1.3.3 1.3.4 Comparison TEST PDF

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1.3.3 Direct Comparison Test Theorem:  Let ∑  𝑘=1 𝑎𝑘 and ∑𝑘=1 𝑏𝑘 be series with non-negative terms, and suppose that 𝑎2 ≤ 𝑏2 ,

𝑎1 ≤ 𝑏1 ,

𝑎3 ≤ 𝑏3 , … … . . , 𝑎𝑘 ≤ 𝑏𝑘 … ….

 a) If the bigger series ∑  𝑘=1 𝑏𝑘 converges, then the smaller series ∑ 𝑘=1 𝑎𝑘 also converges.  b) If the smaller series ∑  𝑘=1 𝑎𝑘 diverges, then the bigger series ∑ 𝑘=1 𝑏𝑘 also diverges.

To help in finding smaller/bigger series,

Principle (i): constant terms in the denominator of 𝑢𝑘 can usually be deleted without

affecting the convergence or divergence of series. Example:

1  For ∑  𝑘=1 ( 𝑘+2) , we examine ∑ 𝑘=1 ( 𝑘 )

For ∑  𝑛=1 (

3𝑛

2𝑛 +1

1

(harmonic, diverges)

) , we examine ∑  𝑛=1 ( 2𝑛 ) 3𝑛

(GS, r >1, diverges)

Principle (ii): for any rational function involving polynomials, all except the leading term can be deleted without affecting the convergence or divergence of series. Example:

3𝑘 +4𝑘  For ∑  𝑘=1 ( 2𝑘 3 +𝑘−1) , we examine ∑ 𝑘=1 ( 2𝑘 3 ) 3𝑘 2

2

(harmonic, diverges)

Example Use Direct Comparison Test to determine whether the series converges or diverges. 1.

∑𝑘=1 (

1

)

1

√𝑘−2

By Principle (i), examine ∑ 𝑘=1 (

which is a p-series with 𝑝 =

By comparison,

1

1 √𝑘− 2

>

1

√𝑘

1 2

1

√𝑘

)=1+

1

1 22

+

1

1 32

+

1

1

42

≤ 1, hence diverges.

+ ⋯ ….

the smaller series diverges, hence.. the bigger series also diverges. Hence, ∑ 𝑘=1 (

1

1

√𝑘− 2

) is a divergent series.

2.

∑𝑘=1 ( 

1

2𝑘 2 +𝑘

)

By Principle (ii), examine ∑ 𝑘=1 (

1

2𝑘 2

) = ∑ 𝑘=1 ( 2 ) = (1 + 22 + 32 + 42 + ⋯ … ) 2 𝑘 1

2

which is a p-series with 𝑝 = 2 > 1, hence converges.

By comparison,

1

2𝑘 2 +𝑘

<

1

1

1

1

1

1

2𝑘 2

the bigger series converges, hence..

the smaller series also converges. Hence, ∑ 𝑘=1 (

3.

∑𝑘=1

1

2𝑘 2 +𝑘

) is a convergent series.

1

3 1 (𝑘+ 2)

By Principle (ii), examine ∑ 𝑘=1 (

1

𝑘3

) = 1 + 23 + 1

1 33

[

1

1 3 (𝑘+ ) 2

=

+ 43 + ⋯ … 1

1 ] 𝑘 3 + ฀𝑘 2 +฀𝑘+ ฀

which is a p-series with 𝑝 = 3 > 1 , hence converges.

By comparison,

1

1 3 (𝑘+ ) 2

<

1 𝑘3

the bigger series converges, hence.. the smaller series also converges. Hence, ∑ 𝑘=1 (

4.

∑𝑘=1 (

1

√𝑘 3 +2𝑘

1

1 3

(𝑘+ ) 2

) is a convergent series.

)

By Principle (ii), examine ∑ 𝑘=1 ( which is a p-series with 𝑝 =

By comparison,

1

√𝑘 3 +2𝑘

<

3 2

1

1 ) √𝑘 3

= ∑𝑘=1 ( 3) = 1 + 1

𝑘2

3 22

+

1

3 32

+

1

3

42

+ ⋯ ….

> 1 , hence converges.

√𝑘 3

the bigger series converges, hence..

the smaller series also converges.

Hence, ∑ 𝑘=1 ( √𝑘 3 +2𝑘 ) is a convergent series. 1

1

5.

∑𝑘=1 (

6.

7.



1

)

(D.I.Y)

(Answer:converges)

∑𝑛=1 ( 3𝑛+1)

(D.I.Y)

(Answer: converges)

∑𝑛=2 (

+ 4𝑛 )

2𝑘 +1 2𝑛

3

√𝑛 (√𝑛−1)

5

(D.I.Y)

(Answer: D+C=Diverges)

EX. 9.4, No. 3-9

1.3.4 Limit Comparison Test Theorem:

 𝑘 Let ∑  𝑘=1 𝑎𝑘 and ∑𝑘=1 𝑏𝑘 be series with positive terms and suppose 𝐿 = lim ( 𝑏 ). 𝑘→∞

If 𝐿 is finite and 𝐿 > 0, then the series are both converge or both diverge.

𝑎

𝑘

Note:

• The theorem is not true for 𝐿 = 0 or 𝐿 = ±∞.

• Normally we let 𝑎𝑘 is the general term for the given series and 𝑏𝑘 is the general term for the guessed series which is known for the convergence/divergence.

Example Use the Limit Comparison Test to determine whether the following series converge or diverge. 1.

∑ 𝑘=1 (

1

√𝑘+1

)

Let 𝑎𝑘 = √𝑘+1 and 𝑏𝑘 = 1

1

√𝑘

Now, 𝐿 = lim ( ) = lim ( 𝑏 𝑘→∞

𝑎𝑘 𝑘

𝑘→∞

1 √𝑘+1 1 √𝑘

(p-series, 𝑝 =

) = lim ( 𝑘→∞

1

1+

1

√𝑘

1 2

< 1, diverges)

)=1

Since 𝐿 is finite and positive, hence ∑  𝑘=1 ( √𝑘+1) diverges. 1

2.

∑𝑘=1 ( 

1

2𝑘2 +𝑘

Let 𝑎𝑘 =

)

1

2𝑘2 +𝑘

and 𝑏𝑘 =

𝑎

(p-series, 𝑝 = 2 > 1, converges)

1 2𝑘2 1 2

2𝑘2

Now, 𝐿 = lim (𝑏 𝑘) = lim ( 2𝑘 1+𝑘 ) = lim ( 2𝑘2+𝑘 ) = lim ( 𝑘→∞ 𝑘→∞

𝑘→∞

𝑘

𝑘→∞

2𝑘2

Since 𝐿 is finite and positive, hence ∑  𝑘=1 (

3.

∑ 𝑘=1 (

3𝑘3 −2𝑘2 +4 ) 𝑘 7−𝑘3 +2

4.

∑ 𝑘=1 (

2 ) 𝑘 3 +𝑒5

Let 𝑎𝑘 =

2

𝑘 3 +𝑒5

1 𝑘

2+

)=1

) converges.

(D.I.Y)

and 𝑏𝑘 =

𝑎

2 𝑘3

(𝑘 Now, 𝐿 = lim (𝑏 𝑘) = lim 𝑘→∞ 𝑘→∞

1

2𝑘2 +𝑘

2

𝑘

2 3+𝑒5 2 𝑘3

(Answer: converges)

(p-series, 𝑝 = 3 > 1, converges) 𝑘3

) = lim ( 𝑘3+𝑒5 ) = lim ( 𝑘→∞

Since 𝐿 is finite and positive, hence ∑  𝑘=1 (

2

𝑘 3 +𝑒5

𝑘→∞

1

𝑒5

1+ 3 𝑘

)=1

) converges.

(D.I.Y: Try to answer this question using Direct Comparison Test)

5.

∑ 𝑛=1 (

2𝑛4 +3 𝑛2 +5)2

Let 𝑎𝑛 = (𝑛2

Now,

2𝑛4 +3

and 𝑏𝑛 =

+5)2

𝑎𝑛

2𝑛 4 𝑛4

𝐿 = lim ( 𝑏 ) = lim ( 𝑛→∞

= lim ( 𝑛→∞

𝑛

2𝑛4 +3

𝑛→∞

2(𝑛 2 +5)2

2𝑛4 +3

= 2 (∑  𝑛=1 𝑏𝑛 = 2 + 2 + 2 + ⋯ . . diverges)

2𝑛4 +3 2 (𝑛2+5)

2

)

)

2𝑛 4 +3

= lim ( 2(𝑛4 +10𝑛2+25)) =

=

=

𝑛→∞ 1

1

3 2𝑛4 + 𝑛4 𝑛4 10𝑛2 + 4 + 254 𝑛4 𝑛 𝑛

lim ( 𝑛4

2 𝑛→∞

1

2𝑛 4 +3

lim ( 𝑛4 +10𝑛2+25)

2 𝑛→∞

lim (

2 𝑛→∞

3

2+ 4 𝑛 10

25

1+ 2 + 4 𝑛 2𝑛

2𝑛4 +3

[ (𝑛2+5)2 = 𝑛4 +10𝑛2+25]

)

)=1

Since 𝐿 is finite and positive, hence ∑  𝑛=1 ( (𝑛2 +5)2 ) diverges. 2𝑛4 +3

(D.I.Y: Try to answer Question 5 using: i) Direct Comparison Test

(test fails)

ii) Divergence Test

(diverges)

EX. 9.4, No. 15-21...