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Numerical Sequences and Series
Written by
MenGen Tsai
email: [email protected]
1. Prove that the convergence of
{
s
n
}
implies convergence of
{
s
n
}
. Is
the converse true?
Solution:
Since
{
s
n
}
is convergent, for any
± >
0, there exists
N
such
that

s
n

s

< ±
whenever
n
≥
N
. By Exercise 1.13 I know that

s
n
  
s
 ≤ 
s
n

s

. Thus,

s
n
  
s

, that is,
{
s
n
}
is convergent.
The converse is not true. Consider
s
n
=(

1)
n
.
2. Calculate lim
n
→∞
(
√
n
2
+
n

n
).
Solution:
√
n
2
+
n

n
=
n
√
n
2
+
n
+
n
=
1
±
1
/n
+ 1 + 1
→
1
2
as
n
→∞
.
3. If
s
n
=
√
2 and
s
n
+1
=
±
2+
√
s
n
(
n
=1
,
2
,
3
,...
)
,
prove that
{
s
n
}
converges, and that
s
n
<
2 for
n
,
2
,
3
.
Proof:
First, I show that
{
s
n
}
is strictly increasing. It is trivial that
s
2
=
±
√
s
1
=
²
±
√
2
>
√
2=
s
1
. Suppose
s
k
>s
k

1
when
1
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View Full Document k < n
. By the induction hypothesis,
s
n
=
±
2+
√
s
n

1
>
±
√
s
n

2
=
s
n

1
By the induction,
{
s
n
}
is strictly increasing. Next, I show that
{
s
n
}
is
bounded by 2. Similarly, I apply the induction again. Hence
{
s
n
}
is
strictly increasing and bounded, that is,
{
s
n
}
converges.
4.
5.
6.
7. Prove that the convergence of
∑
a
n
implies the convergence of
²
√
a
n
n
if
a
n
≥
0.
Proof:
By Cauchy’s inequality,
k
²
n
=1
a
n
k
²
n
=1
1
n
2
≥
k
²
n
=1
a
n
√
a
n
n
for all
n
∈
N
. Also, both
∑
a
n
and
∑
1
n
2
are convergent; thus
∑
k
n
=1
a
n
√
a
n
n
is bounded. Besides,
√
a
n
n
≥
0 for all
n
. Hence
∑
√
a
n
n
is convergent.
8.
9. Find the radius of convergence of each of the following power series:
(
a
)
²
n
3
z
n
,
(
b
)
²
2
n
n
!
z
n
,
(
c
)
²
2
n
n
2
z
n
,
(
d
)
²
n
3
3
n
z
n
.
2
Solution:
(a)
α
n
=(
n
3
)
1
/n
→
1 as
n
→∞
. Hence
R
=1
/α
= 1.
(b)
α
n
= (2
n
/n
!)
1
/n
=2
/
(
n
!)
1
/n
→
0 as
n
. Hence
R
=+
∞
.
(c)
α
n
= (2
n
/n
2
)
1
/n
→
2
/
1 = 2 as
n
. Hence
R
/α
/
2.
(d)
α
n
n
3
/
3
n
)
1
/n
→
1
/
3 as
n
. Hence
R
/α
= 3.
10.
11. Suppose
a
n
>
0,
s
n
=
a
1
+
...
+
a
n
, and
∑
a
n
diverges.
(a) Prove that
∑
a
n
1+
a
n
diverges.
(b) Prove that
a
N
+1
s
N
+1
+
...
+
a
N
+
k
s
N
+
k
≥
1

s
N
s
N
+
k
and deduce that
∑
a
n
s
n
diverges.
(c) Prove that
a
n
s
2
n
≤
1
s
n

1

1
s
n
and deduce that
∑
a
n
s
2
n
converges.
(d) What can be said about
±
a
n
1+
na
n
and
±
a
n
n
2
a
n
?
Proof of (a):
Note that
a
n
a
n
→
0
⇔
1
1
a
n
+1
→
0
⇔
1
a
n
⇔
a
n
→
0
as
n
. If
∑
a
n
1+
a
n
converges, then
a
n
→
0 as
n
. Thus for
some
±
±
= 1 there is an
N
1
such that
a
n
<
1 whenever
n
≥
N
1
. Since
∑
a
n
1+
a
n
converges, for any
± >
0 there is an
N
2
such that
a
m
a
m
+
...
+
a
n
a
n
< ±
3
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View Full Document all
n>m
≥
N
2
. Take
N
= max(
N
1
,N
2
). Thus
±
>
a
m
1+
a
m
+
...
+
a
n
a
n
>
a
m
1+1
+
...
+
a
n
=
a
m
+
...
+
a
n
2
for all
≥
N
. Thus
a
m
+
...
+
a
n
<
2
±
for all
n > m
≥
N
. It is a contradiction. Hence
∑
a
n
1+
a
n
diverges.
Proof of (b):
a
N
+1
s
N
+1
+
...
+
a
N
+
k
s
N
+
k
≥
a
N
+1
s
N
+
k
+
...
+
a
N
+
k
s
N
+
k
=
a
N
+1
+
...
+
a
N
+
k
s
N
+
k
=
s
N
+
k

s
N
s
N
+
k
=1

s
N
s
N
+
k
If
∑
a
n
s
n
converges, for any
± >
0 there exists
N
such that
a
m
s
m
+
...
+
a
n
s
n
< ±
for all
m,n
whenever
n > m
≥
N
. Fix
m
=
N
and let
n
=
N
+
k
.
Thus
±
>
a
m
s
m
+
...
+
a
n
s
n
=
a
N
s
N
+
...
+
a
N
+
k
s
N
+
k
≥
1

s
N
s
N
+
k
4
for all
k
∈
N
. But
s
N
+
k
→∞
as
k
since
∑
a
n
diverges and
a
n
>
0. Take
±
=1
/
2 and we obtain a contradiction. Hence
∑
a
n
s
n
diverges.
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This note was uploaded on 07/15/2008 for the course MATH 118 taught by Professor Ponce during the Spring '08 term at UCSB.
 Spring '08
 Ponce
 Sequences And Series

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