Ans-11 - Answers Assignment#11 Exercises 2.2 1(a False sn =...

Answers, Assignment #11
Exercises 2.2
1. (a) False. s
n
=
1
n
> 0 for all n ; s
n
→ 0.
(b) False. { s
n
} = { 1 , - 1 , 1 , - 1 , · ··} ; { t
n
} = {- 1 , 1 , - 1 , 1 , ·· ·} ; { s
n
+ t
n
} = { 0 , 0 , 0 , . .. } →
0.
(c) False. The example in (b) will work here; { s
n
t
n
} = {- 1 , - 1 , - 1 , . . . } → - 1.
(d) True. Suppose { s
n
} → A and { s
n
+ t
n
} → B . Then t
n
= ( s
n
+ t
n
) - s
n
→ B - A .
(e) False. Let s
n
=
1
n
, t
n
= n . Then { s
n
t
n
} = { 1 , 1 , 1 , . .. } → 1, but { t
n
} diverges.
(f) False. Let s
n
= ( - 1) n n = {- 1 , 2 , - 3 , 4 , - 5 ,. .. } . { s
n
} is not bounded above, { s
n
}
does not diverge to + ∞ .
2. (a) s
n
→ - 2 (b) s
n
→ 0
(c) { s
n
} diverges; for large n, s
n
≈ ±
1
2
(d) s
n
=
2 3 n
3 2 n
=
8
9
n
→ 0
(e) s
n
diverges to + ∞ (f) s
n
diverges to + ∞
3. (a) n 2 + 1 - n = n 2 + 1 - n
√
n 2 + 1 + n
√
n 2 + 1 + n
=
1
√
n 2 + 1 + n
→ 0.
(b) n 2 + n - n = n 2 + n - n
√
n 2 + n + n
√
n 2 + n + n
=
n
√
n 2 + n + n
→
1
2
.
4. (1) Suppose s
n
→ s, t
n
→ t . Let > 0. Then there exists a positive integer N
1
such that
| s
n
- s | <
2
for all n > N
1
.
There exists a positive integer N
2
such that
| t
n
- t | <
2
for all n > N
2
.
Let N = max { N
1
, N
2
} . Then
| ( s
n
+ t
n
) - ( s + t ) | = | ( s
n
- s ) + ( t
n
- t ) | ≤ | s
n
- s | + | t
n
- t | <
2
+
2
=
for all n > N .
(3) Let > 0. Since s
n
→ s , { s
n
} is bounded, say | s
n
| ≤ M for all n . There exists a
positive integer N
1
such that
| s
n
- s | <
2 | t |
for all n > N
1
(assuming t = 0)
There exists a positive integer N
2
such that
| t
n
- t | <
2 M
for all n > N
2
.