Solutions, Homework #4
Section 4.2
6. (a) Counterexample: ( sn) = (1 , - 1 , 1 , - 1 , · · · ); ( tn) = ( - 1 , 1 , - 1 , 1 , · · · ).
(b) Counterexample: the example in part (a) will work here, too.
(c) Proof: Suppose ( sn) → A and ( sn+ tn) → B . Then
tn= ( sn+ tn) - sn→ B - A
(d) Counterexample: ( sn) =
parenleftbigg
1 ,
1
2
,
1
3
, · · · ,
1
n
· · ·
parenrightbigg
, sn→ 0, ( tn) = (( -
1)n) =
( - 1 , 1 , - 1 , 1 , . . . ) does not converge. But ( sntn) =
parenleftbigg
( -
1)n
n
parenrightbigg
→
0.
8. (a) There are lots of examples. Here are two: ( sn) =
parenleftbigg
1
n
parenrightbigg
, ( sn) =
parenleftbigg
n
n2+ 1
parenrightbigg
(b) There are lots of examples. Here are two: ( sn) = ( n ) = (1 , 2 , 3 , 4 , . . . ) ( sn) = parenleftbigg
n2
n + 1
parenrightbigg
9. ( sn) and ( tn ) are sequences such that sn≤ tnfor all n .
(a) Choose any number M ∈ R . Since lim
n →∞
sn= + ∞ , there is a positive
integer N such that sn> M for all n > N . Since tn≥ snfor all n , it
follows that tn> M for all n > N . Therefore limn →∞tn= + ∞ :
(b) Choose any number M ∈ R . Since limn →∞tn= -∞ , there is a positive
integer N such that tn< M for all n > N . Since sn≤ tnfor all n , it
follows that sn< M for all n > N . Therefore limn →∞sn= -∞ :
12. Let epsilon1 >
0. Since ( sn ) is convergent, it is bounded; | sn| ≤ M for all n .
1
| s2
n- s2| = | sn- s | | sn+ s | ≤ | sn- s | ( M + | s | ). Since sn→ s there is a
positive integer N such that | sn- s | <
epsilon1
M + | s |
for all n > N . Therefore
vextendsingle vextendsingle
s2
n- s2 vextendsingle vextendsingle = | sn- s | | sn+ s | ≤ | sn- s | ( M + | s | ) < ( M + | s | )
epsilon1
M + | s |
= epsilon1
for all n > N and it follows that s2
n→ s2.
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)
√
n2+ 1 - n =
parenleftBig √
n2+ 1 - n
parenrightBig
√
n2+ 1 + n
√
n2+ 1 + n
=
1
√
n2+ 1 + n
<
1
2 n
→ 0
(c)
√
n2+ n - n =
parenleftBig √
n2+ n - n
parenrightBig
√
n2+ n + n
√
n2+ n + n
=
n
√
n2+ n + n
→
1
2
17. Let sn= kn/n !.