1.3.7 Absolute AND Conditional Convergence PDF

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1.3.7 ABSOLUTE AND CONDITIONAL CONVERGENCE ABSOLUTE CONVERGENCE 1

1

1

1

1

1

Series of the form 1 − 2 − 3 + 4 + 5 − − 7 + ⋯ has mixed signs, but not alternating 6 series. Definition: A series ∑ ∞ 𝑛=1 𝑢𝑛 = 𝑢1 + 𝑢2 + 𝑢3 + ⋯ + 𝑢𝑛 + ⋯ is said to converge absolutely if the series of the absolute values ∑∞ 𝑛=1|𝑢𝑛 | = |𝑢1 | + |𝑢2 | + |𝑢3 | + ⋯ + |𝑢𝑛 | + ⋯ converges. The series is said to diverge if the series of the absolute value diverges.

Example: Determine if converge absolutely: 1.

1

1

1

1 − 2 − 22 + 23 +

1

24

−⋯

Examine the series of ABSOLUTE values, then it becomes: 1

1

1

1 + 2 + 22 + 23 +

1

24

−⋯

which is a GS with 𝑟 =

1

2

< 1, hence converges.

Since the series of absolute values is converges, hence the given series converge absolutely. 2.

1

1

1

1− 2−3+ +⋯ 4 Examine the series of ABSOLUTE values, then it becomes: 1

1

1

1+ + +4+⋯ 2

3

which is a harmonic series, hence diverges.

Since the series of absolute values is diverges, hence the given series diverges.

CONDITIONAL CONVERGENCE Consider the series: 1

1

1

(1) 1 − − + + ⋯ and 2

(2) −1 −

1

2

3

1

4

1

−3− −⋯ 4

Both having the series of absolute value: 1

1

1

1 + 2 + 3 + + ⋯ of harmonic series. 4 Hence, both series diverge.

However, by Alternating Series Test, it was shown that series (1) converges, since: I. II.

1

1>2> 𝑙𝑖𝑚

1

𝑛→∞ 𝑛

1

3

1

> >. . . . .. (a’s decreasing) 4

=0

and

∞ ∞ If ∑ ∞ 𝑛=1|𝑢𝑛 | diverges but ∑𝑛=1 𝑢𝑛 converges, then ∑ 𝑛=1𝑢𝑛 is said to converge conditionally.

Flow chart for converge absolutely, converge conditionally and diverges.

Start

C Converge Absolutely

Use any suitable test

D

Alternating Series Test

D

Diverges

C Converge Conditionally

Example: Determine if the series converge absolutely, converge conditionally or diverges. 1.

∑

∞

(−1)𝑘+1

𝑘=1

1

3𝑘

The series of absolute values is: ∑

∞

|(−1)𝑘+1

𝑘=1

1

3𝑘

|= ∑

∞

1

𝑘=1

1

= ∑ 3

1

3𝑘

∞

1

𝑘=1 𝑘 1

1

1

= 3 [1 + 2 + 3 + + ⋯ ] 4 which is a constant multiple of harmonic series, hence diverges. ∞

So, ∑

(−1)𝑘+1

𝑘=1

1

3𝑘

diverges because the series of its absolute value

diverges.

By Alternating Series Test, a)

1

3

>

1

1

>9>⋯

6

and

b) 𝑙𝑖𝑚

1

𝑘→∞ 3𝑘

=

1



=0

By AST, the series converges.

Since the series diverges but converges by using AST, therefore ∑

2.

∑

∞

(−1)𝑘+1

𝑘=1 ∞

(−1)𝑛

𝑛=1

1

3𝑘

converge conditionally.

√𝑛

𝑛+1

The series of absolute values is: ∑

∞

|(−1)𝑛

𝑛=1

√𝑛 | 𝑛+1

=∑

∞

√𝑛 𝑛+1

𝑛=1

Limit Comparison Test, √𝑛

Let 𝑎𝑛 = 𝑛+1 and 𝑏𝑛 =

√𝑛 𝑛

=

1

√𝑛

=

1

1 𝑛2

(p-series, 𝑝 =

1

2

< 1, diverges)

Now, 𝐿 = lim (𝑎𝑛 ) = lim 𝑛+1 ( 1 ) 𝑛→∞ 𝑏𝑛 √𝑛

𝑘→∞

√𝑛

√𝑛

√𝑛

= lim ( 𝑛+1 𝑛→∞

𝑛

= lim ( 𝑛+1) 𝑛→∞

= lim ( 𝑛→∞

1

1+

)

1

)=1

1 𝑛

√𝑛 Since 𝐿 is finite and positive, by LCT the series ∑  𝑛=1( 𝑛+1) diverges too.

So, ∑

∞

(−1)𝑛

𝑛=1

√𝑛 𝑛+1

diverges because the series of its absolute value diverges.

By Alternating Series Test, a) ∑

∞

(−1)𝑛

𝑛=1

√𝑛

1

=−2+

𝑛+1

√2 3

√3 4

+⋯

0.5 > 0.47 > 0.43 > ⋯

hence,

√𝑛

√𝑛 = lim ( 𝑛 1 𝑙𝑖𝑚 𝑛→∞ 𝑛+1 𝑛→∞ 1+𝑛

b)

−

) = lim ( 𝑛→∞

1 √𝑛 1 1+ 𝑛

)=

0

1+0

=0

By AST, the series converges.

Since the series diverges but converges by using AST, therefore ∑

3.

∑

∞

(−1)𝑛

𝑛=1 ∞

√𝑛 𝑛+1

(−1)𝑛

𝑛=1

converge conditionally.

𝑛

1+3𝑛

The series of absolute values is: ∑

∞

|(−1)𝑛

𝑛=1

∞

𝑛

|=∑ 1+3𝑛

𝑛=1

𝑛

1+3𝑛

Using Divergence Test, 𝑛

lim ( 1+3𝑛 ) = lim (

𝑛→∞

𝑛→∞

1

1 +3 𝑛

)=

1

3

≠0

By Divergence Test, lim ( 𝑛 ) diverges. 1+3𝑛 ∞

So, ∑

𝑛→∞

𝑛

(−1)𝑛

diverges because the series of its absolute value

1+3𝑛

𝑛=1

diverges.

By Alternating Series Test, a) ∑

∞

(−1)𝑛

𝑛=1

𝑛

1+3𝑛

1

2

=− + − 4 7

3

10

+⋯

0.25 > 0.29 > 0.3 > ⋯ is not true

hence,

Condition a) fails, hence series diverges. ∞

Therefore ∑

4.

∑

∞

(−1)𝑛

𝑛=1

𝑛 1+3𝑛

(−1)𝑛

𝑛=1

diverges.

𝑛+6

𝑛4 +1

The series of absolute values is: ∑

∞

|(−1)𝑛

𝑛=1

𝑛+6

𝑛4 +1

∞

|=∑

𝑛+6

4 𝑛=1 𝑛 +1

Limit Comparison Test, Let 𝑎𝑛 = Now, 𝐿 =

𝑛+6

𝑛4 +1

and 𝑏𝑛 =

𝑎 lim ( 𝑏𝑛 ) 𝑛→∞ 𝑛

𝑛

=

𝑛4

𝑛+6 𝑛4 +1 1 𝑛3

= lim ( 𝑘→∞

𝑛+6

= lim ( 𝑛4 +1 𝑛→∞

𝑛4 +6𝑛3

= lim (

1+ 𝑛

𝑛→∞

𝑛4 +1 6

1

1+ 4 𝑛

1

(p-series, 𝑝 = 3 > 1, converges)

)

𝑛3

= lim ( 𝑛→∞

1

𝑛3

)

)

)=1

Since 𝐿 is finite and positive, by LCT the series ∑  𝑛=1( So, ∑

∞

(−1)𝑛

𝑛=1

𝑛+6

𝑛4 +1

value converges.

𝑛+6

𝑛4 +1

) converges too.

converges absolutely because the series of its absolute

5.

∑

6.

∑

∞

(−1)𝑛

𝑛 2𝑛+1

D.I.Y

(−1)𝑘

√𝑘−1 2𝑘 3 +1

D.I.Y

𝑛=1

∞

𝑘=1

(Answer: diverges)

(Answer: converge absolutely)

Conclusion: ∞ If a series ∑ ∞ 𝑛=1|𝑢𝑛 | converges then the series ∑ 𝑛=1 𝑢𝑛 converges and is said

∞ to converge absolutely. If ∑ ∞ 𝑛=1|𝑢𝑛 | diverges but ∑ 𝑛=1 𝑢𝑛 converges, then

∑∞ 𝑛=1 𝑢𝑛 is said to converge conditionally....