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INFINITE SERIES
Sequence:
If a set of real numbers u1 ,u2 ,……,un occur according to some definite rule, then it is called
a sequence denoted by {Sn }={u1 , u2 ,……,un } if n is finite
Or {Sn }={u1 , u2 ,……,un ,…….} if n is infinite.
Series:
u1+u2+……+un is called a series and is denoted by Sn=∑k=1
n
uk
Infinite Series:
If the number of terms in the series is infinitely large, then it is called infinite series and is denoted by ∑ un=¿u1+u2+……+un+……¿ and the sum of its first n terms be denoted by
Sn=∑k=1
n
uk=u1+u2+……+un.
Convergence:
An infinite series ∑ un is said to be convergent if limn→∞
Sn=k, a definite unique number.
Example: 1+ 12+ 1
4+…… ..
Sn=1+ 12+ 1
22 +…… ..+ 12n−1=
(1−1
2n )(1−1
2 )=2, finite.
Therefore given series is convergent.
Divergence:
limn→∞
Sn tends to either ∞∨−∞ then the infinite series ∑ un is said to be divergent.
Example: ∑ un=1+2+3+……. .
Sn=n (n+1 )
2
limn→∞
Sn=∞
Therefore ∑ un is divergent.
Oscillatory Series:
If limn→∞
Sn tends to more than one limit either finite or infinite, then the infinite series ∑ un is
said to be oscillatory series.
Example:1. ∑ un=1−1+1−1+…… ..:
Sn={1 ,n is odd0 , n is even
Therefore series is oscillatory.
2. ∑ un=1+(−3)+(−3)2+…… ..
Sn=1−(−1)n3n
1+3
limn→∞
Sn={ ∞,n is odd−∞ ,n is even
Properties of infinite series:
1. The convergence or divergence of an infinite series remains unaltered on multiplication of each term by c ≠o.
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 ∑n=0
∞
rn=1+r+r 2+r 3+……+rn+……
a. Converges if |r|<1b. Diverges if r ≥1c. Oscillates finitely if r=−1 and oscillates infinitely if r←1
Proof:
Let Sn be the partial sum of ∑n=0
∞
rn.
Sn=1+r+r2+…+r n−1
Case 1:|r|<1 i.e. −1<r<1
Sn=1−rn
1−r
limn→∞
Sn=1
1−r
Therefore the series is convergent.
Case 2i:r>1 i.e. limn→∞
rn=∞
Sn=rn−1r−1
limn→∞
Sn=∞
Therefore the series is divergent.
Case 2ii: r=1,Sn=1+1+1+1+…….+1=n limn→∞
Sn=∞.
Therefore the series is divergent.
Case 3i:r←1 i.e. Let r=−m
Sn=1−rn
1−r=
1−(−1)nmn
1+m
limn→∞
Sn={ ∞,n is odd−∞ ,n is even
Therefore the series is oscillatory.
Case 3ii:r=−1
i.e. Sn=1−1+1−1+…….
limn→∞
Sn={1 , n is odd0 , n is even
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 f (1 )+ f (2 )+…….+f (n )+……. Where f (n) decreases as n increases,
converges or diverges according as the integral ∫1
∞
f ( x )dx is finite or infinite.
p-series or Harmonic series test:
A positive term series ∑ un=∑ 1
np=1+ 1
2p+ 1
3p+……+ 1
np+…….is
i) Convergent if p>1ii) Divergent if p≤1
Proof:
Let f ( x )= 1
x p
∫1
∞
f ( x )dx=∫1
∞1
x pdx=[ x−p+1
−p+1 ]1
∞
, For p≠1
¿ { ∞ ,if−p+1>01p−1
, if−p+1<0
¿ { ∞ ,if p<11p−1
, if p>1
When p=1, ∫1
∞
f ( x )dx=∫1
∞1xdx= [logx ]1
∞=∞
Thus ∑ 1
np converges if p>1 and diverges if p≤1.
Theorem:
Let ∑ un be a positive term series. If ∑ un is convergent then limn→∞
un=0.
Proof:
If ∑ un is convergent then limn→∞
Sn=k.
un=(u1+u2+……+un )−(u1+u2+……+un−1 )
¿ Sn−Sn−1
limn→∞
Sn−1=k
limn→∞
un= limn→∞
Sn− limn→∞
Sn−1
¿k−k=0
Note:
Converse need not be always true. i.e. Even if limn→∞
un=0, then ∑ unneed not be convergent.
Example 1: ∑ un=1+ 12+ 1
3+ 1
4+…… ....
∑ un=1n
is divergent by integral test. But limn→∞
un= limn→∞
1n=¿0¿
Hence limn→∞
un=0 is a necessary condition but not a sufficient condition for convergence of
∑ un.
Example 2
Test the series for convergence, ∑n=2
∞1
nlogn
Solution: Consider ∫2
∞1
nlogndn=[ log ( logn ) ]2
∞=∞
Therefore ∑ un is divergent by Integral test.
Example 2
Test the series for convergence, ∑ ne−n2
Solution: Let x2=t . Then 2 xdx=dt
∫1
∞
x e−x2
dx=∫1
∞e−t
2dt=[ e−t−2 ]
1
∞
=1
2e
Therefore ∑ un is convergent.
Comparison test:
1. Let ∑ un and ∑ vn be two positive term series. If
a. ∑ vn is convergent
b. un≤vn ,∀n
Then ∑ un is also convergent.
That is if a larger series converges then smaller also converge.
2. Let ∑ un and ∑ vn be two positive term series. If
c. ∑ vn is divergent
d. un≥vn ,∀n
Then ∑ un is also divergent.
That is if a smaller series diverges then larger also diverges.
Example 2
Test the series for convergence, ∑n=2
∞1logn
Solution:
Let un=1
logn and vn=
1n
But ∑ vn=∑ 1n
is a p-series with p=1.
Therefore ∑ vn is divergent.
By comparison test ∑ un is also divergent.
Example 2
Test the series for convergence, ∑ 1
2n+1
Solution:
Let un=1
2n+1 and vn=
1
2n
But ∑ vn=∑ 1
2n is a geometric series with r=
12<1.
Therefore ∑ vn is convergent.
By comparision test ∑ un is also convergent.
Another form of comparison test is
Limit test
Statement: If ∑ un and ∑ vn be two positive term series such that limn→∞
unvn
=k (≠0). Then
∑ un and ∑ vn behave alike.
That is if ∑ un converges then ∑ vn also converge.
If ∑ un diverges then ∑ vn also diverge.
Examples 3.
Test the series for convergence, 1
1.2.3+ 3
2.3.4+ 5
3.4 .5+……….
Solution :
un=2n−1
n (n+1)(n+2)
Choose vn=1
n2 then limn→∞
unvn
=2
But ∑ vn=∑ 1
n2 with p=2>1.
Therefore ∑ vn is convergent. By limit test ∑ un is also convergent.
Examples 4.
Test the series for convergence, ∑n=1
∞
(√n2+1−n )
Solution: un=(√n2+1−n ) (√n2+1+n )(√n2+1+n )
¿n2+1−n2
√n2+1+n
¿1
n (√1+n2+1)
Let ∑ vn=∑ 1n
(p=1 )
limn→∞
unvn
=12
But ∑ vn is divergent. By limit test ∑ un is also divergent.
Examples 5.
Test the series for convergence, ∑ 3√n3+1−n
Solution:
Let ∑ vn=∑ 1
n2 with p=2>1.
limn→∞
unvn
=13
But ∑ vn is convergent. By limit test ∑ un is also convergent.
Example 6.
Test the series for convergence, Solve √2−133−1
+ √3−143−1
+ √4−153−1
+…………
Solution:
Let ∑ vn=∑ 1
n52 with p=
52>1.
limn→∞
unvn
=1
But ∑ vn is convergent. By limit test ∑ un is also convergent.
Example 7
Test the series for convergence, ∑ 1
n3tan
1n
Solution: un=1
n3tan
1n
We know that limn→∞
tan1n
1n
=1
Let ∑ vn=∑ 1
n4 . Then limn→∞
unvn
=1
But ∑ vn is convergent. By limit test ∑ un is also convergent.
Example 8
Test the series for convergence, ∑ 1n− log( n+1
n )
Solution:un=1n− log(1+ 1
n )¿ 1n−[ 1n− 1
2n2+ 1
6n3−……… ]
¿ [ 1
2n2− 1
6n3+………]
Let ∑ vn=∑ 1
n2 . Then limn→∞
unvn
=12
But ∑ vn is convergent. By limit test ∑ un is also convergent.
Exercises
Test for convergence of the series
1. ∑n=0
∞2n3+14n5+1
2. 1+ 22
2!+ 32
3 !+ 42
4 !+¿…… …∞
3.1
1.2.3+ 3
2.3.4+ 5
3.4 .5+¿…….. …∞
4. ∑ √ 3n−1
2n+1
5. ∑ nn
(n+1 )n+1
6.1
12+ 1+2
12+22+ 1+2+3
12+22+32+¿…… …∞
INFINITE SERIES
D’Alembert’s Ratio Test: If ∑ un is a series of positive terms, and ¿ l(a finite value)
then the series is convergent if l<1 , is divergent if l>1 and the test fails if l=1.
If the test fails, one should apply comparison test or the Raabe’s test, as given below:
Raabe’s Test: If ∑ un is a series of positive terms, and
limn→∞
n( unun+1
−1)=l ( finite ) , then the series is convergent if l>1 , is divergent if l<1 and the
test fails if l=1.
Remark: Ratio test can be applied when (i) vn does not have the form 1
np
(ii) nth term has xn , x2n etc.
(iii) nth term has n ! , (n+1 ) ! ,(n!)2 ect.
(iv) the number of factors in numerator and denominator increase steadily, ex: (
13+ 1.2
3.5+ 1.2.3
3.5 .7+…¿
Example : Test for convergence the series
1 +
22
2! +
32
3! +
42
4 ! + ….
>> The given series is of the form
12
1! +
22
2! +
32
3! +
43
4 ! + … whose nth term is un =
n2
n ! .
Therefore un+1 =
(n+1 )2
n+1 !
un+1
un =
(n+1 )2
n+1 !
n !
n2 =
(n+1 )2
n2.
n !(n+1 )(n !) =
n+1
n2
Therefore limn→∞
un+1
un = limn→∞
( n+1
n2 ) =
limn→∞
( 1n+
1
n2) = 0 < 1
Therefore by ratio test, un is convergent.
Example : Discuss the nature of the series
x1. 2 +
x2
2. 3 +
x3
3. 4 + ….
>> un =
xn
n(n+1 )
Therefore un+1 =
xn+1
(n+1 )(n+1+1) =
xn+1
(n+1 )(n+2)
Now
un+1
un =
xn+1
(n+1 )(n+2) .
n(n+1 )xn =
nn+2 x
Therefore limn→∞
un+1
un = limn→∞
nn+2 x =
limn→∞
1(1+2/n ) x = x
Therefore by D’Alembert’s ratio test un is {convergent if x < 1¿ ¿¿¿
And the test fails if x = 1
But when x = 1, un =
1n
n(n+1 ) =
1n(n+1 ) =
1
n2+n
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 +
x2 +
x2
5 +
x3
10 + ….
>> Omitting the first term, the given series can be written in the form
x1
12+1 +
x2
22+1 +
x3
32+1 + … so that un =
xn
n2+1
Therefore un+1 =
xn+1
n2+2n+2 .
n2+1n2+2n+2 x =
limn→∞
n2 (1+1 /n2 )n2(1+2 /n+2/n2 ) .x
That is, limn→∞
un+1
un = x
Hence by ratio test un is {convergent if x < 1¿ ¿¿¿
and the test fails if x = 1.
But when x = 1, un =
1n
n2+1 =
1
n2+1 is of order
1
n2 (p = 2 > 1)
Therefore un is convergent if x 1 and divergent if x > 1.
Example: Find the nature of the series
12√1 +
x2
3√2 +
x 4
4 √3 + …
>> omitting the first term, the general term of the series is given by un =
x2 n
(n+2 )√n+1
Therefore un+1 =
x2 (n+1)(n+1+2 )√(n+1)+1 =
x2 n+2
(n+3 )√n+2
un+1
un =
x2 n+2
(n+3 )√n+2
(n+2 )√n+1
x2 n
=
n+2n+3 √ n+1
n+2x2
=
√(n+2 )(n+1)(n+3 ) x2
limn→∞
un+1
un = limn→∞
√n(1+2/n)n(1+1/n )n(1+3/n ) . x2 = x2
Hence by ratio test un is {convergent if x2< 1 ¿¿¿¿
and the fails if x2 = 1.
When x2 = 1, un =
(1 )n
(n+2 )√n+1 =
1(n+2 )√n+1
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 x2 > 1.
Example : Discus the convergence of the series
x +
x3
2. 3 +
32. 4 +
x5
5 +
3 .52. 4 .6 .
x7
7 + … (x > 0)
>> We shall write the given series in the form
x +
12 .
x3
3 +
1 .32. 4 .
x5
5 +
1 .3 . 52. 4 .6 .
x7
7 + ….
Now, omitting the first term we have
un =
1. 3 .5 . ..(2n−1)2 . 4 . 6 .. .2n .
x2 n+1
2n+1
un+1 =
1. 3 .5 . ..[ 2(n+1 )−1 ]2. 4 .6 . .. 2(n+1 ) .
x2 (n+1)+1
2(n+1)+1
That is, un+1 =
1 .3 . 5. . .(2n+1)2. 4 .6 . .. .(2n+1) .
x2 n+3
2n+3
That is, un+1 =
1. 3 .5 . ..(2n−1)(2n+1)2 .4 .6 . . ..(2n )(2n+2) .
x2 n+3
2n+3
Therefore
un+1
un =
1. 3 .5 . ..(2n−1)(2n+1)2 .4 .6 . . ..(2n )(2n+2) .
x2 n+3
2n+3
2 . 4 . 6 .. .2n1. 3 .5 . ..(2n−1) .
2n+1
x2 n+1
That is,
un+1
un =
(2n+1)(2n+1 )x2
(2n+2)(2n+3 )
Therefore limn→∞
un+1
un = limn→∞
n(2+1/n )n (2+1/n) x2
n(2+2/n )n(2+3 /n) = x2
Hence by ratio test, un is {convergent if x2< 1 ¿¿¿¿
And the test fails if x2 = 1
When x2 = 1,
un+1
un =
(2n+1 )(2n+1 )(2n+2)(2n+3 ) and we shall apply Raabe’s test.
limn→∞ n
( unun+1
−1) =
limn→∞ n
[ (2n+2)(2n+3 )(2n+1)2
−1]
= limn→∞ n
[ (4n2+10n+6 )−(4 n2+4n+1 )(2n+1)2 ]
= limn→∞ n
( 6n+5
(2n+1 )2) =
limn→∞
n2 (6+5 /n)n2(2+1 /n)2
64 =
32 > 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 x2 > 1.
Example : Examine the convergence of
1 +
25 x +
69 x2 +
1417 x3 + … +
2n+1−22n+1+1 xn + ….
>> un =
2n+1−22n+1+1 xn.
Therefore un+1 =
2n+2−22n+2+1 xn+1
un+1
un =
2n+2−22n+2+1 xn+1
2n+1+12n+1−2 .
1
xn
un+1
un =
2n+2(1−2 /2n+2 )2n+2(1+1/2n+2 ) .x.
2n+1(1+1/2n+1 )2n+1(1−2 /2n+1 )
=
(1−1 /2n+1 )(1+1/2n+2 ) .x .
(1+1/2n+1 )(1−1 /2n)
Therefore limn→∞
un+1
un =
(1−0 )(1+0 ) . x .
(1+0 )(1−0 ) = x.
Therefore by ratio test un is {convergent if x < 1¿ ¿¿¿
and the test fails if x = 1.
When x = 1, un =
2n+1−22n+1+1
Therefore limn→∞ un =
limn→∞
2n+1(1−1 /2n)2n+1(1+1/2n+1 ) = 1
Since limn→∞ 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 +
2!
22 +
3!
33 +
4 !
44 + …
>> the first term of the given series can be written as 1!/11 so that we have,
un =
n !
nn and un+1 =
(n+1) !(n+1 )n+1
=
(n+1 )(n !)(n+1)n+1
=
n !
(n+1 )n
Therefore
un+1
un =
n !
(n+1 )n .
nn
n ! =
nn
(n+1 )n =
nn
nn (1+1 /n)n
limn→∞
un+1
un = limn→∞
1
(1+1/n )n =
1e < 1
Hence by ratio test un is convergent.
Cauchy’s Root Test: If ∑ un is a series of positive terms, and
limn→∞
(u¿¿n)1n=l(finite)¿,
then, the series converges if l<1, diverges if l>1 and fails if l=1.
Remark: Root test is useful when the terms of the series are of the form un=[ f (n )]g (n ).
We can note : (i) limn→∞
n1n=1
(ii) limn→∞
(1+ 1n)
1n=e
(iii) ) limn→∞
(1+ xn)
1n=e x
Example : Test for convergence ∑n=1
∞ [1+ 1√n ]
−n3/2
>> un = [1+ 1
√n ]−n3/2
Therefore (un)1/n = {[1+ 1
√n ]−n3 /2
}1 /n
= [1+ 1
√n ]−n1/2
= [1+ 1
√n ]−√n
limn→∞ (un) 1/n =
limn→∞
[1+ 1√n ]
−√n
= limn→∞
1
(1+ 1√n )
√n
=
1e < 1.
Therefore as n , √n also
Therefore by Cauchy’s root test, un is convergent.
Example : Test for convergence ∑n=1
∞
(1−3n )
n2
>> un = (1−3
n )n2
Therefore (un)1/n = [(1−3
n )n2
]1/n
= (1−3
n )n
limn→∞ (un)1/n =
limn→∞
(1+−34 )
n
= e-3.
Therefore limn→∞
(1+ xn )
n
= ex
That is, limn→∞ (un)1/n =
1
e3 < 1, therefore e = 2.7
Hence by Cauchy’s root test, un is convergent.
Example : Find the nature of the series ∑n=1
∞ [1+ 1√n ]
−n3 /2
>> un = [1+ 1
√n ]−n3/2
Therefore (un)1/n = {[1+ 1
√n ]−n3 /2
}1 /n
= [1+ 1
√n ]−n1/2
= [1+ 1
√n ]−√n
limn→∞ (un)1/n =
limn→∞
[1+ 1√n ]
−√n
= limn→∞
1
(1+ 1√n )
√n
=
1e < 1, since as n , √n also
Therefore by Cauchy’s root rest, un is convergent.
Example : Test for convergence ∑n=1
∞
(1−3n )
n2
>> un = (1−3
n )n2
Therefore (un)1/n = [(1−3
n )n2
]1/n
= (1−3
n )n
limn→∞ (un)1/n =
limn→∞
(1+−3n )
n
= e-3, since limn→∞
(1+ xn )
n
= ex
That is, limn→∞ (un)1/n =
1
e3 < 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.,
LEBINITZ’S SERIES
An alternating series converges if
(i) each term is numerically less than its preceding term
(ii)
Note: If then the given series is oscillatory.
Q Test the convergence of
16 -
113 +
120 -
127 + …
Solution: Here un =
17n−1
then un+1 =
17(n=1)−1 =
17n+6
therefore, un – un+1 =
17n−1 -
17n+6
=
(7n+6 )−(7n−1 )(7n−1)(7n+6 ) =
7(7n−1)(7n+6 ) > 0
That is, un – un+1 > 0, un > un+1
Also, limn→∞ un =
limn→∞
17n−1 =
limn→∞
1n
1(7−1/n) = 0
Therefore by Leibnitz test the given alternating series is convergent .
Q Find the nature of the series
(1− 1log 2 )
- (1− 1
log 3 ) +
(1− 1log 4 )
- (1− 1
log 5 ) + …
Solution: Here un = 1 -
1log (n+1) then un+1 = 1 -
1log (n+2)
Therefore, un – un+1 =
1log (n+2) -
1log (n+1)
=
log (n+1)−log (n+2 )log (n+2) log (n+1) < 0.
Since (n + 1) < (n + 2)
un - un+1 < 0 un < un+1
further limn→∞ un =
limn→∞ 1 -
[ 1log(n+1 ) ]
= 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
ABSOLUTELY & CONDITIONALLY CONVERGENT SERIES
An alternating series is said to be absolutely convergent if the positive
series is convergent.
An alternating series is said to be conditionally convergent if
(i) is divergent
(ii) is convergent
Theorem: An absolutely convergent series is convergent. The converse need not be true.
Proof: Let be an absolutely convergent series then is
convergent.
We know,
By comparison test, is convergent.
Q. Show that each of the following series also converges absolutely
(i) an2; (ii)
an2
1+an2
; (iii)
an1+an
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
an2
1+an2
an2
the convergence of an2 implies the convergence of
an2
1+an2
.
(iii) |an
1+an| =
|an||1+an| <
|an|1−|an| .
As |an| converges, |an| 0 as n . Hence for some positive integer N, we have |an| < ½
for all n N.
This gives |an
1+an| < 2|an| for all n N.
Now, by comparison test, |an
1+an| converges.
That is,
an1+an converges absolutely.
Q. Test the convergence
Solution: Here
then
i.e.,
Thus by Lebinitz rule, is convergent.
Also, . Take
Then
Since is divergent, therefore is also divergent.
Thus the given series is conditionally convergent.
POWER SERIES
A series of the form where the ’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
Therefore,
If , then by ratio test, the series (i) converges when and diverges for
other values.
Thus the power series (i) has an interval 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 .
Solution: Here and
Therefore,
By Ratio test the given series converges for and diverges for .
When x=1 the series reduces to , which is an alternating series and is
convergent.
When x=-1 the series becomes , which is divergent (by comparison with
p-series when p=1)
Hence the interval of convergence is .
Q. Show that the series ∑
1
∞
(−1 )n−1
xn
√2n+1 is absolutely convergent for | x | < 1,
conditionally convergent for x = 1 and divergent for x = -1.
Solution. Here un = (-1)n-1
xn
√2n+1
Therefore un+1 =
(−1)n xn+1
√2n+3
limn→∞
|uN+1
un| =
limn→∞
|(−1)n xn+1√2n+1
√2n+3(−1)n−1 xn|
= limn→∞
|(−1 )√ 2n+12n+3
x|
= limn→∞
|(−1 )√ n(2+1/n )n(2+3/n )
x| = | 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
1
√3 -
1
√5 +
1
√7 -
1
√9 + …
Here un =
1
√2n+1 , un+1 =
1
√2n+3
But 2n + 1< 2n + 3 un > un+1
Also limn→∞ un =
limn→∞
1
√2n+3 = 0
Therefore by Leibnitz test the series is convergence when x = 1.
But the absolute series
1
√3 +
1
√5 +
1
√7 + … whose general term is un =
1
√2n+1 and is of
order
1
√n =
1
n1/2 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
−1
√3 -
1
√5 -
1
√7 - ….
= - ( 1
√3+ 1
√5+ 1
√7+.. .)
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
2. Prove that is absolutely convergent
3. For what values of x the following series are convergent
4. Test the nature of convergence
******
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