If
a
n
→
0
,
then
s
n
→
0
.
Proof:
Note first that
a
n
≥
0
for all
n > N
. Since
a
n
→
0,
there exists a positive integer
N
1
such that

a
n

<
/k
. Without loss of generality, assume that
N
1
≥
N
. Then,
for all
n > N
1
,

s
n

0

=

s
n
 ≤
k a
n
< k
k
=
.
14
Therefore,
s
n
→
0.
Corollary
Let
{
s
n
}
and
{
a
n
}
be sequences and let
s
∈
R
.
Suppose that there is a positive
number
k
and a positive integer
N
such that

s
n

s
 ≤
k a
n
for all
n > N.
If
a
n
→
0,
then
s
n
→
s
.
Exercises 2.1
1. True – False.
Justify your answer by citing a theorem, giving a proof, or giving a counter
example.
(a) If
s
n
→
s
,
then
s
n
+1
→
s
.
(b) If
s
n
→
s
and
t
n
→
s
,
then there is a positive integer
N
such that
s
n
=
t
n
for all
n > N
.
(c) Every bounded sequence converges
(d) If to each
>
0
there is a positive integer
N
such that
n > N
implies
s
n
<
,
then
s
n
→
0.
(e) If
s
n
→
s
,
then
s
is an accumulation point of the set
S
=
{
s
1
, s
2
,
· · ·}
.
2. Prove that
lim
3
n
+ 1
n
+ 2
= 3.
3. Prove that
lim
sin
n
n
= 0.
4. Prove or give a counterexample:
(a) If
{
s
n
}
converges, then
{
s
n
}
converges.
(b) If
{
s
n
}
converges, then
{
s
n
}
converges.
5. Give an example of:
(a) A convergent sequence of rational numbers having an irrational limit.
(b) A convergent sequence of irrational numbers having a rational limit.
6. Give the first six terms of the sequence and then give the n
th
term
(a)
s
1
= 1
,
s
n
+1
=
1
2
(
s
n
+ 1)
(b)
s
1
= 1
,
s
n
+1
=
1
2
s
n
+ 1
(c)
s
1
= 1
,
s
n
+1
= 2
s
n
+ 1
7. use induction to prove the following assertions:
(a) If
s
1
= 1
and
s
n
+1
=
n
+ 1
2
n
s
n
,
then
s
n
=
n
2
n

1
.
(b) If
s
1
= 1
and
s
n
+1
=
s
n

1
n
(
n
+ 1)
,
then
s
n
=
1
n
.
15
8. Letrbe a real number,r= 0. Define a sequence{Sn}by
9. Set
a
n
=
1
n
(
n
+ 1)
, n
= 1
,
2
,
3
, . . .
,
and form the sequence
S
1
=
a
1
S
2
=
a
1
+
a
2
S
3
=
a
1
+
a
2
+
a
3
.
.
.
S
n
=
a
1
+
a
2
+
a
3
+
· · ·
+
a
n
.
.
.
Find a formula for
S
n
.
II.2.
LIMIT THEOREMS
THEOREM 4.
Suppose
s
n
→
s
and
t
n
→
t
. Then:
1.
s
n
+
t
n
→
s
+
t
.
2.
s
n

t
n
→
s

t
.
3.
s
n
t
n
→
st
.
Special case:
ks
n
→
ks
for any number
k
.
4.
s
n
/t
n
→
s/t
provided
t
= 0
and
t
n
= 0
for all
n
.
THEOREM 5.
Suppose
s
n
→
s
and
t
n
→
t
. If
s
n
≤
t
n
for all
n
,
then
s
≤
t
.
Proof:
Suppose
s > t
. Let
=
s

t
2
. Since
s
n
→
s
, there exists a positive integer
N
1
such that

s
n

s

<
for all
n > N
1
. This implies that
s

< s
n
< s
+
for all
n > N
1
. Similarly, there
exists a positive integer
N
2
such that
t

< t
n
< t
+
for all
n > N
2
. Let
N
= max
{
N
1
, N
2
}
.
Then, for all
n > N
,
we have
t
n
< t
+
=
t
+
s

t
2
=
s
+
t
2
=
s

< s
n
16
which contradicts the assumption
s
n
≤
t
n
for all
n
.
Corollary
Suppose
t
n
→
t
. If
t
n
≥
0
for all
n
,
then
t
≥
0.
Infinite Limits
Definition 3.
A sequence
{
s
n
}
diverges to
+
∞
(
s
n
→
+
∞
)
if to each real number
M
there
is a positive integer
N
such that
s
n
> M
for all
n > N
.
{
s
n
}
diverges to
∞
(
s
n
→ ∞
)
if to each real number
M
there is a positive integer
N
such that
s
n
< M
for all
n > N
.
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 Fall '08
 Staff
 Real Numbers, Integers, Limit of a sequence, Sn