( x ) = 2
> 0 & ↓ at [1; + ∞ ) .As ∫ 2
: Riemann, p = 2 > 1
⎪
n2 x 1 x
⎪ n =1
⎨
⎪ ∞
1 ⇒ Improper integral converges ⇒ Series given converges.
⎪
⎪
⎩
b/ ∑ n ln n 1
+∞
dx
n=2 b/ f ( x ) = > 0, x ≥ 2 & ↓ at [2; + ∞ ). For ∫ : t = ln x
x ln x 2
x ln x
+∞ +∞
dx dt 1
⇒ ∫ = ∫ : p = 1 ⇒ Diverges ⇒ ∑ : Diverges
2
x ln x ln 2 t n ln n
 IMPORTANT: p – SERIES (RIEMANN SERIES)
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∞
1
For what values of p, the following series is convergent? ∑ np
n =1
+∞
dx 1
For p > 0 → Integral test : ∫ x p
: converges ⇔ p > 1; p ≤ 0 : lim p ≠ 0.
n →∞ n
1
∞
1
The next series is called p – series: ∑ n p
converges ⇔ p > 1
n =1

1 1 1 1 1 1
Study convergence : a/ 1 + + + ... + 3 + ... b/ 1 + 3 + 3 + ... + + ...
3
8 64 n 4 9 n2
1 1 1
Answer : a/ 1 + + + ... + 3 + ... : p - series, p = 3 > 1 → Converges
8 64 n
1 1 1 2
b/ 1 + 3 + 3 + ... + + ... : p - series, p = < 1 → Diverges
4 9 3 2 3
n
 NONNEGATIVE SERIES: COMPARISON TEST 1
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Like improper integral: Nonnegative series → Comparison Test.

∞ ∞
Suppose ∑ u n , ∑ vn : nonnegativ es series with u n ≥ vn ≥ 0 ∀n ≥ N 0
n =1 n =1
∞ ∞
(a) ∑ u n (BIG) converges ⇒ ∑ vn (SMALL) converges.
n =1 n =1
∞ ∞
(b) ∑ vn (SMALL) diverges ⇒ ∑ u n (BIG) diverges.
n =1 n =1

1
Remark : Compare ∑ un to ∑ q n : Geometric or ∑ p
: p - series.
n
∞ ∞ ∞
1 1 1
Example : Study convergence : a/ ∑ b/ ∑ c/ ∑
n =1 3 ⋅ 2 + 1 n =1 2 n − 1 n =1 2 n + 1
n
 ANSWER
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⎧∞ ∞
⎪ ∑ u n converges ⇒ ∑ v n converges.
⎪ n =1 n =1
u n ≥ vn ≥ 0 ∀n ≥ N 0 : ⎨
∞ ∞
⎪ v diverges ⇒
⎪∑ n ∑ u n diverges.
⎩ n =1 n =1

∞ ∞ ∞
1 1 1
Example : Study convergence : a/ ∑ b/ ∑ c/ ∑
n =1 3 ⋅ 2 + 1 n =1 2 n − 1 n =1 2 n + 1
n

∞
1 ∞ ⎛1⎞
n
1 1 1 1
Answer : a/ <
3⋅ 2 +1 3⋅ 2
& ∑ = ∑ ⎜ ⎟
3 n=1 ⎝ 2 ⎠
: q = < 1 → Converges
n =1 3 ⋅ 2
n n n 2
∞
1 1 1 1 1 1
b/ > &∑ : p = < 1 → Diverges c/ < → ???
2 n − 1 2 n n =1 2 n 2 2 n +1 2 n
 NONNEGATIVE SERIES: LIMIT COMPARISON TEST
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∞ ∞ Converge
un
∑ u n , ∑ vn : u n , vn > 0 & nlim
→ ∞ vn
= l ≠ 0, + ∞ ⇒ Same
Diverge
n =1 n =1

1
Choose vn = p
(p - series), using the following rules :
n
⎧ Polynomial ao x n + ... : Substitute by a0 x n (highest degree), x → ∞
⎪
⎪ x2
⎪Trigonomet ric : sin x , tan x : by x , 1 − cos x : by 2 , x → 0
⎨
⎪ Exponentia l : e x − 1, Logarithm ln (1 + x ) : by x , x → 0
⎪
⎪ Power (1 + x )α − 1 : by α x , Root n 1 + ax − 1 : by ax , x → 0
⎩ n
∞ ∞ ∞
1 3 n−4 1
Example : Study convergence : a/ ∑ b/ ∑ 2 c/ ∑ n sin 2
n =1 2 n + 1 n =1 2n − n + 1 n =1 n
 ANSWER
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∞ ∞
un ⎡ converges
u n , vn > 0 & lim
n → ∞ vn
= l ∈ (0; + ∞ ) ⇒ ∑ u n , ∑ vn : same ⎢⎣diverges
n =1 n =1
⎧ Polynomial ao x n + ... → Keep highest degree, x → ∞
⎪
Rule find vn : ⎨ x2
⎪sin x ⇒ Substitute by x for x → 0, 1 − cos x : as ...
⎩ 2
∞ ∞ ∞
1 3 n−4 1
Example : Study convergence : a/ ∑ b/ ∑ 2 c/ ∑ n sin 2
n =1 2 n + 1 n =1 2n − n + 1 n =1 n
∞
un 1 1 1
Answer : a/vn = : lim = 1& ∑ : p = < 1 ⇒ Diverges
2 n n → ∞ vn n =1 2 n 2
3 n un 1 un
b/vn = . lim
2 n →∞ v
= 1 ⇒ converges c/v n = n ⋅ . lim
2 n →∞ v
= 1 ⇒ Diverges.
2n n n n
 ABSOLUTE CONVERGENCE TEST
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Given any series Σan, we can consider the corresponding series

∞
∑ an = a1 + a2 + a3 + K
n =1

whose terms are the absolute values of the terms of the Σan.

Theorem: If Σ|an| is convergent then Σan is convergent

In this case the series Σan is called absolutely convergent

1 1 1 1 1 1
Example: 1 − − + − − + K ⇒ ∑ an = ∑ 2 : converges
4 9 16 25 36 n
The inverse if false! When Σ|an| is divergent, Σan can be
divergent or convergent (conditionally convergent). Ex.: ∑
( − 1)n
n
 THE RATIO TEST (D’ALAMBERT TEST)
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Given any series Σan let consider the following limit:

⎡ d < 1 ⇒ ∑ a n is absolutely convergent (and
⎢
a n +1 ⎢ therefo re convergent )
lim =⎢
n→∞ an d > 1 or ∞ ⇒ ∑ a n is divergent
⎢
⎢⎣1 : no conclusion ( ∑ a n can converge or diverge)

an+1
If the series Σan is positive: an > 0 ⇒ We evaluate simply lim
n →∞ a n

Advise: If an is given with factorial sign (!) ⇒ Apply the ratio test.

n! 5 n (n!)2
Example: Study Σan: a / a n = (− 1)n b / an =
nn (2 n )!
 ANSWER
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⎡ d < 1 ⇒ ∑ a n is absolutely convergent (and

⎢
a n +1 ⎢ therefo re convergent )
lim =⎢
n→∞ an d > 1 or ∞ ⇒ ∑ a n is divergent
⎢
⎢⎣1 : no conclusion ( ∑ a n can converge or diverge)

n! 5 n (n!)2
Example: Study Σan: a / a n = (− 1) n
b / an =
nn (2 n )!
an+1 (n + 1)! n n 1 1
Answer : a/ lim = lim ⋅ = lim = <1
n→∞ an n→∞ (n + 1)n +1 n! n→∞ ⎛ 1 ⎞ n e
⎜1 + ⎟
⎝ n⎠
an+1 5(n + 1)2 5
⇒ Converges. b/ lim = lim = > 1 ⇒ Diverges.
n →∞ a n n→∞ (2n + 1)(2n + 2 ) 4
 THE ROOT TEST (CAUCHY TEST)
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Given any series Σan let consider the following limit:

⎡ q < 1 ⇒ ∑ a n is absolutely convergent (and
⎢
⎢ therefo re convergent )
lim n an = ⎢
n→∞ q > 1 or ∞ ⇒ ∑ a n is divergent
⎢
⎢⎣1 : no conclusion ( ∑ a n can converge or diverge)

If the series Σan is positive: an > 0 ⇒ We evaluate simply lim n an

n →∞

Advise: If an is given in power form of n ⇒ Apply the root test.

∞ n ∞ n2
n +1 ⎛ n +1 ⎞ 1 ⎛ 1⎞
Study the convergence of: ∑ (− 1) ⎜ ⎟
⎝ 2n + 5 ⎠
∑ n⎜
1+ ⎟
2 ⎝ n⎠
n =1 n =1
 EXAMPLE
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⎡ q < 1 ⇒ ∑ a n is absolutely convergent (and

⎢
⎢ therefo re convergent )
lim n an = ⎢
n→∞ q > 1 or ∞ ⇒ ∑ a n is divergent
⎢
⎢⎣1 : no conclusion ( ∑ a n can converge or diverge)

∞ n ∞ n2
n +1 ⎛ n +1 ⎞ 1 ⎛ 1⎞
Study the convergence of: ∑ (− 1) ⎜ ⎟
⎝ 2n + 5 ⎠
∑ n⎜
1+ ⎟
2 ⎝ n⎠
n =1 n =1
n
⎛ n +1 ⎞ n +1 1
Answer : a/ lim n an = lim ⎜
n ⎟ = lim = < 1 ⇒ Converges.
n→∞ n →∞ ⎝ 2 n + 5 ⎠ n →∞ 2 n + 5 2

n2 n
1 ⎛ 1⎞ 1 ⎛ 1⎞ e
b/ lim n an = lim n
n⎜
1+ ⎟ = lim ⎜1 + ⎟ = > 1 ⇒ Diverges.
n →∞ n →∞ 2 ⎝ n⎠ 2 n →∞ ⎝ n ⎠ 2
 ALTERNATIVE SERIES – ALTERNATING TEST
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The terms are alternately positive and negative. Examples:

1 1 1 ∞
(− 1)n −1 1 1 1 ∞
(− 1)n
1− + − +K = ∑ −1+ − + + K = ∑ 2
2 3 4 n =1 n 4 9 16 n =1 n
∞ ∞
General alternative series: ∑ ( ) ∑ ( )
n −1
− 1 bn or − 1 n
bn , bn > 0
n =1 n =1

∞
If the alternating series ∑ ( − 1)n −1
bn = b1 − b2 + K , bn > 0
n =1

⎧⎪bn +1 ≤ bn ∀ n ⇔ {bn } is decreasing

satisfies two conditions:
⎨ lim b = 0
⎪⎩ n → ∞ n

then the series is convergent

 EXAMPLE
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1 1 ∞
( − 1)n−1 1 ⎧bn :↓
Example 1: 1 − + − K = ∑ ⇒ bn = : ⎨
2 3 n =1 n n ⎩bn → 0
This test applied only for ALTERNATING. Positive Σ1/n: Wrong.

Example 2: Study absolute convergence, conditional convergence:

a/∑
∞
(− 1)n+1 n 2 b/ ∞
(− 1)n 3n
n +1
3 ∑ 4n − 1
n =1 n =1
∞
n2 1
Answer : a/ Absolute convergence : ∑ an = ∑ 3 & vn = ⇒ Not Abso.
n =1 n + 1 n
n2
Alternating & bn = 3
x2
⇒ lim bn = 0; f ( x ) = 3 : f / =
x 2 − x 3
<0
( )
n + 1 n →0 x +1 x3 + 1
2
( )
∀ x > 3 2 ⇒ bn ↓, n ≥ 2 ⇒ Conditional conver. b/ lim an ≠ 0 ⇒ Diverges
n →∞
 NUMERIC SERIES REVIEW
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∞
Evaluate sum of series: S = ∑ a n = lim S n , S n = a1 + a 2 + K + a n
n =1 n→∞
∞ ∞
1 1 ⎡ p > 1 ⇒ Convergenc e
Geomet. : ∑ q = , q < 1; p - series
n
∑ np :⎢
n=0 1− q n =1 ⎣ p ≤ 1 ⇒ Divergence

Divergence Test: lim a n ≠ 0 or ∃ lim a n ⇒ ∑ a n is divergent

n→∞ n→∞

⎡∞ ∞
⎡∞
Positive ⎢ ∑ f (n ), f ↓→ ∫ f ( x )dx ⎢∑ ( − ) bn : bn ↓ & → 0
n
1
⎢ n =1 1 ⎢ n =1
Series: ⎢ ⎢
a (small ) ≤ bn (big) ⎡∑ an
⎢ n ⎢
Σan, an ⎢ ⎢
lim
an
∈ (0; + ∞ ) ⎢∑ an : ⎢ a n +1
≥0 ⎢ ⎢ lim or lim n a n
⎢⎣
n → ∞ bn ⎢n→∞ a n→∞
⎣ ⎣ n