1.3.1 1.3.2 TEST FOR Convergence PDF

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1.3 TEST FOR CONVERGENCE 1.3.1 The Divergence Test Theorem: i. If lim 𝑎𝑛  0, then the series diverges. 𝑛→∞

If lim 𝑎𝑛 = 0, then the series may either converges or diverges.

ii.

𝑛→∞

Example for Theorem i:  ( ∑ 𝑛=1

𝑛

𝑛+1

)=

1

𝑛→∞

𝑛+1

lim 𝑎𝑛 = lim (

𝑛→∞

2

2

3

4

+ 3 + 4 + 5 + ⋯…+

𝑛

)

1

𝑛

𝑛+1

+⋯

= lim ( 1⁄ ) 1+ 𝑛→∞

𝑛

 0 =1 Since lim 𝑎𝑛  0, then the series diverges. 𝑛→∞

Example for Theorem ii: 1

1

1

1. ∑𝑛=1 ( 2𝑛 ) = 2 +

22

1

1

1 2

lim 𝑎𝑛 = lim ( 𝑛 ) = 2

𝑛→∞

𝑛→∞

1

1

+ 23 + 24 + ⋯ . . + 2𝑛 + ⋯ .. =0

hence, series either converges or diverges

(Remember: the 0 only tells that the series is either converges or diverges. It is NOT the limit or sum of the series) The series is actually a GS with 𝑎 =

1

2

1

and 𝑟 = 2.

Since |r| < 1, the series converges and has the sum of 𝑆=

𝑎 1−𝑟

1

= 1

1 2

1

1−2 1

=1 1

1

1 2. ∑ 𝑛=1 ( 𝑛) = 1 + 2 + 3 + 4 + ⋯ . . + 𝑘 + ⋯ .. 1

lim 𝑎𝑛 = lim ( ) = 𝑛

𝑛→∞

𝑛→∞

1



=0

hence, series either converges or diverges.

But, since it is a harmonic series, then it is diverges.

Due to this phenomena, Theorem ii earlier can be extended to another theorem, iii. If the series ∑ 𝑛=1 𝑢𝑛 converges, then lim 𝑢𝑛 = 0 𝑛→∞

Example 1. Apply Divergence Test and state what it tells about the series. 𝑘 a) ∑𝑘=1 ( 𝑘) 𝑒 𝑘

1

lim ( 𝑘) = lim ( 𝑘) = 0 , 𝑒 𝑒

𝑘→∞

𝑘→∞

series either converges or diverges

b) ∑𝑘=1 ln 𝑘 lim (ln 𝑘) = ln  =   0 , series diverges 𝑘→∞

1

c) ∑𝑘=1 (√𝑘) 1

1

lim ( ) = lim ( √𝑘

𝑘→∞

√

𝑘→∞

𝑘+1

d) ∑𝑘=1 (𝑘+2) lim (

𝑘→∞

𝑘+1

1+1 ⁄

) = lim ( 1+2 ⁄𝑘 ) = 1 0 , 𝑘+2

2. Given 𝑎𝑛 = a) {𝑎𝑛 }

1

) =  = 0 , series either converges or diverges

𝑘→∞

(𝑛+1)2 . 𝑛(𝑛+2)

𝑘

Determine if the following converges or diverges.

∞ 𝑛=1

lim (𝑎𝑛 ) = lim (

𝑛→∞

𝑛→∞

= lim (

=

series diverges

(𝑛+1)2

)

𝑛(𝑛+2) 𝑛2 +2𝑛+1

)

𝑛2 +2𝑛 1 + 2 ⁄𝑛 + 1⁄ 2 𝑛 ) lim ( 1+2 ⁄𝑛 𝑛→∞ 𝑛→∞

=1 

Hence, the sequence converges.

b) ∑𝑘=1 𝑎𝑛 By Divergence Test,

(𝑛+1)2

lim (𝑎𝑛 ) = lim ( 𝑛(𝑛+2))

𝑛→∞

𝑛→∞

𝑛2 +2𝑛+1 ) 𝑛2 +2𝑛 𝑛→∞ 1 + 2 ⁄𝑛 + 1⁄ 2 𝑛 ) lim ( 1+2 ⁄𝑛 𝑛→∞

= lim (

=

=1 0

Hence, the series diverges.

EX. 9.2: No. 59-75

1.3.2 The Integral Test Let ∑𝑛=1 𝑢𝑛 be a series with positive terms. If f is a function that is decreasing + and continuous on the interval [a, +), and 𝑢𝑛 = 𝑓(𝑛), then ∑𝑛=1 𝑢𝑛 and ∫𝑎 𝑓(𝑥)𝑑𝑥 will be both converge or both diverge. Example Use Integral Test to determine whether the series converges or diverges. 1

 1. ∑𝑘=1 (𝑘)

1

Let 𝑓(𝑥) = 𝑥 +

∫𝑎

𝑐

𝑓(𝑥)𝑑𝑥 = lim ∫1 𝑓(𝑥)𝑑𝑥 𝑐→∞

𝑐1

= lim ∫1 𝑥 𝑑𝑥 𝑐→∞

𝑐 = lim [ln 𝑥]1 𝑐→∞

= lim (ln 𝑐 − ln 1) 𝑐→∞

= ln 

=

Since the integral diverges, so the series also diverges. 1

 2. ∑ 𝑘=1 (𝑘 2 )

1

Let 𝑓(𝑥) = 𝑥2 +

∫𝑎

𝑐

𝑓(𝑥)𝑑𝑥 = lim ∫1 𝑓(𝑥)𝑑𝑥 𝑐→∞

𝑐 1

= lim ∫1 𝑥2 𝑑𝑥 𝑐→∞

1

= lim [− 𝑥] 𝑐→∞

1

𝑐 1

1

= lim (− 𝑐 + 1) 𝑐→∞ 1

= − + 1

=1

Since the integral converges, so the series also converges.

 3. ∑ 𝑛=1 (

1

𝑛2 +4

)

1

Let 𝑓(𝑥) = 𝑥2 +4 +

∫𝑎

𝑐

𝑓(𝑥)𝑑𝑥 = lim ∫1 𝑓(𝑥)𝑑𝑥 𝑐→∞

𝑐

1

= lim ∫1 𝑥2 +4 𝑑𝑥 𝑐→∞

𝑐 1 𝑥 = lim [ 2 𝑡𝑎𝑛−1 (2)] 1 𝑐→∞ 1

1

𝑐

= 2 lim 𝑡𝑎𝑛−1 ( 2) − 𝑡𝑎𝑛−1 (2) 1

𝑐→∞

1

1

= 2 lim 𝑡𝑎𝑛−1  − 𝑡𝑎𝑛−1 ( 2) 𝑐→∞

𝑡𝑎𝑛−1 ( ) = 26.565  = 0.46𝑟𝑎𝑑 2

1 𝜋

= 2 ( 2 − 0.46) = 0.555

Since the integral converges, so the series also converges.  4. ∑ 𝑛=1 (

𝑛

𝑛2 +7)3

𝑥

Let 𝑓(𝑥) = (𝑥2 +7)3 +

∫𝑎

𝑐

𝑓(𝑥)𝑑𝑥 = lim ∫1 𝑓(𝑥)𝑑𝑥 𝑐→∞

𝑐

𝑥

= lim ∫1 2 3 𝑑𝑥 (𝑥 +7) 𝑐→∞

𝑐 2 +7 𝑥 𝑑𝑢 (𝑢)3 2𝑥

= lim ∫8 𝑐→∞ 1

𝑐 2 +7 1

= 2 lim ∫8 𝑐→∞

1

(𝑢 )3

= lim [− 4 𝑢 −2 ] 𝑐→∞ 1

Let 𝑢 = 𝑥 2 + 7, 𝑑𝑢 = 2𝑥 𝑑𝑥, 𝑑𝑥 =

= 256

𝑑𝑢

𝑐2 + 7 8

𝑐→∞

Since the integral converges, so the series also converges.

Alternative way to deal with the substitution: 𝑥

Let 𝑓(𝑥) = (𝑥2 +7)3 +

∫𝑎

𝑐

𝑓(𝑥)𝑑𝑥 = lim ∫1 𝑓(𝑥)𝑑𝑥 𝑐→∞

2𝑥

when 𝑥 = 1 → 𝑢 = 8, when 𝑥 = 𝑐→ 𝑢 = 𝑐 2 + 7

= − 4 lim[(𝑐 2 + 7)−2 − 8−2] 1

𝑑𝑢

𝑐

= lim ∫1 2 𝑥 3 𝑑𝑥 (𝑥 +7) 𝑐→∞

Let 𝑢 = 𝑥 2 + 7, 𝑑𝑢 = 2𝑥 𝑑𝑥, 𝑑𝑥 =

∫(

=

1

(𝑥2 +7)

= 2 lim [ 𝑐→∞ 1

−2

−2

]

𝑐

1

= − 4 lim[(𝑥 2 + 7)−2] 𝑐→∞

1

=

𝑥 𝑑𝑢 𝑢)3 2𝑥

1

2

1 2

1

∫ (𝑢)3 𝑑𝑢

∫ 𝑢 −3 𝑑𝑢 =

𝑢−2 −2

𝑐 1

= − 4 lim[(𝑐 2 + 7)−2 − (12 + 7)−2] 𝑐→∞

1

= − 4 ((∞ + 7)−2 − 8−2 ) 1

1

1

= − 4 ( (∞+7)2 − 64) 1

1

1

= − ( ∞ − 64) 1

4

= 256

Since the integral converges, so the series also converges.

 5. ∑ 𝑘=1 (

2 𝑒 ⁄𝑘

𝑘2

)

Let 𝑓(𝑥) = +

∫𝑎

2

𝑒 ⁄𝑥 𝑥2

𝑐

𝑓(𝑥)𝑑𝑥 = lim ∫1 𝑓(𝑥)𝑑𝑥 𝑐→∞

2 𝑐 𝑒 ⁄𝑥

= lim ∫1 𝑐→∞

𝑑𝑥

𝑥2

Let 𝑢 =

2 𝑥

= 2𝑥 −1

𝑑𝑢 = −2𝑥 −2 = −

𝑑𝑥 = − 2

= lim ∫2𝑐 𝑐→∞

𝑒𝑢

𝑥2

2

= − lim ∫2𝑐 𝑐→∞

𝑑𝑢

𝑥2

𝑒𝑢 2

2

1

2

𝑑𝑢

= − lim [𝑒 𝑐 − 𝑒 2 ] 2 𝑐→∞

2

𝑑𝑥

(− ) 𝑑𝑢 2

= − 2 lim [𝑒 𝑢 ] 𝑐 𝑐→∞ 2 1

𝑥2

2

𝑥2

when 𝑥 = 1 → 𝑢 = 2, when 𝑥 = 𝑐 → 𝑢 =

2 𝑐

𝑑𝑢

2𝑥

1

= − lim [𝑒 0 − 𝑒 2 ] =

2 𝑐→∞ 1 − 2 (1 −

𝑒 2)

Since the integral converges, so the series also converges.

Alternative way to deal with the substitution: Let 𝑓(𝑥) = +

∫𝑎

2

𝑒 ⁄𝑥 𝑥2

𝑐

𝑓(𝑥)𝑑𝑥 = lim ∫1 𝑓(𝑥)𝑑𝑥 𝑐→∞

2 𝑐 𝑒 ⁄𝑥

= lim ∫1 𝑐→∞

𝑥2

𝑑𝑥

Let 𝑢 =

2

𝑥

= 2𝑥 −1

𝑑𝑢 = −2𝑥 −2 = −

𝑒𝑢

𝑥2

𝑑𝑥 =

𝑥2 − 2

∫ 𝑥2 (− 2 ) 𝑑𝑢 = − ∫ =

2

𝑒𝑥 − lim [ 2 ] 𝑐→∞ 1

1

2

𝑑𝑥

𝑑𝑢 = −

𝑒𝑢 2

𝑐

1

2 𝑥

= − lim [𝑒 ] 2 𝑐→∞

𝑑𝑢

𝑒𝑢

2

𝑥2

2 𝑐

𝑐

1

2

= − lim (𝑒 − 𝑒1 ) 2 1

𝑐→∞ 2

= − (𝑒 ∞ − 𝑒 2 ) 2

1

= − (𝑒 0 − 𝑒 2 ) 2

1

= − (1 − 𝑒 2 ) 2

Since the integral converges, so the series also converges.

EX. 9.3, No. 1-7

1.3.2.1 p Series Theorem A p-series which is of the form 1

1

 ∑𝑘=1 (𝑘 𝑝 ) = 1 +

-

1

+ 3𝑝 + ⋯ . . +

2𝑝

will converge if 𝑝 > 1

1

𝑘𝑝

+ ⋯ ..

(𝑝 > 0)

will diverges if 0 < 𝑝 ≤ 1

For example: 1

1

1

 ∑ 𝑘=1 (𝑘 ) = 1 + 2 + 1 𝑘

1 22

√𝑘

√2

 ∑ 𝑘=1 ( 2) = 1 + 1

 ∑ 𝑘=1 ( )=1+ 4

1

  ∑ 𝑘=1 𝑘 − 3 = ∑ 𝑛=1 3

3

1

+ ⋯..+ + ⋯… 𝑘 1

+ 32 + ⋯ . . + +

1

√𝑘 4

1

√3

1

𝑘2

+ ⋯..+

=1+

3

1

√24

1

(p=1, diverges)

+ ⋯ ..

√𝑘

+3

(p=2, converges)

+ ⋯..

1

√34

(p=1/2, diverges)

…..+ 3

1

√𝑘 4

+ ⋯.

(p=4/3, converges)

Note: Do not get confused with Geometric Series 1

 ( ∑ 𝑘=1 )= 𝑝𝑘 1

1 𝑝

1

1

1 2𝑝

+ 3𝑝 +

+ 𝑝2 + 𝑝3 + ⋯ . . …

 ( ∑ 𝑘=1 )= 1+ 𝑘𝑝

1

1

4𝑝

is a GS with 𝑟 =

…..…

is a p-series

1.3.2.2 Harmonic Series A harmonic series is a series of the form: 1

1

 ∑ 𝑘=1 (𝑘 ) = 1 + 2 +

1

3

1

1

+4 … . . + + ⋯ … 𝑘

and it is a divergent series.

Example 9

1. Show that ∑ 𝑘=1 ( ) is a divergent series. 𝑘 1

9   ( ) ∑ 𝑘=1 (𝑘) = 9 ∑𝑘=1 𝑘 1

1

1

1

= 9(1 + 2 + 3 + 4 … . . + 𝑘 + ⋯ … )

1

𝑝

which is a harmonic series, hence divergent. 1

 2. Determine whether ∑𝑛=1 ( 𝑛3 + 1

1

 ∑𝑛=1 (𝑛3 +

3𝑛

1

1

3𝑛

) is a convergent or divergent series. 1

  ( 𝑛) ) = ∑𝑛=1 ( 3 ) + ∑ 𝑛=1 𝑛 3

1

p-series, 𝑝 = 3 > 1

GS, 𝑟 =

hence, converges

hence, converges

3

1, hence converges 1

 ( ) b) ∑𝑛=1 5𝑛

1

1

 This is harmonic series as 5 ∑ 𝑛=1 ( ), hence diverges 𝑛

2 𝑛

 c) ∑𝑛=1 (5)

2

This is GS, 𝑟 = < 1, hence converges 5

 ( 5. Determine whether ∑𝑛=1  ∑𝑛=1 (

2√𝑛 𝑛

2√𝑛

2

𝑛

−

2

5𝑛

 − 5𝑛 ) = ∑ 𝑛=1 (

 ( ∑ 𝑛=1

) is a convergent or divergent series

2√𝑛 𝑛

2√𝑛 𝑛

)

2

 ( 𝑛) ) − ∑ 𝑛=1 5 2

 ∑𝑛=1 ( 𝑛) 5

1

 ( = 2 ∑𝑛=1 1) 𝑛2

p-series, 𝑝 =

1

2

≤1

hence, diverges

1

 ( ) = 2 ∑ 𝑛=1 5𝑛

GS, 𝑟 =

1

5...