Math 5333 Solns-Assignment4 - Solutions Homework#4 Section...

Solutions, Homework #4
Section 4.2
6. (a) Counterexample: ( s
n
) = (1 , - 1 , 1 , - 1 , · · · ); ( t
n
) = ( - 1 , 1 , - 1 , 1 , · · · ).
(b) Counterexample: the example in part (a) will work here, too.
(c) Proof: Suppose ( s
n
) → A and ( s
n
+ t
n
) → B . Then
t
n
= ( s
n
+ t
n
) - s
n
→ B - A
(d) Counterexample: ( s
n
) =
parenleftbigg
1 ,
1
2
,
1
3
, · · · ,
1
n
· · ·
parenrightbigg
, s
n
→ 0, ( t
n
) = (( - 1)
n
) =
( - 1 , 1 , - 1 , 1 , . . . ) does not converge. But ( s
n
t
n
) =
parenleftbigg
( - 1) n
n
parenrightbigg
→ 0.
8. (a) There are lots of examples. Here are two: ( s
n
) =
parenleftbigg
1
n
parenrightbigg
, ( s
n
) =
parenleftbigg
n
n
2
+ 1
parenrightbigg
(b) There are lots of examples. Here are two: ( s
n
) = ( n ) = (1 , 2 , 3 , 4 , . . . ) ( s
n
) =
parenleftbigg
n 2
n + 1
parenrightbigg
9. ( s
n
) and ( t
n
) are sequences such that s
n
≤ t
n
for all n .
(a) Choose any number M ∈ R . Since lim
n →∞
s
n
= + ∞ , there is a positive
integer N such that s
n
> M for all n > N . Since t
n
≥ s
n
for all n , it
follows that t
n
> M for all n > N . Therefore lim
n →∞
t
n
= + ∞ :
(b) Choose any number M ∈ R . Since lim
n →∞
t
n
= -∞ , there is a positive
integer N such that t
n
< M for all n > N . Since s
n
≤ t
n
for all n , it
follows that s
n
< M for all n > N . Therefore lim
n →∞
s
n
= -∞ :
12. Let epsilon1 > 0. Since ( s
n
) is convergent, it is bounded; | s
n
| ≤ M for all n .
1

| s 2
n
- s 2 | = | s
n
- s | | s
n
+ s | ≤ | s
n
- s | ( M + | s | ). Since s
n
→ s there is a
positive integer N such that | s
n
- s | <
epsilon1
M + | s |
for all n > N . Therefore
vextendsingle
vextendsingle
s
2
n
- s
2
vextendsingle
vextendsingle
= | s
n
- s | | s
n
+ s | ≤ | s
n
- s | ( M + | s | ) < ( M + | s | )
epsilon1
M + | s |
= epsilon1
for all n > N and it follows that s
2
n
→ s
2
.
15. (a)
√
n + 1 -
√
n =
parenleftBig
√
n + 1 -
√
n
parenrightBig
√
n + 1 +
√
n
√
n + 1 +
√
n
=
1
√
n + 1 +
√
n
<
1
2
√
n
→
0
(b)
√
n
2
+ 1 - n =
parenleftBig
√
n
2
+ 1 - n
parenrightBig
√
n 2 + 1 + n
√
n
2
+ 1 + n
=
1
√
n
2
+ 1 + n
<
1
2 n
→ 0
(c)
√
n 2 + n - n =
parenleftBig
√
n 2 + n - n
parenrightBig
√
n
2
+ n + n
√
n
2
+ n + n
=
n
√
n
2
+ n + n
→
1
2
17. Let s
n
= k n /n !.