Comparison Test SP Handout

7/24/2019 Comparison Test SP Handout

1/11

SUMMARY ON THE

COMPARISON TEST FORSERIES WITHTHEORETICAL SOLVED

PROBLEMS

Mika Seppl: Solved Problems on Comparison Test

COMPARISON TEST

Let 0 !ak!bkfor all k.

bk

k=1

!

" converges # akk=1

!

" converges.

ak

k=1

!

" diverges # bkk=1

!

" diverges.


P-SERIES

By Integral Test, we proved that1

kpk=1!

"

converges if and only ifp >1.

The series of the form1

k

pk=1

!

" are called

p-series. They are used quite often for

comparison purposes.


1

OVERVIEW OF PROBLEMS

Let 0 ! ak ! b

kfor all k. Assume that

the series akk=1

"

# and bkk=1"

# converge.

Show that the series akbkk=1

"

# converges.

Show that if ak > 0 for all kand lim

k!"

kak > 0,

then the series akk=1

"

# diverges.

2


2/11


3

4

Show that if ak > 0 for all kand lim

k!"k

2ak = 1,

then the seriesakk=1

"

# converges.


Let akand b

kbe positive for all k. Assume

that the series akk=1

!

" converges and that

limk#!

bk

ak

= L < !. Show that the seris bkk=1

!

"

converges.


6

7

8

9

Determine whether the series converge ordiverge.

9n

1 +10n

n=1

!

"

4 +2n

3n

n=1

!

"

3n2+ 5n

3n

n2+1( )n=1

!

"

2 + cos(n)

3nn=1

!

"

ln 1 +1

2n

!

"#$

%&n=1

'

(

5

1

k3+ kk=1

!

"10



1

k2+ 4k=1

!

"

k2

k3+1k=1

!

"

1

k2k! 5( )k=3

"

#

1

2k+13

k=1

!

"

k! 1

k3!1k=2

"

#


12

13

14

15

11

sin1

k

!

"#$

%&k=1

'

(16



ln k( )k

4

k=1

!

"

sin1

k2

!"#

$%&k=1

'(

8k2! 7

ekk+1

( )

2

k=1

"

#


17

18

19



3/11


Problem 1

Let ak,bk > 0 for all k. Assume that the series

akk=1

!

" and bkk=1!

" converge. Show that the

series akbkk=1

!

" converges.


Solution

This means, !" > 0 #m"

$! : k > m"

% bk < ".

!m

1: k > m

1" b

k 0, a

kk=1

!

" # bkk=1!

" converge

akbkk=1

!

" converges.Claim

Since b

kk

=

1

!

" converges, limk#!

bk = 0.

! = 1.Let


Solution(contd)

ak,b

k > 0, a

kk=1

!

" # bkk=1!

" converge

akbkk=1

!

" converges.Claim

k > m

1! 0 < a

kbk < a

k.

ak > 0 ! bk > 0 "k,Since

akbkk=m

1

!

"

Hence, by The Comparison Test,

converges.


akbk

k=1

!

" = akbkk=1

m1#1

" + akbkk=m

1

!

" ,

Solution(contd)

ak,b

k > 0, a

kk=1

!

" # bkk=1!

" converge

akbkk=1

!

" converges.Claim

akbkk=m

1

!

" The convergence of

implies the convergence of

akbk

k=1

m1!1

"because is a finite sum.


4/11


Problem 2

Show that if limk!"

kak > 0,


"

# diverges.


Solution

k > m

1! a

k >

b

2k.

limk!"

kak > 0# a

kk=1

"

$Problem 2

Assume that limk!"

kak = b > 0.

Since b> 0, !m1such that k > m1 " kak >b

2.

b

2kk=m

1

!

" The series diverges

since the Harmonic Series diverges.

=

b

2

1

kk=m

1

!

"

diverges.


Solution

limk!"

kak > 0# a

kk=1

"

$Problem 2 diverges.

b2k

k=m1

!

" Since diverges,

ak >

b

2k> 0,and since

also

akk=1

!

" diverges.


Problem 3

Show that if limk!"

k2ak = 1,


"

# converges.


5/11


Solution

limk!"

k2ak =1# a

kk=1

"

$ converges.Problem 3

limk!"

k2ak = 1 # $m

1: k > m

1 # 0 < k2a

k < 2.

k > m

1! 0 < a

k m

1 #

bk

ak

< L + 1.

! k > m

1! b

k < L +1( )ak.

akk=1

!

" converges! L +1( )akk=1

"

# converges.


Solution(contd)

akk=1

!

" converges, limk#!

bk

ak

= L < !

bkk=1

!

" Claim: converges.

Problem 4

L + 1( )akk=1

!

" converges! b

k

k=m1

"

# converges.

bk = b

kk=1

m1!1

" +k=1

#

" bkk=m1

#

" Hence also

converges because bkk=1

m1!1

" is a finite sum.

bk < L + 1( )ak.


6/11


9n

1 +10n

n=1

!

"Problem 5

Solution

Does converge?

!n" 1: 0< 9n

1 +10n< 9

n

10n= 9

10#$%

&'(

n

.

9

10

!

"#$

%&

n

n=1

'( is a convergent geometric series.

!9n

1 +10n

n=1

"

#

converges by the Comparison Theorem.


Problem 6

4 + 2n

3n

n=1

!

"

Solution

4 +2n

3n

= 4

3n+ 2

n

3n= 4 ! 1

3"#$

%&'

n

+ 2

3"#$

%&'

n

.

1

3

!

"#$

%&

n

n=1

'( , 23

!

"#$

%&

n

n=1

'( both converge.

!4 +2

n

3nn=1

"

# converges.

Does converge?


Problem 7

Solution

ln 1 +1

2n

!

"#$

%&n=1

'

(

fx( ) =x! ln 1 +x( ),

Does converge?

Hence f is increasing, and f(x) > 0 forx"1.

f 1( ) = 1 ! ln2 > 0,

!f x( ) = 1"

1

1 +x> 0,x > 0.

Conclude: 0 < ln 1 +1

2n

!

"#$

%& 0.

!f x( ) = 1"

1

x

> 0 forx >1.

Hence f is strictly increasing forx !1.

I.e. lnx < x for x !1.

lnk

k4 1.


Problem 18

Solution

lnk

k4

k=2

!

"Does converge?

lnk

k4 k

1"

8k2# 7

k+1( )2


11/11


Problem 19

Solution

k > k1!

8k2" 7

ek

k+1( )2 k1! 8k

2

" 7

ek k+1( )2 < 10

ek,

10

ek

k=1

!

" converges, also

Since

and since

8k2! 7

ek

k+1( )2

k=k1

"

# converges. Hence the wholeseries converges.

8k2! 7

ek

k+1( )2

k=1

"

#Does converge?

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Comparison Test SP Handout