Post on 09-Mar-2018
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INFINITE SERIES
Sequence:
If a set of real numbers occur according to some definite rule, then it is called
a sequence denoted by * + * + if n is finite
Or * + * + if n is infinite.
Series:
is called a series and is denoted by ∑
Infinite Series:
If the number of terms in the series is infinitely large, then it is called infinite series and is
denoted by ∑ and the sum of its first n terms be denoted
by ∑ .
Convergence:
An infinite series ∑ is said to be convergent if , a definite unique number.
Example:
.
/
.
/ , finite.
Therefore given series is convergent.
Divergence:
tends to either then the infinite series ∑ is said to be divergent.
Example: ∑
( )
Therefore ∑ is divergent.
Oscillatory Series:
If tends to more than one limit either finite or infinite, then the infinite series ∑
is said to be oscillatory series.
Example:1. ∑ :
{
Therefore series is oscillatory.
2. ∑ ( ) ( )
( )
{
Properties of infinite series:
1. The convergence or divergence of an infinite series remains unaltered on
multiplication of each term by .
2. The convergence or divergence of an infinite series remains unaltered by addition or
removal of a finite number of its terms.
Positive term series:
An infinite series in which all the terms after some particular term are positive is called a
positive term series.
Geometric Series test:
The series ∑
a. Converges if | |
b. Diverges if
c. Oscillates finitely if and oscillates infinitely if
Proof:
Let be the partial sum of ∑ .
Case 1: | | i.e.
Therefore the series is convergent.
Case 2i: i.e.
Therefore the series is divergent.
Case 2ii: ,
. Therefore the series is divergent.
Case 3i: i.e.
( )
{
Therefore the series is oscillatory.
Case 3ii:
i.e.
{
Therefore the series is oscillatory.
Note: If a series in which all the terms are positive is convergent, the series remains
convergent even when some or all of its terms are negative.
Integral Test:
A positive term series ( ) ( ) ( ) Where ( ) decreases as n
increases, converges or diverges according as the integral ∫ ( )
is finite or infinite.
p-series or Harmonic series test:
A positive term series ∑ ∑
is
i) Convergent if
ii) Divergent if
Proof:
Let ( )
∫ ( ) ∫
[
]
{
{
When , ∫ ( ) ∫
, -
Thus ∑
converges if and diverges if .
Theorem:
Let ∑ be a positive term series. If ∑ is convergent then .
Proof:
If ∑ is convergent then .
( ) ( )
Note:
Converse need not be always true. i.e. Even if , then ∑ need not be
convergent.
Example 1: ∑
.
∑
is divergent by integral test. But
Hence is a necessary condition but not a sufficient condition for convergence
of ∑ .
Example 2
Test the series for convergence, ∑
Solution: Consider ∫
, ( )-
Therefore ∑ is divergent by Integral test.
Example 2
Test the series for convergence, ∑
Solution: Let . Then
∫ ∫
[
]
Therefore ∑ is convergent.
Comparison test:
1. Let ∑ and ∑ be two positive term series. If
a. ∑ is convergent
b.
Then ∑ is also convergent.
That is if a larger series converges then smaller also converge.
2. Let ∑ and ∑ be two positive term series. If
c. ∑ is divergent
d.
Then ∑ is also divergent.
That is if a smaller series diverges then larger also diverges.
Example 2
Test the series for convergence, ∑
Solution:
Let
and
n n
log n n
1 1
log n n
u v
But ∑ ∑
is a p-series with .
Therefore ∑ is divergent.
By comparison test ∑ is also divergent.
Example 2
Test the series for convergence, ∑
Solution:
Let
and
n n
n n
n n
2 2 1
1 1
2 2 1
v u
But ∑ ∑
is a geometric series with
.
Therefore ∑ is convergent.
By comparision test ∑ is also convergent.
Another form of comparison test is
Limit test
Statement: If ∑ and ∑ be two positive term series such that
( ).
Then ∑ and ∑ behave alike.
That is if ∑ converges then ∑ also converge.
If ∑ diverges then ∑ also diverge.
Examples 3.
Test the series for convergence,
( )( )
Choose
then
But ∑ ∑
with .
Therefore ∑ is convergent. By limit test ∑ is also convergent.
Examples 4.
Test the series for convergence, ∑ (√ )
Solution: (√ )(√ )
(√ )
√
(√ )
Let ∑ ∑
( )
But ∑ is divergent. By limit test ∑ is also divergent.
Examples 5.
Test the series for convergence, ∑ √
Solution:
13 3
n
3 3 2 2
3 3
2 2
3 313 3
n 2 13 3 23 3
2 13 3
2
3 3
u n 1 n
a b (a b)(a ab b )
a ba b
a ab b
n 1 nu n 1 n
n 1 (n 1) n n
1
1 1n 1 1 1
n n
Let ∑ ∑
with .
But ∑ is convergent. By limit test ∑ is also convergent.
Example 6.
Test the series for convergence, Solve √
√
√
Solution:
n 3 3
3
3
1 1n 1
n nn 1 1u
n 2 1 2 1n 1
n n
Let ∑ ∑
⁄ with
.
But ∑ is convergent. By limit test ∑ is also convergent.
Example 7
Test the series for convergence, ∑
Solution:
We know that
Let ∑ ∑
. Then
But ∑ is convergent. By limit test ∑ is also convergent.
Example 8
Test the series for convergence, ∑
.
/
Solution:
.
/
[
]
[
]
Let ∑ ∑
. Then
But ∑ is convergent. By limit test ∑ is also convergent.
Exercises
Test for convergence of the series
1. ∑
2.
…… …
3.
…….. …
4. ∑√
5. ∑
( )
6.
…… …
INFINITE SERIES
D’Alembert’s Ratio Test: If ∑ is a series of positive terms, and
nlim
n
n
u
u 1
( )
then the series is convergent if , is divergent if and the test fails if .
If the test fails, one should apply comparison test or the Raabe’s test, as given below:
Raabe’s Test: If ∑ is a series of positive terms, and
.
/ ( ) then the series is convergent if , is divergent if
and the test fails if .
Remark: Ratio test can be applied when (i) does not have the form ⁄
(ii) nth
term has etc.
(iii) nth
term has ( ) ( ) ect.
(iv) the number of factors in numerator and denominator increase steadily, ex: (
)
Example : Test for convergence the series
1 + !2
2 2
+ !3
32
+ !4
4 2
+ ….
>> The given series is of the form !1
12
+ !2
2 2
+ !3
32
+ !4
3 + … whose n
th term is un =
!
2
n
n.
Therefore un+1 = !1
)1( 2
n
n
n
n
u
u 1 = !1
)1( 2
n
n2
!
n
n =
2
2)1(
n
n .
)!)(1(
!
nn
n
=
2
1
n
n
Therefore n
lim n
n
u
u 1 = n
lim
2
1
n
n =
nlim
2
11
nn = 0 < 1
Therefore by ratio test, un is convergent.
Example : Discuss the nature of the series
2.1
x +
3.2
2x +
4.3
3x + ….
>> un = )1( nn
x n
Therefore un+1 = )11)(1(
1
nn
x n
= )2)(1(
1
nn
x n
Now n
n
u
u 1 = )2)(1(
1
nn
x n
. nx
nn )1( =
2n
n x
Therefore n
limn
n
u
u 1 = n
lim 2n
nx =
nlim
)/21(
1
nx = x
Therefore by D’Alembert’s ratio test un is
1 x if divergent
1 x if convergent
And the test fails if x = 1
But when x = 1, un = )1(
1
nn
n
= )1(
1
nn =
nn 2
1
un is of order 1/n2 (p = 2 > 1) and hence un is convergent (when x = 1). Hence we
conclude that un is convergent x 1 and divergent if x > 1
Example : Find the nature of series 1 + 2
x +
5
2x +
10
3x + ….
>> Omitting the first term, the given series can be written in the form
112
1
x +
122
2
x +
132
3
x + … so that un =
12 n
x n
Therefore un+1 = 222
1
nn
x n
. 22
12
2
nn
nx =
nlim
)/2/21(
)/11(22
22
nnn
nn
.x
That is, n
lim n
n
u
u 1 = x
Hence by ratio test un is
1 x if divergent
1 x if convergent
and the test fails if x = 1.
But when x = 1, un = 1
12 n
n
= 1
12 n
is of order 2
1
n (p = 2 > 1)
Therefore un is convergent if x 1 and divergent if x > 1.
Example: Find the nature of the series 12
1 +
23
2x +
34
4x + …
>> omitting the first term, the general term of the series is given by un = 1)2(
2
nn
x n
Therefore un+1 = 1)1()21(
)1(2
nn
nx =
2)3(
22
nn
x n
n
n
u
u 1 = 2)3(
22
nn
x n
nx
nn2
1)2(
= 3
2
n
n
2
2
1x
n
n
=
)3(
)1)(2(
n
nn x
2
nlim
n
n
u
u 1 = n
lim )/31(
)/11()/21(
nn
nnnn
. x
2 = x
2
Hence by ratio test un is
1 xif divergent
1 xif
2
2convergent
and the fails if x2 = 1.
When x2 = 1, un =
1)2(
)1(
nn
n
= 1)2(
1
nn
un is of order 1/n3/2
(p = 3/2 > 1) and hence un is convergent.
Therefore un is convergent if x2 1 and divergent if x
2 > 1.
Example : Discus the convergence of the series
x + 3.2
3x +
4.2
3 +
5
5x +
6.4.2
5.3 .
7
7x + … (x > 0)
>> We shall write the given series in the form
x + 2
1.
3
3x +
4.2
3.1.
5
5x +
6.4.2
5.3.1.
7
7x + ….
Now, omitting the first term we have
un = n
n
2...6.4.2
)12...(5.3.1 .
12
12
n
x n
un+1 = )1(2...6.4.2
]1)1(2...[5.3.1
n
n.
1)1(2
1)1(2
n
x n
That is, un+1 = )12....(6.4.2
)12...(5.3.1
n
n .
32
32
n
x n
That is, un+1 = )22)(2....(6.4.2
)12)(12...(5.3.1
nn
nn .
32
32
n
x n
Therefore n
n
u
u 1 = )22)(2....(6.4.2
)12)(12...(5.3.1
nn
nn.
32
32
n
x n
)12...(5.3.1
2...6.4.2
n
n.
12
12
nx
n
That is, n
n
u
u 1 = )32)(22(
)12)(12( 2
nn
xnn
Therefore n
lim n
n
u
u 1 = n
lim )/32()/22(
)/12()/12( 2
nnnn
xnnnn
= x
2
Hence by ratio test, un is
1 xif divergent
1 xif
2
2convergent
And the test fails if x2 = 1
When x2 = 1,
n
n
u
u 1 = )32)(22(
)12)(12(
nn
nn and we shall apply Raabe’s test.
nlim n
11n
n
u
u =
nlim n
1
)12(
)32)(22(2n
nn
= n
lim n
2
22
)12(
)144()6104(
n
nnnn
= n
lim n
2)12(
56
n
n =
nlim
22
2
)/12(
)/56(
nn
nn
4
6 =
2
3 > 1
Therefore un is convergent (when x2 = 1) by Rabbe’s test.
Hence we conclude that, un is convergent if x2 1 and divergent if x
2 > 1.
Example : Examine the convergence of
1 + 5
2x +
9
6 x
2 +
17
14x
3 + … +
12
221
1
n
n
xn + ….
>> un = 12
221
1
n
n
xn.
Therefore un+1 = 12
222
2
n
n
xn+1
n
n
u
u 1 = 12
222
2
n
n
xn+1
22
121
1
n
n
. nx
1
n
n
u
u 1 = )2/11(2
)2/21(222
22
nn
nn
.x. )2/21(2
)2/11(211
11
nn
nn
= )2/11(
)2/11(2
1
n
n
.x . )2/11(
)2/11( 1
n
n
Therefore n
lim n
n
u
u 1 = )01(
)01(
. x .
)01(
)01(
= x.
Therefore by ratio test un is
1 x if divergent
1 x if convergent and the test fails if x = 1.
When x = 1, un = 12
221
1
n
n
Therefore n
lim un = n
lim)2/11(2
)2/11(211
1
nn
nn
= 1
Since n
lim un = 1 0, un is divergent (when x = 1)
Hence un is convergent if x < 1 and divergent if x 1.
Example : test for convergence of the infinite series
1 + 22
!2 +
33
!3 +
44
!4 + …
>> the first term of the given series can be written as 1!/11 so that we have,
un = nn
n! and un+1 =
1)1(
)!1(
nn
n =
1)1(
)!)(1(
nn
nn =
nn
n
)1(
!
Therefore n
n
u
u 1 = nn
n
)1(
!
.
!n
n n
= n
n
n
n
)1( =
nn
n
nn
n
)/11(
nlim
n
n
u
u 1 = n
lim nn)/11(
1
=
e
1 < 1
Hence by ratio test un is convergent.
Cauchy’s Root Test: If ∑ is a series of positive terms, and
( ) ⁄ ( ),
then, the series converges if , diverges if and fails if
Remark: Root test is useful when the terms of the series are of the form , ( )- ( ).
We can note : (i) ⁄
(ii) ( ⁄ ) ⁄
(iii) ) ( ⁄ ) ⁄
Example : Test for convergence
1
2/3
11
n
n
n
>> un =
2/3
11
n
n
Therefore (un)1/n
=
nn
n
/12/3
11
=
2/1
11
n
n
=
n
n
11
nlim (un)
1/n =
nlim
n
n
11
= n
lim n
n
11
1 =
e
1 < 1.
Therefore as n , n also
Therefore by Cauchy’s root test, un is convergent.
Example : Test for convergence
1
2
31
n
n
n
>> un =
2
31
n
n
Therefore (un)1/n
=
nn
n
/12
31
=
n
n
31
nlim (un)
1/n =
nlim
n
4
31 = e
-3.
Therefore n
lim
n
n
x
1 = e
x
That is, n
lim (un)1/n
= 3
1
e < 1, therefore e = 2.7
Hence by Cauchy’s root test, un is convergent.
Example : Find the nature of the series
2/3
1
11
n
n n
>> un =
2/3
11
n
n
Therefore (un)1/n
=
nn
n
/12/3
11
=
2/1
11
n
n
=
n
n
11
nlim (un)
1/n =
nlim
n
n
11
= n
lim n
n
11
1 =
e
1 < 1, since as n , n also
Therefore by Cauchy’s root rest, un is convergent.
Example : Test for convergence
1
2
31
n
n
n
>> un =
2
31
n
n
Therefore (un)1/n
=
nn
n
/12
31
=
n
n
31
nlim (un)
1/n =
nlim
n
n
31 = e
-3, since
nlim
n
n
x
1 = e
x
That is, n
lim (un)1/n
= 3
1
e < 1, since e = 2.7.
Hence by Cauchy’s root test, un is convergent.
ALTERNATING SERIES
A series in which the terms are alternatively positive or negative is called an alternating
series.
i.e., 1
1 2 3 4
1
... 1n
n
n
u u u u u
LEBINITZ’S SERIES
An alternating series 1
1 2 3 4
1
... 1n
n
n
u u u u u
converges if
(i) each term is numerically less than its preceding term
(ii) lim 0nn
u
Note: If lim 0nn
u
then the given series is oscillatory.
Q Test the convergence of 6
1 -
13
1 +
20
1 -
27
1 + …
Solution: Here un = 17
1
n
then un+1 = 1)1(7
1
n =
67
1
n
therefore, un – un+1 = 17
1
n -
67
1
n
= )67)(17(
)17()67(
nn
nn =
)67)(17(
7
nn > 0
That is, un – un+1 > 0, un > un+1
Also, n
lim un = n
lim 17
1
n =
nlim
n
1
)/17(
1
n = 0
Therefore by Leibnitz test the given alternating series is convergent .
Q Find the nature of the series
2log
11 -
3log
11 +
4log
11 -
5log
11 + …
Solution: Here un = 1 - )1log(
1
n then un+1 = 1 -
)2log(
1
n
Therefore, un – un+1 = )2log(
1
n -
)1log(
1
n
= )1log()2log(
)2log()1log(
nn
nn < 0.
Since (n + 1) < (n + 2)
un - un+1 < 0 un < un+1
further n
lim un = n
lim 1 -
)1log(
1
n = 1 – 0 = 1 0.
Both the conditions of the Leibnitz test are not satisfied. So, we conclude that the series
oscillates between - and + .
Problems:
Test the convergence of the following series
1
1
1
2
2
1 1 11 ....
2 3 4
1 1 1 1....
log 2 log3 log 4 log5
1( )
1
1 0 1
1
1
1
n
n
n n
n
i
ii
niii
n
xiv for x
n n
vn
ABSOLUTELY & CONDITIONALLY CONVERGENT SERIES
An alternating series 1
1 1
1n
n n
n n
a u
is said to be absolutely convergent if the positive
series 1 2 3 4 ... na a a a a is convergent.
An alternating series 1
1 1
1n
n n
n n
a u
is said to be conditionally convergent if
(i) na is divergent
(ii) 1
1 1
1n
n n
n n
a u
is convergent
Theorem: An absolutely convergent series is convergent. The converse need not be true.
Proof: Let 1
1 1
1n
n n
n n
a u
be an absolutely convergent series then na is
convergent.
We know, 1 2 3 4 1 2 3 4... ...a a a a a a a a
By comparison test, 1
n
n
a
is convergent.
Q. Show that each of the following series also converges absolutely
(i) an2; (ii)
2
2
1 n
n
a
a
; (iii)
n
n
a
a
1
Solution: (i) Since an converges, we have an 0 as n . Hence for some positive
integer N, |an| < 1 for all n N. This gives an2 |an| for all n N. As |an| is convergent it
follows an2 converges.
(as an2 is a positive termed series, convergence and absolute convergence are identical).
(ii) As 1 + an2 1 for all n, we get
2
2
1 n
n
a
a
an
2
the convergence of an2 implies the convergence of
2
2
1 n
n
a
a
.
(iii) n
n
a
a
1 =
|1|
||
n
n
a
a
<
||1
||
n
n
a
a
.
As |an| converges, |an| 0 as n . Hence for some positive integer N, we have |an| < ½
for all n N.
This gives n
n
a
a
1 < 2|an| for all n N.
Now, by comparison test, n
n
a
a
1 converges.
That is, n
n
a
a
1converges absolutely.
Q. Test the convergence 3 3 3 3
1 1 1 11 2 1 2 3 1 2 3 4 ...
2 3 4 5
Solution: Here
1 1 1
3 2
1 2 ...1 1 1
1 2 1
n n n
n n
n na u
n n
then
2
1 2 2
1 10
2 1 2n n
n nu u
n n
i.e., 1 &lim 0n n n
nu u u
Thus by Lebinitz rule, na is convergent.
Also, 2
1
2 1n
na
n
. Take
1nv
n
Then 1
lim 02
n
nn
a
v
Since is nv divergent, therefore na is also divergent.
Thus the given series is conditionally convergent.
POWER SERIES
A series of the form 2
0 1 2 ... ... ( )n
na a x a x a x i where the ia ’s are
independent of x, is called a power series in x. Such a series may converge for some or all
values of x.
INTERVAL OF CONVERGENCE
In the power series (i) we have n
n nu a x
Therefore, 1 1lim limn n
n nn n
u ax
u a
If 1lim n
nn
al
a
, then by ratio test, the series (i) converges when
1x
l and diverges for
other values.
Thus the power series (i) has an interval 1 1
xl l
within which it converges and diverges
for values of x outside the interval. Such interval is called the interval of convergence of the
power series.
Q. Find the interval of convergence of the series 2 3 4 5
....2 3 4 5
x x x xx .
Solution: Here 1
1n
n
n
xu
n
and
1
1 11
nn
n
xu
n
Therefore, 1lim lim1
n
n nn
u nx x
u n
By Ratio test the given series converges 1x for and diverges for 1x .
When x=1 the series reduces to 1 1 1
1 ...2 3 4
, which is an alternating series and is
convergent.
When x=-1 the series becomes1 1 1
1 ...2 3 4
, which is divergent (by comparison with
p-series when p=1)
Hence the interval of convergence is 1 1x .
Q. Show that the series
1
1)1( n
12 n
x n
is absolutely convergent for | x | < 1,
conditionally convergent for x = 1 and divergent for x = -1.
Solution. Here un = (-1)n-1
12 n
x n
Therefore un+1 = 32
)1( 1
n
x nn
nlim
n
N
u
u 1 = n
lim nn
nn
xn
nx1
1
)1(32
12)1(
= n
lim xn
n
32
12)1(
= n
lim xnn
nn
)/32(
)/12()1(
= | x |
Therefore by generalized D’ Alembert’s test the series is absolutely convergent if
| x | < 1, not convergent if | x | > 1 and the test fails if | x | = 1.
Now for |x | = 1, x can be +1 or – 1.
If x = 1 the given series becomes 3
1 -
5
1+
7
1 -
9
1 + …
Here un =12
1
n, un+1 =
32
1
n
But 2n + 1< 2n + 3 un > un+1
Also n
lim un = n
lim 32
1
n = 0
Therefore by Leibnitz test the series is convergence when x = 1.
But the absolute series 3
1 +
5
1 +
7
1 + … whose general term is un =
12
1
n and is of
order n
1 =
2/1
1
n and hence un is divergent
Since the alternating series is convergent and the absolute series is divergent when x = 1, the
series is conditionally convergent when x = 1.
If x = -1, the series becomes 3
1 -
5
1 -
7
1 - ….
= -
...
7
1
5
1
3
1 where the series of positive terms is divergent as shown already.
Therefore the given series is divergent when x = -1.
Thus we have established all the results.
Problems:
1. Test the conditional convergence of
1 1
1 2
1 1( )
1
n n
n n
ni ii
nn
2. Prove that 3 3 3
sin sin 2 sin3....
1 2 3
x x x is absolutely convergent
3. For what values of x the following series are convergent
2 2
2 3 4
2 3 4
2 2 2
2 3 4
4 4 9 9
....2 3 4
....2 3 4
....1.2 2.3 3.4 4.5
3 3 3 .... 3 ...n n
x x xi x
x x xii x
x x x xiii
iv x x x x
4. Test the nature of convergence
11
n
n n
******