Post on 20-Feb-2018
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.
Mika Seppl: Solved Problems on Comparison Test
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.
Mika Seppl: Solved Problems on Comparison Test
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
7/24/2019 Comparison Test SP Handout
2/11
Mika Seppl: Solved Problems on Comparison Test
3
4
Show that if ak > 0 for all kand lim
k!"k
2ak = 1,
then the seriesakk=1
"
# converges.
OVERVIEW OF PROBLEMS
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.
Mika Seppl: Solved Problems on Comparison Test
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
OVERVIEW OF PROBLEMS
Mika Seppl: Solved Problems on Comparison Test
1
k2+ 4k=1
!
"
k2
k3+1k=1
!
"
1
k2k! 5( )k=3
"
#
1
2k+13
k=1
!
"
k! 1
k3!1k=2
"
#
Determine whether the series converge ordiverge.
12
13
14
15
11
sin1
k
!
"#$
%&k=1
'
(16
OVERVIEW OF PROBLEMS
Mika Seppl: Solved Problems on Comparison Test
ln k( )k
4
k=1
!
"
sin1
k2
!"#
$%&k=1
'(
8k2! 7
ekk+1
( )
2
k=1
"
#
Determine whether the series converge ordiverge.
17
18
19
OVERVIEW OF PROBLEMS
7/24/2019 Comparison Test SP Handout
3/11
Mika Seppl: Solved Problems on Comparison Test
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.
Mika Seppl: Solved Problems on Comparison Test
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
Mika Seppl: Solved Problems on Comparison Test
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.
Mika Seppl: Solved Problems on Comparison Test
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.
7/24/2019 Comparison Test SP Handout
4/11
Mika Seppl: Solved Problems on Comparison Test
Problem 2
Show that if limk!"
kak > 0,
then the series akk=1
"
# diverges.
Mika Seppl: Solved Problems on Comparison Test
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.
Mika Seppl: Solved Problems on Comparison Test
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.
Mika Seppl: Solved Problems on Comparison Test
Problem 3
Show that if limk!"
k2ak = 1,
then the series akk=1
"
# converges.
7/24/2019 Comparison Test SP Handout
5/11
Mika Seppl: Solved Problems on Comparison Test
Solution
limk!"
k2ak =1# a
kk=1
"
$ converges.Problem 3
limk!"
k2ak = 1 # $m
1: k > m
1 # 0 < k2a
k < 2.
k > m
1! 0 < a
k m
1 #
bk
ak
< L + 1.
! k > m
1! b
k < L +1( )ak.
akk=1
!
" converges! L +1( )akk=1
"
# converges.
Mika Seppl: Solved Problems on Comparison Test
Solution(contd)
akk=1
!
" converges, limk#!
bk
ak
= L < !
bkk=1
!
" Claim: converges.
Problem 4
L + 1( )akk=1
!
" converges! b
k
k=m1
"
# converges.
bk = b
kk=1
m1!1
" +k=1
#
" bkk=m1
#
" Hence also
converges because bkk=1
m1!1
" is a finite sum.
bk < L + 1( )ak.
7/24/2019 Comparison Test SP Handout
6/11
Mika Seppl: Solved Problems on Comparison Test
9n
1 +10n
n=1
!
"Problem 5
Solution
Does converge?
!n" 1: 0< 9n
1 +10n< 9
n
10n= 9
10#$%
&'(
n
.
9
10
!
"#$
%&
n
n=1
'( is a convergent geometric series.
!9n
1 +10n
n=1
"
#
converges by the Comparison Theorem.
Mika Seppl: Solved Problems on Comparison Test
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?
Mika Seppl: Solved Problems on Comparison Test
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.
Mika Seppl: Solved Problems on Comparison Test
Problem 18
Solution
lnk
k4
k=2
!
"Does converge?
lnk
k4 k
1"
8k2# 7
k+1( )2
7/24/2019 Comparison Test SP Handout
11/11
Mika Seppl: Solved Problems on Comparison Test
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?