2.sequences & series -...
TRANSCRIPT
2.Sequences & Series
2.1 Introduction
Infinite series occurs so frequently in all types of problems that the necessity of studying
their convergence or divergence is very important. Unless a series employed in an investigation
is convergent, it may lead to absurd conclusions. Hence it is essential that the students of
engineering begin by acquiring and intelligent grasp of this subject.
2.2 Sequences
(1) Sequence
An ordered set of real numbers ,......,......,,, 321 naaaa is called sequence and is denoted
by ( )na . If the number of terms unlimited, then the sequence is said to be an infinite sequence and
na is its general term.
Ex: i) ,....1
,,.........3
1,
2
1,1
n
ii) 1,2,4,……..2n,……
iii) 1,-1,1,-1,…….(-1)n-1
,…….
(2) Bounded Sequence
A Sequence ( )na is said to be bounded if there exists a number M such that an <M for
all n.
(3) Monotonically Increasing Sequence
The Sequence (an) is said to be monotonically increasing sequence if 1+≤ nn aa for all n.
(4) Monotonically Decreasing Sequence
The Sequence (an) is said to be monotonically decreasing sequence if 1+≥ nn aa for all n.
(5) Monotonic Sequence
A Monotonically increasing sequence or Monotonically decreasing sequence is called
monotonic sequence.
(6) Limit
A sequence ( )na is said to tend to a limit l , if for every ε >0 , a value N of n can be
found such hat ε<− lan for n Nn ≥ .
(7) Convergence
If a sequence ( )na has a finite & limit , then it is called a convergent sequence. Otherwise
it is called divergent.
In the above examples , (i) is a convergent sequence, while (ii) & (iii) are divergent sequences.
Note :
1) If lalt nn
=∞→
is finite & unique , then the sequence is convergent.
2) If nn
alt∞→
is infinite ( )∞± , then the sequence is divergent.
3) If nn
alt∞→
is not unique, then the sequence is oscillatory.
Example 2.1
Examine the following sequences are converges.
i) 1+n
n ii)
1−n
n iii)
2
1
+n
n iv)
2
1
−n
n v) [ ]2
)1( nn −+ vi)
n
n)1(1
−+
vii) n3 viii) n5 ix) ( )n1− x) ( )n
17 −+
Solution:
i) Let 1+
=n
nan
.11
1
11
1
)1
1(1
==
+
=
+
=+
=∞→∞→∞→∞→
n
lt
nn
nlt
n
nltalt
nnnn
n
=
∞0
1∵
Since the limit is finite and unique, 1+n
n is converges.
ii) Let 1−
=n
nan
.11
1
11
1
)1
1(1
==
−
=
−
=−
=∞→∞→∞→∞→
n
lt
nn
nlt
n
nltalt
nnnn
n
=
∞0
1∵
Since the limit is finite and unique, 1−n
n is converges.
iii) Let
2
1
+=
n
nan
.11
1
11
1
)1
1(1
22
2
==
+
=
+
=
+=
∞→∞→∞→∞→
n
lt
nn
nlt
n
nltalt
nnnn
n
=
∞0
1∵
Since the limit is finite and unique,
2
1
+n
n is converges.
iv) Let
2
1
−=
n
nan
.11
1
11
1
)1
1(1
22
2
==
−
=
−
=
−=
∞→∞→∞→∞→
n
lt
nn
nlt
n
nltalt
nnnn
n
=
∞0
1∵
Since the limit is finite and unique,
2
1
−n
n is converges.
v) Let [ ] 1)1(
−−+= n
n na
[ ][ ]
.01
)1(
1)1(
1=
∞=
−+=−+=
∞→
−
∞→∞→ nn
n
nn
n nltnltalt
=
∞0
1∵
Since the limit is finite and unique, [ ] 1)1(
−−+
nn is converges.
vi) Let ( )
−+=
na
n
n
11
( ).1
11 =
−+=
∞→∞→ nltalt
n
nn
n
=
∞0
1∵
Since the limit is finite and unique, ( )
n
n1
1−
+ is converges.
vii) Let nan 3=
.3 ∞==∞→∞→
nltaltn
nn
Since the limit is infinite, n3 is diverges.
viii) Let n
na 5=
.5 ∞==∞→∞→
n
nn
nltalt
Since the limit is infinite, n5 is diverges.
ix) Let n
na )1(−=
( ) 1.1 −=−=∞→∞→
n
nn
nltalt when n is odd
= + 1 when n is even.
Since the limit is not unique, ( )n1− is oscillates.
x) Let ( )n
na 17 −+=
( ) 617 =−+=∞→∞→
n
nn
nltalt when n is odd.
= 8 when n is even.
Since the limit is not unique, ( )n17 −+ is oscillates.
Exercise 2.1
i) 2n ii) 1/2n iii) n
n
21
13
+
− iv)
n
n)1(1
−+
[ Ans : (i) diverges, (ii),(iii) & (iv) converges. ]
2.3 Series
(1) Series
If ,......,......,,, 321 naaaa is infinite sequence of real numbers, then ∞++++ .........21 naaa
is called an infinite series. An infinite series is denoted by ∑ na and the sum of its first n terms
is denoted by sn.
Note:
Consider the infinite series ∞++++=∑ ............21 nn aaaa and let the sum of first n terms
be nn aaas +++= .......21 .Therefore,
4) If lslt nn
=∞→
is finite & unique , then the series ∑ na is convergent.
5) If nn
slt∞→
is infinite ( ∞± ), then the series ∑ na is divergent.
6) If nn
slt∞→
is not unique, then the series ∑ na is oscillatory.
7) If the series ∞++++=∑ ............21 nn aaaa is convergent, then .0=∞→
nn
alt
8) If ,0≠∞→
nn
alt then the series ∑ na is divergent.
9) If ,0≠∞→
nn
alt then the series ∑ na is divergent.
10) p – test : .1&1
1≤>∑ pifdivergentpifconvergentis
np
(2) Necessary condition for convergent sequence
If the series ∞++++=∑ ............21 nn aaaa is convergent, then .0=∞→
nn
alt
The above condition is not sufficient to guarantee the convergence of
∞++++=∑ ......21 nn aaaa
ie) If 0=∞→
nn
alt , then the series ∞++++=∑ ............21 nn aaaa need not be
convergent, but If
,0≠∞→
nn
alt then the series ∑ na is divergent.
Which is the contrapositive form of necessary condition for convergence.
Example:2.2
Examine convergence for the following series.
i) ∞+++++ ........321 n
Solution: Let ∞+++++= ........321 nsn
= 2
)1( +nn
∞=
∞→n
nslt
Hence the series is not convergent (or divergent).
ii) ∑ +
+
)3(10
210 n
n
Let )3(10
210 +
+=
n
nan
.010
1
)3
1(10
)2
1(
)3(10
210
1010
≠=
+
+
=+
+=
∞→∞→∞→
nn
nn
ltn
nltalt
nnn
n
Hence the given series not convergent or divergent.
iii) ∞++++ .......
2
1.....
2
1
2
12 n
Let nns
2
1.....
2
1
2
12
+++=
1
2
11
)2
11(
2
1
=
−
−
=∞→
n
nn
slt
Hence the given series is convergent.
iv) ∞+++++=∑ ....
1......
3
1
2
1
1
11
nn
.1&11
, ≤>∑ pifdivergentpifconvergentisn
WKTp
Here p = 1.
∑ ∞+++++= .....1
......3
1
2
1
1
11divergesis
nnHence
v) ∞+++++=∑ ....
1......
3
1
2
1
1
1122222
nn
.1&11
, ≤>∑ pifdivergentpifconvergentisn
WKTp
Here p = 2.
.....
1......
3
1
2
1
1
1122222
convergesisnn
Hence ∞+++++=∑
vi) 1-1+1-1+1-…..∞
termsnsn −+−+−= ...1111
....11111 −+−+−=∞→
nn
slt
= 0 or 1 according as the number of terms is even or odd.
Hence the series is oscillatory.
vii) Show that the series(geometric series) ∞++++ ......1 32rrr
1) converges if 1<r
2) diverges if 1≥r
and 3) oscillates if 1−≤r
Solution: Let n
n rrrrs +++++= ......1 32
Case1 : When 1<r
n
n rrrrs +++++= ......1 32
= r
rn
−
−
1
)1(1
rslt n
n −=
∞→ 1
1
0=
∞→
n
nrlt∵
Hence the series is convergent.
Case 2: i) When r >1,
1
1
1
1
)1(1
......132
−−
−=
−
−=
+++++=
rr
r
r
r
rrrrs
n
n
n
n
∞=∞=
∞→∞→
n
nn
nrltslt ∵
Hence the series is divergent.
ii) When r =1,
∞=
=++++=
∞→n
n
n
slt
ntimesns )(1......111
Hence the series is divergent.
Case 3 : i) When r = -1,
.1
0
)1(......111
oddisnif
evenisnifslt
s
nn
n
n
=
=
−+−+−=
∞→
Hence the series is oscillatory.
ii)When r < -1,
Let r = -ρ , so that ρ >1 and nnnr ρ)1(−=
∞=∞±=
−
−−=
−
−=
∞→∞→
n
nn
n
nn
n
n
ltoddorevenisnasaccordingslt
r
rs
ρ
ρ
ρ
∵.
1
)1(1
1
1
Hence the series oscillates.
viii) 5-4-1+5-4-1+5-4-1+…….
Solution ; Let ......145145145 +−−+−−+−−=ns
.23,13,315,0 ++=∞→
mmmistermsifnumbertheasaccordingorslt nn
Clearly sn does not tend to a unique limit. Hence the series is oscillatory.
Exercise : 2.2
i) ...8
1
4
1
2
11 ++++ iii) .....
3
1
3
1
3
11
32+−+−
ii) 6-10+4+6-10+…. iv) ......4.3
1
3.2
1
2.1
1+++
[ Ans : i) , iii) & iv) converges ii) oscillatory ]
Various Test for convergence of a series
2.4 Comparison Test
1. If two positive term series ∑ ∑ nn banda be such that i) ∑ nb converges,
ii) nn ba ≤ for all values of n, then ∑ na also converges.
2. If two positive term series ∑ ∑ nn banda be such that i) ∑ nb diverges,
ii) nn ba ≥ for all values of n, then ∑ na also diverges.
3. If two positive term series ∑ ∑ nn banda be such that ),0(≠=∞→
finiteb
alt
n
n
nthen
∑ ∑ nn banda converges or diverges together.
Example : 2.3
i) Test for convergence the series ∞+++ .......5.4.3
5
4.3.2
3
3.2.1
1
Solution: Let the nth
term of the given series be
)2)(1(
12
++
−=
nnn
nan
and let as assume 2
1
nbn = and wkt
nb∑ converges.
0
2
)2
1)(1
1(
)1
2(
1)
21)(
11(
)1
2(
1)2)(1(
12
2
2
2
≠
=
++
−
=
×
++
−
=
×++
−=
∞→
∞→
∞→∞→
nn
nlt
n
nnnn
nn
lt
n
nnn
nlt
b
alt
n
n
nn
n
n
Which is finite and non zero and since ∑ nb converges , then the given
series ∑ na is also converges.
ii) Test for convergence the series ∞+++ .......16.13.10
9
13.10.7
4
10.7.4
1
Solution: Let the nth
term of the given series be
)73)(43)(13(
2
+++=
nnn
nan
and let as assume n
bn
1= and wkt
nb∑ diverges.
0
27
1
)7
3)(4
3)(1
3(
1
1)
73)(
43)(
13(
1)73)(43)(13(
3
2
2
≠
=
+++
=
×
+++
=
×+++
=
∞→
∞→
∞→∞→
nnn
lt
n
nnnn
nlt
n
nnn
nlt
b
alt
n
n
nn
n
n
Which is finite and non zero and since ∑ nb diverges , then the given series
∑ na is also diverges.
iii) Test for convergence the series ∞++++ .......4
3
3
2
2
11
4
3
3
2
2
Solution: Let the nth
term of the given series be
1)1(
++=
n
n
nn
na
and let as assume n
bn
1= and wkt
nb∑ diverges.
0
)1
1(1
1
)1
1(
1
)1
1(
1
)1
()1(
1)
1(
)1(
1
1)1( 1
≠
=+×=
+
×
+
=
+×
+=
×+
×+
=
×+
=
∞→
∞→∞→
∞→∞→
∞→
+∞→∞→
en
lte
n
lt
n
lt
n
nlt
n
nlt
n
n
n
nlt
n
n
nlt
b
alt
n
n
nnn
n
nn
n
n
n
n
nn
n
n
∵
Which is finite and non zero and since ∑ nb divergent , then the given
series ∑ na is also divergent.
iv) Test for convergence the series ∑∞
= ++1 1
1
n nn
Solution: Let the nth
term of the given series be
}1....)8
1
2
11{(
}1)1
1{(
1
1
1
)()1(
1
1
1
1
1
2
2
1
22
−+−+=
−+=
−+=
−+
−+=
−+
−+=
−+
−+×
++=
nnn
nn
nn
nn
nn
nn
nn
nn
nn
nnan
....}
8
1
2
1{
1
....}8
1
2
1{
+−=
+−=
nn
nn
n
and let as assume n
bn
1= and wkt
nb∑ diverges.
0
2
1
....}8
1
2
1{
1....}
8
1
2
1{
1
≠
=
+−=
×+−=
∞→
∞→∞→
nlt
n
nnlt
b
alt
n
nn
n
n
Which is finite and non zero and since ∑ nb divergent , then the given
series ∑ na is also divergent.
v) Test for convergence the series ∑∞
= +
−
1 12
13
nn
n
Solution: Let the nth term of the given series be
)2
11(
)3
11(
)2
3(
)2
11(2
)3
11(3
12
13
2
n
nn
n
n
n
n
n
n
na
+
−
=
+
−
=
+
−=
and let as assume
2
3
)2
3( 2
==
=
rwherer
b
n
n
n
∑ nb is a geometric series with r > 1.
Therefore ∑ nb divergent
.
0
1
)2
11(
)3
11(
)2
3(
)2
11(
)3
11(
)2
3(
2
2
≠
=
+
−
=
+
−
=
∞→
∞→∞→
n
n
n
n
n
nn
nn
n
n
lt
ltb
alt
Which is finite and non zero and since ∑ nb divergent , then the given series
∑ na is also divergent.
vi) Test for convergence the series ∞+−
−+
−
−+
−
−......
15
14
14
13
13
12333
Solution: Let the nth term of the given series be
}1
)2
1{(
)11
1(
}1
)2
1{(
)11
1(
1)2(
11
3
32
5
3
33
3
nnn
nn
nnn
nnn
n
nan
−+
−+
=
−+
−+
=
−+
−+=
and let as assume 2
5
1
n
bn =
& Wkt ∑ nb convergent.
0
1
}1
)2
1{(
)11
1(
1}
1)
21{(
)11
1(
3
3
2
5
3
32
5
≠
=
−+
−+
=
×
−+
−+
=
∞→
∞→∞→
nn
nnlt
n
nnn
nnlt
b
alt
n
nn
n
n
Which is finite and non zero and since ∑ nb convergent , then the given
series ∑ na is also convergent.
vii) Test for convergence the series nn
1sin
1∑
Solution: Let the nth term of the given series be
−+−=
−+−=
−+−=
=
......!5
1
!3
1
1
11
.....!5!3
sin......!5
1
!3
111
1sin
1
422
53
53
nnn
xxxx
nnnn
nnan
∵
and let as assume 2
1
nbn =
Wkt ∑ nb convergent.
0
1
......!5
1
!3
1
1
1
1......
!5
1
!3
1
1
11
42
2
422
≠
=
−+−=
×
−+−=
∞→
∞→∞→
nnlt
n
nnnlt
b
alt
n
nn
n
n
Which is finite and non zero and since ∑ nb convergent , then the given
series ∑ na is also convergent.
Exercise : 2.3
Test the following series for convergence
i) .:.......7.5
3
5.3
2
3.1
1DivergesAns∞+++
ii) ConvergesAns:.......!4
4
!3
3
!2
21
222
∞+++
iii) .:)2)(1(
2DivergesAns
nn
nn∑
++
iv) .:14
525
3
ConvergesAnsn
n∑
+
+
v) .:.......321
321
21
21
1
1222222
DivergesAns∞+++
+++
+
++
2.5 Integral Test
1) A positive term series where f(n) decreases as n increases , is converges ( or diverges) if
∫∞
1
)( dxxf is finite (or infinite).
Example :2.4
i) Show that the p – series ∑∞
=
∞+++=1
..........3
1
2
1
1
11
npppp
n i) converges for p > 1 ii)
diverges for .1≤p
Solution : Given series is ∑∞
=
∞+++=1
..........3
1
2
1
1
11
npppp
n is a positive term series.
Here pppp
xxfand
nnfff
1)(
1)(.......,,
2
1)2(,
1
1)1( ====
Clearly f(n) decreases when n increases.
Apply integral Test,
[ ]
∞=
=
==
∞=
+−
−∞=<
−=
+−
−=>
+−
−=
+−=
=
=
=
∞
∞∞
∞
∞
+−
∞→
+−
∞→
−
∞→
∞→
∞∞
∫∫
∫
∫
∫
∫
∫∫
1
11
1
1
1
1
1
1
1
11
log
1)(,1
1
1)(,1
1
1
1
10)(,1
1
1
1
1
1)(
x
dxx
dxxfpIf
pdxxfpIf
p
pdxxfpIf
p
mlt
p
xlt
dxxlt
dxx
lt
dxx
dxxf
p
m
mp
m
m
p
m
m
pm
p
.inf)(1)(,1
11
initeisdxxfpifandfiniteisdxxfpifHence ∫∫∞∞
≤>
ii) Deterrmine the nature of the series ∑∞
=2 log
1
n nn .
Solution : Given series is ∑∞
=2 log
1
n nn is a positive term series and decreases as n increases after
n =2.
Here xx
xfandnn
nffflog
1)(
log
1)(.......,,
3log3
1)3(,
2log2
1)2( ====
Clearly f(n) decreases when n increases.
Apply integral Test,
( )
[ ]
.inf)(
log
)(log)(
)()log(log
log
1
log
1)(
2
'
2
2
22
initeisdxxf
xfdxxf
xfx
dxx
x
dxxx
dxxf
m
∫
∫
∫
∫∫
∞
∞
∞∞
∴
∞=∞∞=
==
=
=
∵
∵
.log
1int
2
divergesisnn
seriesgiventhetestegralbyHence ∑∞
iii) Determine the nature of the series ∑∞
>2
)0(,)(log
1p
nnp
Solution : Given series is ∑
∞
=2 )(log
1
np
nn is a positive term series and decreases as n increases
after n =2.
Here pppp xx
xfandnn
nfff)(log
1)(
)(log
1)(.......,,
)3(log3
1)3(,
)2(log2
1)2( ====
Clearly f(n) decreases when n increases.
Apply integral Test,
( )∫
∫∫
∞
∞∞
=
=
2
22
)(log
1
)(log
1)(
dxx
x
dxxx
dxxf
p
p
2log,2
1
,,log
===
∞=∞==
txdxx
dt
txwhenxtPut
∞=
+−
−∞=
=
=<
∞=
−∞=
=
=
==
=
=
+−
∞
+−
+−
∞
−
∞
∞
∞
∞
−
∞
∞
−
∞∞
∫∫
∫
∫∫
∫
∫∫
1
)2(log
)(,1
)2log(loglog
)log(
1
)(,1
)(
1
2log
1
1
2log2
2log
2log
2log
1
2
2log
2log2
p
t
dttdxxfpIf
t
dtt
dttdxxfpIf
dtt
t
dtdxxf
p
p
p
p
p
p
finite
k
kt
k
tdxxf
kkpLet
ppSince
t
dttdxxfpIf
k
k
k
p
p
p
=
−−
∞=
−=
−=⇒
>−=−
<−>
−−−
=
=>
∞
∞−∞
∞
+−
+−
∞
−
∞
∫
∫∫
2log
11
1
)()1(
.0,1
&01,1
)1(
)(,1
2log
2log2
2log
1
1
2log2
∴The given series ∑∞
=2 )(log
1
np
nn is convergent for p > 1and divergent for .1≤p
Exercise : 2.4
Test the convergence for the following series
i) ∞++++ ......4.3
1
3.2
1
2.1
11
ii) In a positive term series ∑ 2)(log
1
nn
[ Ans : i) & ii) Converges ]
2.6 D’ Alembert’s Ratio Test
1) In a positive term series ∑ na , if α=+
∞→1n
n
n a
alt , then the series
i) converges for 1>α
ii) diverges for 1<α
iii) the test fails for 1=α
Example :2.5
i) Test for convergence the series ∞++++ ........45342312
1 642xxx
Solution : The given series is ∞++++=∑ ........45342312
1 642xxx
an
and 1)2(
,)1(
2
1
22
++=
+= +
−
nn
xa
nn
xa
n
n
n
n
2
2
2
2
22
1
)1
1(
11)
21(
1
)1(
1)2(1
1)2(
)1(
−
∞→
∞→
−
∞→
+∞→
=
+
++
×=
+
++×=
++×
+=
=
x
nn
n
nn
nn
xlt
nn
nn
xlt
x
nn
nn
xlt
a
alt
n
n
n
n
n
n
n
n
α
α
1,1)
1,11)
2
22
<>
><>∴−
α
α
xwhenii
xorxwheni
.111 222><>∴
−∑ xfordivergesandxorxforconvergesan
finite
n
nn
lt
n
nnlt
b
alt
n
bserieseconvergenctheletandnn
axwheniii
n
nn
n
n
nn
=
=
×
+
=
×+
=
=+
==
→∞
→∞→∞
1
1)
11(
1
1)1(
1
1
)1(
1,1)
2
3
2
3
2
∑∑→∞
convergesatestcomparisonofformitbythenfiniteisb
altandconvergesbSince n
n
n
nn lim,
ii) Test for convergence the series .0,........17
14
9
6
5
21 32
>∞++++ xxxx
Solution : The given series is .0,........17
14
9
6
5
21 32
>∞++++ xxxx
and n
n
n
n
n
n
n
n xaxa12
22,
12
221
1
1
1
+
−=
+
−=
+
+
+
−
1
1
1
1
1
1
1
)2
22(2
)2
12(2
)2
11(2
)2
21(2
22
12
12
22
−
∞→
−
+
+
∞→
+∞→
=
=
×
−
+×
+
−=
×−
+×
+
−=
=
x
x
xlt
x
xlt
a
alt
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
α
α
1,1)
1,11)
1
11
<>
><>∴−
α
α
xwhenii
xorxwheni
.111,'' 1 ><>∴ −∑ xfordivergesandxorxforconvergesaTestRatiosAlembertDBy n
iii) when x = 1, 12
22
+
−=
n
n
na
0
1
)2
11(2
)2
21(2
12
22
≠
=
+
−=
+
−=
→∞
→∞→∞
n
n
n
n
n
n
n
nn
n
lt
ltalt
WKT
∑≠→∞
divergentisaseriesthethenaltIf nnn
,0
iii) Test for convergence the series ∑ n
n
3
2
Solution : The given series is ∑ n
n
3
2
and 1
2
1
2
3
)1(,
3 ++
+==
nnnn
na
na
13
3
)1
1(
)1(
3
3
22
2
2
12
1
>=
×
+
=
+×=
=
∞→
+
∞→
+∞→
α
α
nn
nlt
n
nlt
a
alt
n
n
nn
n
n
n
.1111 ≥<>∴ −∑ xfordivergesandxorxforconvergesan
.,'' convergesaTestRatiosAlembertDBy n∑∴
iv) Test for convergence the series ∞++++ ........4
!4
3
!3
2
!21
332
Solution : The given series is ∑ ∞++++= ........4
!4
3
!3
2
!21
332na
and 11)1(
)!1(,
!++
+
+==
nnnnn
na
n
na
1
)1
1(
)1(
!.)1(
)1(!
)!1(
)1(!
1
1
1
>=
+=
+=
+
+×=
+
+×=
=
→∞
→∞
+
→∞
+
→∞
+→∞
e
nlt
n
nlt
nn
n
n
nlt
n
n
n
nlt
a
alt
n
n
n
n
n
n
nn
n
nn
n
n
n
α
α
v) Test for convergence the series
∞++++
++++
++
+++
+
++ ........
)13)(12)(1(
)13)(12)(1(
)12)(1(
)12)(1(
1
11
bbb
aaa
bb
aa
b
a
Solution : The given series is ∞++++
++++
++
+++
+
++=∑ ........
)13)(12)(1(
)13)(12)(1(
)12)(1(
)12)(1(
1
11
bbb
aaa
bb
aa
b
aan
and )1)1)((1).....(12)(1(
)1)1)((1).....(12)(1(,
)1).....(12)(1(
)1).....(12)(1(1
+++++
+++++=
+++
+++= +
bnnbbb
annaaaa
nbbb
naaaa nn
.,'' convergesaTestRatiosAlembertDBy n∑∴
a
b
ana
bnblt
annaaa
bnnbbb
nbbb
naaalt
a
alt
n
n
n
n
n
=
++
++=
+++++
+++++×
+++
+++=
=
∞→
∞→
+∞→
α
α
)1(
)1(
)1)1)((1).....(12)(1(
)1)1)((1).....(12)(1(
)1).....(12)(1(
)1).....(12)(1(
1
1,1)
1,1)
<><
><>∴
α
α
baora
bwhenii
baora
bwheni
iii)when a = b, .,......1111 divergentiswhichan ∞++++=∑
.0101 >≥≤<<>∴ ∑ baora
bfordivergesandbaor
a
bforconvergesan
vi) Test for convergence the series .........111 3
3
2
2
∞++
++
++ x
x
x
x
x
x
Solution : The given series is ∞++
++
++
=∑ ........111 3
3
2
2
x
x
x
x
x
xan
and 1
1
11
,1 +
+
++
=+
=n
n
nn
n
nx
xa
x
xa
1
1
1
1
1
1
1
1
11
1
1
+
+
∞→
+
∞→
+
+
∞→
+∞→
+
+=
+
+×=
+×
+=
=
n
n
n
n
n
n
n
n
n
n
n
n
n
n
xx
xlt
x
x
xlt
x
x
x
xlt
a
altα
.11,'' baora
bfordivergesandbaor
a
bforconvergesaTestRatiosAlembertDBy n ><<>∴ ∑
],01
[1
)1(
)1
1(
,1
],0[1
,1
1
1
1
1
1
1
∞→→=
+
+
=>
∞→→=<
+
+
+
+
+
∞→
+
nasx
x
xx
xx
ltxIf
nasxx
xIf
n
n
n
n
n
n
n
∵
∵
α
α
1,1)
1,1)
<>
><∴
α
α
xwhenii
xwheni
iii) when x = 1, .,..........2
1
2
1
2
1divrgentiswhichan ∞+++=∑
Exercise : 2.5
Test for convergence the following series.
1) ]11:[.....32
32
≥<∞+++ xforDivergesandxforConvergesAnsxx
x
2) ]11:[)2(
1≥<
+
+∑ xforDivergesandxforConvergesAnsx
nn
n n
3) ]:[3!
1
ConvergesAnsn
n
nn
n
∑∞
=
4) ]11:[......1
.....52
12
2
≥<∞++
++++ xforDivergesandxforConvergesAnsn
xxxn
5) ]11:[......531 2 ≥<∞+++ xforDivergesandxforConvergesAnsxx
.11,'' ><∴ ∑ xfordivergesandxforconvergesaTestRatiosAlembertDBy n
.11 ≥<∴ ∑ xfordivergesandxforconvergesan
2.7 Alternating Series
1) A series in which the terms are alternatively positive or negative is called an
alternative series.
2) Leibnitz’s Test
An alternating series ......,4321 +−+− aaaa converges if i) each term is
numerically less than its preceding term (ie) 01 <−+ nn aa and ii) .0=→∞
nn
alt
If ,0≠∞→
nn
alt then the given series is oscillatory.
Example : 2.6
i) Discuss the convergence of the series ∞+−+− ........4
1
3
1
2
11
Solution : The terms of the given series alternatively positive and negative and each term is
numerically less than its preceding term.
.01
1
1&
1
1,
1) 11 <−
+=−
+== ++
nnaa
na
naie nnnn
Also 01
==∞→∞→ n
ltaltn
nn
∴ By Leibnitz’s Test, the given series is convergent.
ii) Discuss the convergence of the series ∞+−+− ........8
11
6
9
4
7
2
5
Solution : The terms of the given series alternatively positive and negative and each term is
numerically less than its preceding term.
012
)3
2(
2
32
1,02
32
22
52&
22
52,
2
32)
1
1
≠=
+
=+
=
><+
−+
+=−
+
+=
+=
∞→∞→∞→
+
+
n
nn
ltn
nltalt
nn
n
n
naa
n
na
n
naie
nnn
n
nn
nn
∴ By Leibnitz’s Test, the given series is oscillatory.
iii) Examine the character of the series ∑∞
=
−
−
−
1
1
12
)1(
n
n
n
n
Solution : The given series is ∞+−+−=−
−=∑∑
∞
=
−∞
=
......7
4
5
3
3
21
12
)1(
1
1
1 n
n
n
nn
na
The terms of the given series alternatively positive and negative and each term is
numerically less than its preceding term.
02
1
)1
2(12
0)12)(12(
1
1212
1&
12
1,
12)
1
1
≠=
∞−
=−
=
<−+
−=
−−
+
+=−
+
+=
−=
→∞→∞→∞
+
+
n
nlt
n
nltaltAlso
nn
n
n
n
naa
n
na
n
naie
nnn
n
nn
nn
∴ By Leibnitz’s Test, the given series is oscillatory.
iv) Examine the character of the series ∑∞
=
−
−
−
2
1
)1(
)1(
n
nn
nn
x
Solution : The given series is ∞−+−+−=−
−=∑∑
∞
=
−∞
=
......201262)1(
)1( 5432
2
1
1
xxxx
nn
xa
n
nn
n
n
The terms of the given series alternatively positive and negative and each term is
numerically less than its preceding term.
]10,2[0)1(
]10,2[0
)1()1(
)]1()1[(
)1()1(&
)1(,
)1()
1
1
1
1
<<≥=−
=
<<≥<
−+
+−−=
−−
+=−
+=
−=
∞→∞→
+
+
+
+
xnnn
xltaltAlso
xn
nnn
nxnx
nn
x
nn
xaa
nn
xa
nn
xaie
n
nn
n
n
nn
nn
n
n
n
n
∵
∵
∴ By Leibnitz’s Test, the given series is convergent.
v) Examine the character of the series
∞−
−+
−−
− .......
4log
1
2
1
3log
1
2
1
2log
1
2
1
Solution : The given series is ∞−
−+
−−
−=∑
∞
=
.......4log
1
2
1
3log
1
2
1
2log
1
2
1
1n
na
The terms of the given series alternatively positive and negative and each term is
numerically less than its preceding term.
)]1log()2log([0)2log()1log(
)1log()2log(
)2log(
1
)1log(
1
)1log(
1
2
1
)2log(
1
2
1&
)2log(
1
2
1,
)1log(
1
2
1)
1
1
+>+>++
+−+=
+−
+=
+−−
+−=−
+−=
+−=
+
+
nnnn
nn
nn
nnaa
na
naie
nn
nn
∵
∴ Leibnitz’s Test cannot be apply for the given series.
Now, 02
1
)1log(
1
2
1≠=
+−=
→∞→∞ nltalt
nn
n
Since 0≠∞→
nn
alt , the given series is divergent.
Exercise:2.6
Discuss the convergence of the following series
1) ]:[1
)1(
1
cos
12
12
tConverrgenAnsn
orn
n
n
n
n
∑∑∞
=
∞
= +
−
+
π
2) ]:[.......4log
1
2
1
3log
1
2
1
2log
1
2
1ConvergentAns∞−
−+
−−
−
3) ]:[1
)1(
22
1
ConvergentAnsn
n
n
n
∑∞
=
−
+
−
4) ]:[)2
1(,..........4321 32
convergentAnsxxxx <∞+−+−
5) ]:[.......!6
1
!4
1
!2
11 ConvergentAns∞+−+−
6) ]:[!
)1(
0
ConvergemtAnsnn
n
∑∞
=
−
2.8 Series of Positive and Negative Terms
The series of positive and negative terms and the alternating series are special types of
these series with arbitrary sign.
Let ∑ na be the series of real numbers. If ∑ na converges, then ∑ na is called
absolutely convergent. If ∑ na converges and ∑ na diverges, then ∑ na is called
conditionally convergent.
For Example 1) ....2
1)1(....
2
1
2
1
2
11
1
1
32+−++−+−=
−
−∑ n
n
na
The above series converges by Leibnitz’s Test.
Also, ....2
1....
2
1
2
1
2
11
132++++++=
−∑ nna
Which is a geometric series and it is converges.
Hence ∑ na is absolutely convergent.
2) ....1
)1(....4
1
3
1
2
11 1 +−++−+−= +∑
na
n
n
The above series converges by Leibnitz’s Test.
But , ....1
....4
1
3
1
2
11
1++++++==∑∑
nnan
The above series diverges by p test.
Hence ∑ na is conditionally convergent.
Note:
11) An absolutely convergent series necessarily convergent, but
12) the convergent series need not be absolutely convergent.
13) The series∑ na is absolutely convergent if 11
>+
∞→n
n
n a
alt and divergent if .1
1
<+
→∞n
n
n a
alt
Example : 2.7
i) Test the given series ∞+−−++−−+ ......8
1
7
1
6
1
5
1
4
1
3
1
2
11
2222222is absolutely
convergent or not.
Solution : The given series ∞+−−++−−+=∑ ......8
1
7
1
6
1
5
1
4
1
3
1
2
11
2222222na
and ∞++++++++==∑∑ ......8
1
7
1
6
1
5
1
4
1
3
1
2
11
122222222
nan
Which converges by p test. ∑ ≤>⇒ ]1&11
[ pifdivergespifconvergesn
Testpp
∴ The given series ∑ na is absolutely convergent and hence it is convergent.
ii) Test the given series ∞++++−++++− ......)4321(5
1)321(
4
1)21(
3
1
2
13333
is
absolutely convergent or not.
Solution : The given series ∞++++−++++−=∑ ......)4321(5
1)321(
4
1)21(
3
1
2
13333na
and ∞++++++++++=∑ ......)4321(5
1)321(
4
1)21(
3
1
2
13333na
0
)2
1(2)1(2)1(2
0])1()2(
)2()1([
2
1
)1(2)2(2
)1(
)2(2
)1(
)2(2
)2)(1(
)2(
)]1(..321[
,)1(2)1(2
)1(
)1(
)..321(
2222
22
23
221
2331
233
=
+
=+
=+
=
<++
+−+=
+−
+
+=−
+
+=
+
++=
+
++++++=
+=
+
+=
+
++++=
∞→∞→∞→∞→
+
+
nn
nlt
n
nlt
n
nltalt
nn
nnn
n
n
n
naa
n
n
n
nn
n
nna
n
n
n
nn
n
na
nnnn
n
nn
n
n
∴ By Leibnitz Test, the series ∑ na is convergent.
iii) Test the given series ∑∞
=
−
−
−
1
1
12
)1(
n
n
nis absolutely convergent or not.
Solution : The given series ∑∑∞
=
−
−
−=
1
1
12
)1(
n
n
nn
a and ∑∑∞
= −=
1 12
1
n
nn
a
012
1
0)12(2
212
12
1
2
1
2
1,
12
1
1
1
=−
=
<−
−−=
−−=−
=−
=
→∞→∞
+
+
nltalt
nn
nn
nnaa
na
na
nn
n
nn
nn
∴ By Leibnitz Test, the given series ∑ na is convergent.
Also, Let n
bn
1= & WKT which is a divergent series.
02
1
1)1(2 2≠=×
+=
∞→∞→
n
n
nlt
b
alt
nn
n
n
∑ ∑ .,sin divergesalsoisadivergesbce nn
∑∑ ∑
∴
∴
.
.
convergentllyconditionaisa
divergesabutconvergesa
n
nn
iv) Test the given series ∑∞
=
−
22)(log
)1(
n
n
nnis absolutely convergent or not.
Solution : The given series ∑∑∞
=
∞
=
−=
22
2 )(log
)1(
n
n
n
nnn
a and ∑∑∞
=
∞
=
=2
22 )(log
1
nn
nnn
a
0)(log
1
0)(log
1
))1)(log(1(
1
))1)(log(1(
1,
)(log
1
2
221
212
==
<−++
=−
++==
∞→∞→
+
+
nnltalt
nnnnaa
nna
nna
nn
n
nn
nn
∴ By Leibnitz Test, the given series ∑∞
=2n
na is convergent.
Also,
..,2log
1
,,2log,21
1,log
1
)(log
1
2log
2log
2
2
2
finiteiswhich
txwhentxwhent
dxx
dtxtPutdtt
dxxx
=
∞=∞===
−=
===
∞
∞∞
∫∫
∴By Integral Test, the ∑∞
2
na is converges.
Since .,&222
absolutelyconvergesaconvergesareaa nnn ∑∑∑∞∞∞
v) Show that the given series ∑∞
=
−
+
−
1
1
12
)1(
n
nn
n
xis absolutely convergent for 1<x .
Solution : The given series ∑∑∞
=
−∞
= +
−=
1
1
1 12
)1(
n
nn
n
nn
xa and ∑∑
∞
=
∞
= +=
11 12n
n
n
nn
xa
x
nn
nn
xlt
n
n
xlt
x
n
n
xlt
a
alt
n
xa
n
xa
n
n
n
n
n
n
n
n
n
n
n
n
1
)1
2(
)3
2(1
12
321
1)1(2
12
,1)1(2
,12
1
1
1
1
=
+
+
×=
+
+×=
++×
+=
=
++=
+=
∞→
∞→
+∞→
+∞→
+
+
α
.&1,111
1
convrgentabsolutelyisaseriesgiventhethenxorx
Ifn
n∑∞
=
><> α
vi) Show that the given series ∞−+− .......3
3sin
2
2sin
1
sin333
xxxis converges absolutely.
Solution : The given series ∞−+−=∑∞
=
.......3
3sin
2
2sin
1
sin333
1
xxxa
n
n and
∞+++=∑∞
=
......3
3sin
2
2sin
1
sin333
1
xxxa
n
n
∑==∴ .&1sin
33testpbyconvegesbwkt
nbletand
n
nxa nnn
Also, 33
1sin
nn
nx≤
x
nn
nn
xlt
n
n
xlt
x
n
n
xlt
a
alt
n
xa
n
xa
n
n
n
n
n
n
n
n
n
n
n
n
1
)1
2(
)3
2(1
12
321
1)1(2
12
,1)1(2
,12
1
1
1
1
=
+
+
×=
+
+×=
++×
+=
=
++=
+=
∞→
∞→
+∞→
+∞→
+
+
α
.&1,111
1
convrgentabsolutelyisaseriesgiventhethenxorx
Ifn
n∑∞
=
><> α
Exercise : 2.7
Test the given series is absolutely convergent or not.
1) ConvergesAbsolutelyAnsxx :..........13
1
12
1
11
1 2
333∞−
++
+−
+
2) ConvergesllyConditionaAns :..........14
1
13
1
12
1∞−
++
+−
+