Mathematics  >  Mathematics for IIT JAM, CSIR NET, UGC NET  >  Convergence Tests - Sequences and Series, CSIR-NET Mathematical Sciences

Convergence Tests - Sequences and Series, CSIR-NET Mathematical Sciences - Notes | Study Mathematics for IIT JAM, CSIR NET, UGC NET - Mathematics

Download, print and study this document offline
 Page 1


6.7 Convergence Tests
De¯nition 6.8 We say that a series
P
a
n
satis¯es the Cauchy criterion if its se-
quence of partial sums is a Cauchy sequence.
This means that for each ² > 0 there exists a number N such that if m;n > N
then jS
n
¡ S
m
j < ². Nothing is lost in this de¯nition if we impose the restriction
n¸ m. Moreover, it is only a notational matter to work with m¡1 where m· n
instead of m where m < n. That means that the de¯nition is equivalent to for each
² > 0 there exists a number N such that if n ¸ m > N then jS
n
¡ S
m¡1
j < ².
The reason for doing this is that S
n
¡S
m¡1
=
P
n
k=m
a
k
. Then this condition can be
rewritten as for each ² > 0 there exists a number N such that if n ¸ m > N then
j
P
n
k=m
j<².
Theorem 6.17 A series converges if and only if it satis¯es the Cauchy criterion.
Corollary 6.1 If a series
P
a
n
converges, then lima
n
=0.
It is often easier to prove that a limit exists or that a series converges than it is
to determine its exact value. As an example consider the following.
Theorem 6.18 (Comparison Test) Let
P
a
n
be a series where a
n
¸ 0 for all n.
(i) If
P
a
n
converges and jb
n
j·a
n
for all n, then
P
b
n
converges.
(ii) If
P
a
n
=+1 and b
n
¸a
n
for all n, then
P
b
n
=+1.
Proof: (i) For n¸m we have
¯
¯
¯
¯
¯
n
X
k=m
b
k
¯
¯
¯
¯
¯
·
n
X
k=m
jb
k
j·
n
X
k=m
a
k
where the ¯rst inequality follows from the Triangle Inequality. Since
P
a
n
converges, it satis¯es the Cauchy criterion. It then follows from the above that
P
b
n
also satis¯es the Cauchy criterion and hence it also converges.
(ii) LetfS
n
g andfT
n
g be the sequences of partial sums for
P
a
n
and
P
b
n
, respec-
tively. Since b
n
¸a
n
for all n we then clearly have that T
n
¸S
n
for all n. Since
limS
n
=+1, we conclude that limT
n
=+1, and
P
b
n
=+1.
Corollary 6.2 Absolutely convergent series are convergent.
Proof: Suppose that
P
a
n
is absolutely convergent. This means that
P
b
n
con-
verges where b
n
= ja
n
j for all n. Then ja
n
j · b
n
, so that
P
a
n
converges by the
Comparison Test.
Page 2


6.7 Convergence Tests
De¯nition 6.8 We say that a series
P
a
n
satis¯es the Cauchy criterion if its se-
quence of partial sums is a Cauchy sequence.
This means that for each ² > 0 there exists a number N such that if m;n > N
then jS
n
¡ S
m
j < ². Nothing is lost in this de¯nition if we impose the restriction
n¸ m. Moreover, it is only a notational matter to work with m¡1 where m· n
instead of m where m < n. That means that the de¯nition is equivalent to for each
² > 0 there exists a number N such that if n ¸ m > N then jS
n
¡ S
m¡1
j < ².
The reason for doing this is that S
n
¡S
m¡1
=
P
n
k=m
a
k
. Then this condition can be
rewritten as for each ² > 0 there exists a number N such that if n ¸ m > N then
j
P
n
k=m
j<².
Theorem 6.17 A series converges if and only if it satis¯es the Cauchy criterion.
Corollary 6.1 If a series
P
a
n
converges, then lima
n
=0.
It is often easier to prove that a limit exists or that a series converges than it is
to determine its exact value. As an example consider the following.
Theorem 6.18 (Comparison Test) Let
P
a
n
be a series where a
n
¸ 0 for all n.
(i) If
P
a
n
converges and jb
n
j·a
n
for all n, then
P
b
n
converges.
(ii) If
P
a
n
=+1 and b
n
¸a
n
for all n, then
P
b
n
=+1.
Proof: (i) For n¸m we have
¯
¯
¯
¯
¯
n
X
k=m
b
k
¯
¯
¯
¯
¯
·
n
X
k=m
jb
k
j·
n
X
k=m
a
k
where the ¯rst inequality follows from the Triangle Inequality. Since
P
a
n
converges, it satis¯es the Cauchy criterion. It then follows from the above that
P
b
n
also satis¯es the Cauchy criterion and hence it also converges.
(ii) LetfS
n
g andfT
n
g be the sequences of partial sums for
P
a
n
and
P
b
n
, respec-
tively. Since b
n
¸a
n
for all n we then clearly have that T
n
¸S
n
for all n. Since
limS
n
=+1, we conclude that limT
n
=+1, and
P
b
n
=+1.
Corollary 6.2 Absolutely convergent series are convergent.
Proof: Suppose that
P
a
n
is absolutely convergent. This means that
P
b
n
con-
verges where b
n
= ja
n
j for all n. Then ja
n
j · b
n
, so that
P
a
n
converges by the
Comparison Test.
Theorem 6.19 (Limit Comparison Test) Suppose
P
a
n
and
P
b
n
are two in¯-
nite series. Suppose also that r = limja
n
=b
n
j exists, and 0 < r < +1. Then
P
a
n
converges absolutely if and only if
P
b
n
converges absolutely.
Proof: Since r = limja
n
=b
n
j exists, and r is between 0 and +1, there exist con-
stants c and C, 0<c<C < +1 such that for some N > 1 we have that if n>N
c<
¯
¯
¯
¯
a
n
b
n
¯
¯
¯
¯
<C:
Assume that
P
a
n
converges absolutely. For n > N we have that cjb
n
j < ja
n
j.
Therefore,
P
b
n
converges absolutely by the Comparison Test.
Now assume that
P
b
n
converges absolutely. From the above inequality we have
that ja
n
j < Cjb
n
j for n > N. But since the series C
P
b
n
also converges absolutely,
we can use again the Comparison Test to see that
P
a
n
must converge absolutely.
Theorem 6.20 (Cauchy Condensation Test) Suppose fa
n
g is a decreasing se-
quence of positive terms. Then the series
P
a
n
converges if and only if the series
P
2
k
a
2
k converges.
Proof: Assume that
1
X
n=1
a
n
converges. Sincefa
n
g is a decreasing sequence, we have
that
2
k¡1
a
2
k = a
2
k +a
2
k +a
2
k +:::+a
2
k
· a
2
k¡1
+1
+a
2
k¡1
+2
+¢¢¢+a
2
k¡1
+2
k¡2 +a
2
k =
2
k
X
m=2
k¡1
+1
a
m
:
Therefore, we have that
N
X
k=1
2
k¡1
a
2
k ·
N
X
k=1
0
@
2
k
X
m=2
k¡1
+1
a
m
1
A
=
2
N
X
m=2
a
m
:
Now the partial sums on the right are bounded, by assumption. Hence the partial
sums on the left are also bounded. Since all terms are positive, the partial sums
now form an increasing sequence that is bounded above, hence it must converge.
Multiplying the left sequence by 2 will not change convergence, and hence the series
P
N
k=1
2
k
a
2
k converges.
Now, assume that
P
N
k=1
2
k
a
2
k converges: We have
2
N
X
m=2
k¡1
+1
a
m
= a
2
k¡1
+1
+a
2
k¡1
+2
+¢¢¢+a
2
k¡1
+2
k¡2 +a
2
k
· a
2
k¡1 +a
2
k¡1 +a
2
k¡1 +:::+a
2
k¡1 =2
k¡1
a
2
k¡1
Page 3


6.7 Convergence Tests
De¯nition 6.8 We say that a series
P
a
n
satis¯es the Cauchy criterion if its se-
quence of partial sums is a Cauchy sequence.
This means that for each ² > 0 there exists a number N such that if m;n > N
then jS
n
¡ S
m
j < ². Nothing is lost in this de¯nition if we impose the restriction
n¸ m. Moreover, it is only a notational matter to work with m¡1 where m· n
instead of m where m < n. That means that the de¯nition is equivalent to for each
² > 0 there exists a number N such that if n ¸ m > N then jS
n
¡ S
m¡1
j < ².
The reason for doing this is that S
n
¡S
m¡1
=
P
n
k=m
a
k
. Then this condition can be
rewritten as for each ² > 0 there exists a number N such that if n ¸ m > N then
j
P
n
k=m
j<².
Theorem 6.17 A series converges if and only if it satis¯es the Cauchy criterion.
Corollary 6.1 If a series
P
a
n
converges, then lima
n
=0.
It is often easier to prove that a limit exists or that a series converges than it is
to determine its exact value. As an example consider the following.
Theorem 6.18 (Comparison Test) Let
P
a
n
be a series where a
n
¸ 0 for all n.
(i) If
P
a
n
converges and jb
n
j·a
n
for all n, then
P
b
n
converges.
(ii) If
P
a
n
=+1 and b
n
¸a
n
for all n, then
P
b
n
=+1.
Proof: (i) For n¸m we have
¯
¯
¯
¯
¯
n
X
k=m
b
k
¯
¯
¯
¯
¯
·
n
X
k=m
jb
k
j·
n
X
k=m
a
k
where the ¯rst inequality follows from the Triangle Inequality. Since
P
a
n
converges, it satis¯es the Cauchy criterion. It then follows from the above that
P
b
n
also satis¯es the Cauchy criterion and hence it also converges.
(ii) LetfS
n
g andfT
n
g be the sequences of partial sums for
P
a
n
and
P
b
n
, respec-
tively. Since b
n
¸a
n
for all n we then clearly have that T
n
¸S
n
for all n. Since
limS
n
=+1, we conclude that limT
n
=+1, and
P
b
n
=+1.
Corollary 6.2 Absolutely convergent series are convergent.
Proof: Suppose that
P
a
n
is absolutely convergent. This means that
P
b
n
con-
verges where b
n
= ja
n
j for all n. Then ja
n
j · b
n
, so that
P
a
n
converges by the
Comparison Test.
Theorem 6.19 (Limit Comparison Test) Suppose
P
a
n
and
P
b
n
are two in¯-
nite series. Suppose also that r = limja
n
=b
n
j exists, and 0 < r < +1. Then
P
a
n
converges absolutely if and only if
P
b
n
converges absolutely.
Proof: Since r = limja
n
=b
n
j exists, and r is between 0 and +1, there exist con-
stants c and C, 0<c<C < +1 such that for some N > 1 we have that if n>N
c<
¯
¯
¯
¯
a
n
b
n
¯
¯
¯
¯
<C:
Assume that
P
a
n
converges absolutely. For n > N we have that cjb
n
j < ja
n
j.
Therefore,
P
b
n
converges absolutely by the Comparison Test.
Now assume that
P
b
n
converges absolutely. From the above inequality we have
that ja
n
j < Cjb
n
j for n > N. But since the series C
P
b
n
also converges absolutely,
we can use again the Comparison Test to see that
P
a
n
must converge absolutely.
Theorem 6.20 (Cauchy Condensation Test) Suppose fa
n
g is a decreasing se-
quence of positive terms. Then the series
P
a
n
converges if and only if the series
P
2
k
a
2
k converges.
Proof: Assume that
1
X
n=1
a
n
converges. Sincefa
n
g is a decreasing sequence, we have
that
2
k¡1
a
2
k = a
2
k +a
2
k +a
2
k +:::+a
2
k
· a
2
k¡1
+1
+a
2
k¡1
+2
+¢¢¢+a
2
k¡1
+2
k¡2 +a
2
k =
2
k
X
m=2
k¡1
+1
a
m
:
Therefore, we have that
N
X
k=1
2
k¡1
a
2
k ·
N
X
k=1
0
@
2
k
X
m=2
k¡1
+1
a
m
1
A
=
2
N
X
m=2
a
m
:
Now the partial sums on the right are bounded, by assumption. Hence the partial
sums on the left are also bounded. Since all terms are positive, the partial sums
now form an increasing sequence that is bounded above, hence it must converge.
Multiplying the left sequence by 2 will not change convergence, and hence the series
P
N
k=1
2
k
a
2
k converges.
Now, assume that
P
N
k=1
2
k
a
2
k converges: We have
2
N
X
m=2
k¡1
+1
a
m
= a
2
k¡1
+1
+a
2
k¡1
+2
+¢¢¢+a
2
k¡1
+2
k¡2 +a
2
k
· a
2
k¡1 +a
2
k¡1 +a
2
k¡1 +:::+a
2
k¡1 =2
k¡1
a
2
k¡1
Therefore, similar to above, we get:
2
N
X
m=2
a
m
=
N
X
k=1
0
@
2
k
X
m=2
k¡1
+1
a
m
1
A
·
N
X
k=1
2
k¡1
a
2
k:
Now the sequence of partial sums on the right is bounded, by assumption. There-
fore, the left side forms an increasing sequence that is bounded above, and therefore
must converge.
Corollary 6.3 For a positive number p
1
X
n=1
1
n
p
converges if and only if p> 1:
Proof: If p < 0 then the sequence
½
1
n
p
¾
diverges to in¯nity. Hence, the series
diverges by the Divergence Test.
If p> 0 then consider the series
1
X
n=1
2
n
a
2
n =
1
X
n=1
2
n
1
(2
n
)
p
=
1
X
n=1
(2
1¡p
)
n
:
This right-hand side is a geometric series. Thus, we know that
² if 0<p· 1 then 2
1¡p
¸ 1, hence the right-hand series diverges;
² if p> 1 then 2
1¡p
< 1 and the right-hand series converges,
Now the result follows from the Cauchy Condensation Test .
Theorem 6.21 (Root Test) Let
P
a
n
be a series and let ® = limsupja
n
j
1=n
. The
series
P
a
n
(i) converges absolutely if ®< 1,
(ii) diverges if ®> 1.
(iii) Otherwise ® =1 and the test gives no information.
Although this Root Test is more di±cult to apply, it is better than the Ratio Test
in the following sense. There are series for which the Ratio Test give no information,
yet the Root Test will be conclusive. We will use the Root Test to prove the Ratio
Test, but you cannot use the Ratio Test to prove the Root Test. It is important to
remember that when the Root Test gives 1 as the answer for the limsup, then no
conclusion at all is possible.
The use of the limsup rather than the regular limit has the advantage that we
do not have to be concerned with the existence of a limit. On the other hand, if
the regular limit exists, it is the same as the limsup, so that we are not giving up
anything using the limsup.
Page 4


6.7 Convergence Tests
De¯nition 6.8 We say that a series
P
a
n
satis¯es the Cauchy criterion if its se-
quence of partial sums is a Cauchy sequence.
This means that for each ² > 0 there exists a number N such that if m;n > N
then jS
n
¡ S
m
j < ². Nothing is lost in this de¯nition if we impose the restriction
n¸ m. Moreover, it is only a notational matter to work with m¡1 where m· n
instead of m where m < n. That means that the de¯nition is equivalent to for each
² > 0 there exists a number N such that if n ¸ m > N then jS
n
¡ S
m¡1
j < ².
The reason for doing this is that S
n
¡S
m¡1
=
P
n
k=m
a
k
. Then this condition can be
rewritten as for each ² > 0 there exists a number N such that if n ¸ m > N then
j
P
n
k=m
j<².
Theorem 6.17 A series converges if and only if it satis¯es the Cauchy criterion.
Corollary 6.1 If a series
P
a
n
converges, then lima
n
=0.
It is often easier to prove that a limit exists or that a series converges than it is
to determine its exact value. As an example consider the following.
Theorem 6.18 (Comparison Test) Let
P
a
n
be a series where a
n
¸ 0 for all n.
(i) If
P
a
n
converges and jb
n
j·a
n
for all n, then
P
b
n
converges.
(ii) If
P
a
n
=+1 and b
n
¸a
n
for all n, then
P
b
n
=+1.
Proof: (i) For n¸m we have
¯
¯
¯
¯
¯
n
X
k=m
b
k
¯
¯
¯
¯
¯
·
n
X
k=m
jb
k
j·
n
X
k=m
a
k
where the ¯rst inequality follows from the Triangle Inequality. Since
P
a
n
converges, it satis¯es the Cauchy criterion. It then follows from the above that
P
b
n
also satis¯es the Cauchy criterion and hence it also converges.
(ii) LetfS
n
g andfT
n
g be the sequences of partial sums for
P
a
n
and
P
b
n
, respec-
tively. Since b
n
¸a
n
for all n we then clearly have that T
n
¸S
n
for all n. Since
limS
n
=+1, we conclude that limT
n
=+1, and
P
b
n
=+1.
Corollary 6.2 Absolutely convergent series are convergent.
Proof: Suppose that
P
a
n
is absolutely convergent. This means that
P
b
n
con-
verges where b
n
= ja
n
j for all n. Then ja
n
j · b
n
, so that
P
a
n
converges by the
Comparison Test.
Theorem 6.19 (Limit Comparison Test) Suppose
P
a
n
and
P
b
n
are two in¯-
nite series. Suppose also that r = limja
n
=b
n
j exists, and 0 < r < +1. Then
P
a
n
converges absolutely if and only if
P
b
n
converges absolutely.
Proof: Since r = limja
n
=b
n
j exists, and r is between 0 and +1, there exist con-
stants c and C, 0<c<C < +1 such that for some N > 1 we have that if n>N
c<
¯
¯
¯
¯
a
n
b
n
¯
¯
¯
¯
<C:
Assume that
P
a
n
converges absolutely. For n > N we have that cjb
n
j < ja
n
j.
Therefore,
P
b
n
converges absolutely by the Comparison Test.
Now assume that
P
b
n
converges absolutely. From the above inequality we have
that ja
n
j < Cjb
n
j for n > N. But since the series C
P
b
n
also converges absolutely,
we can use again the Comparison Test to see that
P
a
n
must converge absolutely.
Theorem 6.20 (Cauchy Condensation Test) Suppose fa
n
g is a decreasing se-
quence of positive terms. Then the series
P
a
n
converges if and only if the series
P
2
k
a
2
k converges.
Proof: Assume that
1
X
n=1
a
n
converges. Sincefa
n
g is a decreasing sequence, we have
that
2
k¡1
a
2
k = a
2
k +a
2
k +a
2
k +:::+a
2
k
· a
2
k¡1
+1
+a
2
k¡1
+2
+¢¢¢+a
2
k¡1
+2
k¡2 +a
2
k =
2
k
X
m=2
k¡1
+1
a
m
:
Therefore, we have that
N
X
k=1
2
k¡1
a
2
k ·
N
X
k=1
0
@
2
k
X
m=2
k¡1
+1
a
m
1
A
=
2
N
X
m=2
a
m
:
Now the partial sums on the right are bounded, by assumption. Hence the partial
sums on the left are also bounded. Since all terms are positive, the partial sums
now form an increasing sequence that is bounded above, hence it must converge.
Multiplying the left sequence by 2 will not change convergence, and hence the series
P
N
k=1
2
k
a
2
k converges.
Now, assume that
P
N
k=1
2
k
a
2
k converges: We have
2
N
X
m=2
k¡1
+1
a
m
= a
2
k¡1
+1
+a
2
k¡1
+2
+¢¢¢+a
2
k¡1
+2
k¡2 +a
2
k
· a
2
k¡1 +a
2
k¡1 +a
2
k¡1 +:::+a
2
k¡1 =2
k¡1
a
2
k¡1
Therefore, similar to above, we get:
2
N
X
m=2
a
m
=
N
X
k=1
0
@
2
k
X
m=2
k¡1
+1
a
m
1
A
·
N
X
k=1
2
k¡1
a
2
k:
Now the sequence of partial sums on the right is bounded, by assumption. There-
fore, the left side forms an increasing sequence that is bounded above, and therefore
must converge.
Corollary 6.3 For a positive number p
1
X
n=1
1
n
p
converges if and only if p> 1:
Proof: If p < 0 then the sequence
½
1
n
p
¾
diverges to in¯nity. Hence, the series
diverges by the Divergence Test.
If p> 0 then consider the series
1
X
n=1
2
n
a
2
n =
1
X
n=1
2
n
1
(2
n
)
p
=
1
X
n=1
(2
1¡p
)
n
:
This right-hand side is a geometric series. Thus, we know that
² if 0<p· 1 then 2
1¡p
¸ 1, hence the right-hand series diverges;
² if p> 1 then 2
1¡p
< 1 and the right-hand series converges,
Now the result follows from the Cauchy Condensation Test .
Theorem 6.21 (Root Test) Let
P
a
n
be a series and let ® = limsupja
n
j
1=n
. The
series
P
a
n
(i) converges absolutely if ®< 1,
(ii) diverges if ®> 1.
(iii) Otherwise ® =1 and the test gives no information.
Although this Root Test is more di±cult to apply, it is better than the Ratio Test
in the following sense. There are series for which the Ratio Test give no information,
yet the Root Test will be conclusive. We will use the Root Test to prove the Ratio
Test, but you cannot use the Ratio Test to prove the Root Test. It is important to
remember that when the Root Test gives 1 as the answer for the limsup, then no
conclusion at all is possible.
The use of the limsup rather than the regular limit has the advantage that we
do not have to be concerned with the existence of a limit. On the other hand, if
the regular limit exists, it is the same as the limsup, so that we are not giving up
anything using the limsup.
Proof:
(i) Suppose that ® < 1. Then choose an ² > 0 so that ® + ² < 1. Then by the
de¯nition of the limit superior there is a natural number N such that
®¡²< lubfja
n
j
1=n
jn>Ng;®+²:
In particular, we have ja
n
j
1=n
<®+² for n>N, so
ja
n
j< (®+²)
n
for n>N:
Since 0 <®+²< 1, the geometric series
P
1
n=N+1
(®+²)
n
converges. Thus, the
Comparison Test shows that
P
1
n=N+1
a
n
converges. This means that
P
a
n
also
converges.
(ii) If ®> 1, then there is a subsequence ofja
n
j
1=n
that has limit ®> 1. That means
thatja
n
j > 1 for in¯nitely many choices of n. In particular, the sequence fa
n
g
cannot converge to 0, so the series
P
a
n
cannot converge.
(iii) For the series
P
1
n
and for the series
P
1
n
2
, ® turns out to be 1. Since the
harmonic series diverges and the series
P
1
n
2
converges, the equality ® = 1
cannot guarantee either convergence or divergence of the series.
Theorem 6.22 (Ratio Test) A series
P
a
n
of nonzero series
(i) converges absolutely if limsupja
n+1
=a
n
j< 1,
(ii) diverges is liminfja
n+1
=a
n
j> 1.
(iii) Otherwise liminfja
n+1
=a
n
j· 1· limsupja
n+1
=a
n
j and the test gives no infor-
mation.
Proof:
(i)Supposethatlimsupja
n+1
=a
n
j< 1. Thenchoosean²> 0sothatja
n+1
=a
n
j+²< 1.
Then by the de¯nition of the limit superior there is a natural number N such that
for n>N
¯
¯
¯
¯
a
n+1
a
n
¯
¯
¯
¯
< 1¡²:
Multiplying both sides by ja
n
j we get
ja
n+1
j< (1¡²)ja
n
j for n>N:
Therefore, we also have
ja
n+2
j< (1¡²)ja
n+1
j< (1¡²)
2
ja
n
j for n>N:
Page 5


6.7 Convergence Tests
De¯nition 6.8 We say that a series
P
a
n
satis¯es the Cauchy criterion if its se-
quence of partial sums is a Cauchy sequence.
This means that for each ² > 0 there exists a number N such that if m;n > N
then jS
n
¡ S
m
j < ². Nothing is lost in this de¯nition if we impose the restriction
n¸ m. Moreover, it is only a notational matter to work with m¡1 where m· n
instead of m where m < n. That means that the de¯nition is equivalent to for each
² > 0 there exists a number N such that if n ¸ m > N then jS
n
¡ S
m¡1
j < ².
The reason for doing this is that S
n
¡S
m¡1
=
P
n
k=m
a
k
. Then this condition can be
rewritten as for each ² > 0 there exists a number N such that if n ¸ m > N then
j
P
n
k=m
j<².
Theorem 6.17 A series converges if and only if it satis¯es the Cauchy criterion.
Corollary 6.1 If a series
P
a
n
converges, then lima
n
=0.
It is often easier to prove that a limit exists or that a series converges than it is
to determine its exact value. As an example consider the following.
Theorem 6.18 (Comparison Test) Let
P
a
n
be a series where a
n
¸ 0 for all n.
(i) If
P
a
n
converges and jb
n
j·a
n
for all n, then
P
b
n
converges.
(ii) If
P
a
n
=+1 and b
n
¸a
n
for all n, then
P
b
n
=+1.
Proof: (i) For n¸m we have
¯
¯
¯
¯
¯
n
X
k=m
b
k
¯
¯
¯
¯
¯
·
n
X
k=m
jb
k
j·
n
X
k=m
a
k
where the ¯rst inequality follows from the Triangle Inequality. Since
P
a
n
converges, it satis¯es the Cauchy criterion. It then follows from the above that
P
b
n
also satis¯es the Cauchy criterion and hence it also converges.
(ii) LetfS
n
g andfT
n
g be the sequences of partial sums for
P
a
n
and
P
b
n
, respec-
tively. Since b
n
¸a
n
for all n we then clearly have that T
n
¸S
n
for all n. Since
limS
n
=+1, we conclude that limT
n
=+1, and
P
b
n
=+1.
Corollary 6.2 Absolutely convergent series are convergent.
Proof: Suppose that
P
a
n
is absolutely convergent. This means that
P
b
n
con-
verges where b
n
= ja
n
j for all n. Then ja
n
j · b
n
, so that
P
a
n
converges by the
Comparison Test.
Theorem 6.19 (Limit Comparison Test) Suppose
P
a
n
and
P
b
n
are two in¯-
nite series. Suppose also that r = limja
n
=b
n
j exists, and 0 < r < +1. Then
P
a
n
converges absolutely if and only if
P
b
n
converges absolutely.
Proof: Since r = limja
n
=b
n
j exists, and r is between 0 and +1, there exist con-
stants c and C, 0<c<C < +1 such that for some N > 1 we have that if n>N
c<
¯
¯
¯
¯
a
n
b
n
¯
¯
¯
¯
<C:
Assume that
P
a
n
converges absolutely. For n > N we have that cjb
n
j < ja
n
j.
Therefore,
P
b
n
converges absolutely by the Comparison Test.
Now assume that
P
b
n
converges absolutely. From the above inequality we have
that ja
n
j < Cjb
n
j for n > N. But since the series C
P
b
n
also converges absolutely,
we can use again the Comparison Test to see that
P
a
n
must converge absolutely.
Theorem 6.20 (Cauchy Condensation Test) Suppose fa
n
g is a decreasing se-
quence of positive terms. Then the series
P
a
n
converges if and only if the series
P
2
k
a
2
k converges.
Proof: Assume that
1
X
n=1
a
n
converges. Sincefa
n
g is a decreasing sequence, we have
that
2
k¡1
a
2
k = a
2
k +a
2
k +a
2
k +:::+a
2
k
· a
2
k¡1
+1
+a
2
k¡1
+2
+¢¢¢+a
2
k¡1
+2
k¡2 +a
2
k =
2
k
X
m=2
k¡1
+1
a
m
:
Therefore, we have that
N
X
k=1
2
k¡1
a
2
k ·
N
X
k=1
0
@
2
k
X
m=2
k¡1
+1
a
m
1
A
=
2
N
X
m=2
a
m
:
Now the partial sums on the right are bounded, by assumption. Hence the partial
sums on the left are also bounded. Since all terms are positive, the partial sums
now form an increasing sequence that is bounded above, hence it must converge.
Multiplying the left sequence by 2 will not change convergence, and hence the series
P
N
k=1
2
k
a
2
k converges.
Now, assume that
P
N
k=1
2
k
a
2
k converges: We have
2
N
X
m=2
k¡1
+1
a
m
= a
2
k¡1
+1
+a
2
k¡1
+2
+¢¢¢+a
2
k¡1
+2
k¡2 +a
2
k
· a
2
k¡1 +a
2
k¡1 +a
2
k¡1 +:::+a
2
k¡1 =2
k¡1
a
2
k¡1
Therefore, similar to above, we get:
2
N
X
m=2
a
m
=
N
X
k=1
0
@
2
k
X
m=2
k¡1
+1
a
m
1
A
·
N
X
k=1
2
k¡1
a
2
k:
Now the sequence of partial sums on the right is bounded, by assumption. There-
fore, the left side forms an increasing sequence that is bounded above, and therefore
must converge.
Corollary 6.3 For a positive number p
1
X
n=1
1
n
p
converges if and only if p> 1:
Proof: If p < 0 then the sequence
½
1
n
p
¾
diverges to in¯nity. Hence, the series
diverges by the Divergence Test.
If p> 0 then consider the series
1
X
n=1
2
n
a
2
n =
1
X
n=1
2
n
1
(2
n
)
p
=
1
X
n=1
(2
1¡p
)
n
:
This right-hand side is a geometric series. Thus, we know that
² if 0<p· 1 then 2
1¡p
¸ 1, hence the right-hand series diverges;
² if p> 1 then 2
1¡p
< 1 and the right-hand series converges,
Now the result follows from the Cauchy Condensation Test .
Theorem 6.21 (Root Test) Let
P
a
n
be a series and let ® = limsupja
n
j
1=n
. The
series
P
a
n
(i) converges absolutely if ®< 1,
(ii) diverges if ®> 1.
(iii) Otherwise ® =1 and the test gives no information.
Although this Root Test is more di±cult to apply, it is better than the Ratio Test
in the following sense. There are series for which the Ratio Test give no information,
yet the Root Test will be conclusive. We will use the Root Test to prove the Ratio
Test, but you cannot use the Ratio Test to prove the Root Test. It is important to
remember that when the Root Test gives 1 as the answer for the limsup, then no
conclusion at all is possible.
The use of the limsup rather than the regular limit has the advantage that we
do not have to be concerned with the existence of a limit. On the other hand, if
the regular limit exists, it is the same as the limsup, so that we are not giving up
anything using the limsup.
Proof:
(i) Suppose that ® < 1. Then choose an ² > 0 so that ® + ² < 1. Then by the
de¯nition of the limit superior there is a natural number N such that
®¡²< lubfja
n
j
1=n
jn>Ng;®+²:
In particular, we have ja
n
j
1=n
<®+² for n>N, so
ja
n
j< (®+²)
n
for n>N:
Since 0 <®+²< 1, the geometric series
P
1
n=N+1
(®+²)
n
converges. Thus, the
Comparison Test shows that
P
1
n=N+1
a
n
converges. This means that
P
a
n
also
converges.
(ii) If ®> 1, then there is a subsequence ofja
n
j
1=n
that has limit ®> 1. That means
thatja
n
j > 1 for in¯nitely many choices of n. In particular, the sequence fa
n
g
cannot converge to 0, so the series
P
a
n
cannot converge.
(iii) For the series
P
1
n
and for the series
P
1
n
2
, ® turns out to be 1. Since the
harmonic series diverges and the series
P
1
n
2
converges, the equality ® = 1
cannot guarantee either convergence or divergence of the series.
Theorem 6.22 (Ratio Test) A series
P
a
n
of nonzero series
(i) converges absolutely if limsupja
n+1
=a
n
j< 1,
(ii) diverges is liminfja
n+1
=a
n
j> 1.
(iii) Otherwise liminfja
n+1
=a
n
j· 1· limsupja
n+1
=a
n
j and the test gives no infor-
mation.
Proof:
(i)Supposethatlimsupja
n+1
=a
n
j< 1. Thenchoosean²> 0sothatja
n+1
=a
n
j+²< 1.
Then by the de¯nition of the limit superior there is a natural number N such that
for n>N
¯
¯
¯
¯
a
n+1
a
n
¯
¯
¯
¯
< 1¡²:
Multiplying both sides by ja
n
j we get
ja
n+1
j< (1¡²)ja
n
j for n>N:
Therefore, we also have
ja
n+2
j< (1¡²)ja
n+1
j< (1¡²)
2
ja
n
j for n>N:
Repeating this procedure, we get that
ja
k
j< (1¡²)
k¡N
ja
N
j for k >N:
These terms form a convergent geometric series. Thus, the Comparison Test shows
that
P
a
n
also converges.
Theothertwopartsareproveninmuchthesamefashionastheprevioustheorem.
The next lemma is used to prove Abel's Convergence Test. It is computational in
nature.
Lemma 6.5 (Summation by Parts) Consider the two sequences fa
n
g and fb
n
g.
Let S
n
=
n
X
k=1
a
n
be the n-th partial sum. Then for any 0·m·n we have
n
X
j=m
a
j
b
j
=[S
n
b
n
¡S
m¡1
b
m
]+
n+1
X
j=m
S
j
(b
j
¡b
j+1
):
Proof: Just be careful with the subscripts:
n
X
j=m
a
j
b
j
=
n
X
j=m
(S
j
¡S
j¡1
)b
j
=
n
X
j=m
S
j
b
j
¡
n
X
j=m
S
j¡1
b
j
=
n
X
j=m
S
j
b
j
¡
n¡1
X
j=m¡1
S
j
b
j+1
=
n¡1
X
j=m
S
j
(b
j
¡b
j+1
+(S
n
b
n
¡S
m¡1
b
m
)
Theorem 6.23 (Abel's Test) Consider the series
P
a
n
b
n
. Suppose that
(i) the partial sums S
N
=
P
N
n=1
a
n
form a bounded sequence,
(ii) the sequence fb
n
g is decreasing,
(iii) limb
n
=0.
Then the series
P
a
n
b
n
converges.
Read More
559 videos|198 docs
Download as PDF

Download free EduRev App

Track your progress, build streaks, highlight & save important lessons and more!

Related Searches

CSIR NET

,

Sample Paper

,

CSIR-NET Mathematical Sciences - Notes | Study Mathematics for IIT JAM

,

past year papers

,

CSIR-NET Mathematical Sciences - Notes | Study Mathematics for IIT JAM

,

Previous Year Questions with Solutions

,

mock tests for examination

,

pdf

,

practice quizzes

,

ppt

,

CSIR NET

,

Convergence Tests - Sequences and Series

,

Viva Questions

,

shortcuts and tricks

,

Free

,

Extra Questions

,

CSIR-NET Mathematical Sciences - Notes | Study Mathematics for IIT JAM

,

UGC NET - Mathematics

,

Convergence Tests - Sequences and Series

,

Important questions

,

Summary

,

study material

,

CSIR NET

,

UGC NET - Mathematics

,

MCQs

,

Convergence Tests - Sequences and Series

,

Exam

,

Objective type Questions

,

UGC NET - Mathematics

,

video lectures

,

Semester Notes

;