If s n converges then s n is bounded proof suppose s

If { s
n
} converges, then { s
n
} is bounded.
THEOREM 3. Let { s
n
} and { a
n
} be sequences and suppose that there is a positive number k
and a positive integer N such that
| s
n
| ≤ k a
n
for all n > N.
If a
n
→ 0 , then s
n
→ 0 .
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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
.
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8. Let r be a real number, r = 0. Define a sequence { S
n
} by
S
1
= 1
S
2
= 1 + r
S
3
= 1 + r + r
2
.
.
.
S
n
= 1 + r + r
2
+ · · · + r
n - 1
.
.
.
(a) Suppose r = 1. What is S
n
for S
n
= 1 , 2 , 3 , . . . ?
(b) Suppose r = 1. Find a formula for S
n
.
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.
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 .
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