Partial solutions for second homework

Solutions for Homework #2
Math 451(Section 3, Fall 2014)
8.2a) Claim: lim n2n+1 = 0.
Proof: Given > 0, let N = 1 . If n > N , then
n
n
1
n
1
n2 + 1 − 0 = n2 + 1 < n2 = n < N = Given > 0, we have exhibited N so that if n > N , then | n2n+1 −0| < . Therefore, ( n2n+1 ) converges
to 0.
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8.2d) Claim: lim 2n+4
5n+2 = 5 .
Proof: Given > 0, let N =
16−10
25 .
If n > N , then
2n + 4
16
2 16
5n + 2 − 5 = 25n + 10 < 25( 16−10 ) + 10 = .
25
2
To summarize, given > 0, we have exhibited N so that if n > N , then 2n+4
5n+2 − 5 < . Therefore,
2
( 2n+4
5n+2 ) converges to 5 .
Alternate Proof: Let sn = 2n+4
5n+2 . Notice that
16
16
0 < 25n+10
< 16n
= n1 if n ∈ N.
Given > 0, let N = 1 . If n > N , then
2n+4
5n+2
−
2
5
=
16
25n+10
and that
2
16
1
1
|sn − | =
< <
= .
5
25n + 10
n
N
2
To summarize, given > 0, we have exhibited N so that if n > N , then 2n+4
5n+2 − 5 < . Therefore,
2
( 2n+4
5n+2 ) converges to 5 .
8.4) Claim: If (tn ) is a bounded sequence and (sn ) is a sequence such that lim sn = 0, then
lim sn tn = 0.
Proof: Since (tn ) is bounded there exists M ≥ 0 such that |tn | ≤ M for all n ∈ N. Let > 0.
Since lim sn = 0 and M+1 > 0, there exists N such that if n > N ,
then |sn | = |sn − 0| < M+1 . So, if n > N ,
|sn tn − 0| = |sn ||tn | <
M < .
M +1
Therefore, given any > 0, there exists N such that if n > N , then |sn tn − 0| < , so (sn tn )
converges to 0 as claimed.
8.5a) Claim: Suppose that (an ), (bn ) and (sn ) are three sequences and that
an ≤ sn ≤ bn for all n ∈ N. If lim an = lim bn = s, then (sn ) converges and lim sn = s.
Proof: Let > 0. Since lim an = s, there exists N1 such that if n > N1 , then |an − s| < .
In particular, if n > N1 , then an > s − . Similarly, since lim bn = s, there exists N2 such that if
n > N2 , then |bn − s| < . In particular, if n > N2 , then bn < s + .
Let N = max{N1 , N2 }. If n > N , then both an > s − and bn < s + . Since, an ≤ sn ≤ bn ,
this implies that s − < an ≤ sn ≤ bn < s + . Hence, if n > N then |sn − s| < . Thus, given any
> 0, there exists N such that if n > N , then |sn − s| < . It follows that lim sn = s.
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b) Suppose that (sn ) and (tn ) are sequences such that |sn | ≤ tn for all n and (tn ) converges to 0.
Claim: (sn ) converges to 0.
Proof: Let > 0. Since lim tn = 0, there exists N such that n > N implies that |tn | = |tn − 0| < .
But, by assumption, |sn | ≤ tn , so for all n > N ,
|sn − 0| = |sn | ≤ tn ≤ |tn | < .
Thus, given any > 0, there exists N such that if n > N , then |sn − 0| < . Therefore, (sn )
converges to 0.
Remark: If we were allowed to use the scalar multiplication limit law, we could first use it to show
that lim −tn = 0. (Alternatively, we could simply prove directly that lim −tn = 0.) Then, since
−tn ≤ sn ≤ tn
for all n ∈ N and lim −tn = lim tn = 0, the squeeze principle (Exercise 8.5a) would imply that
lim tn = 0.
8.6a) Let (sn ) be a sequence in R.
Claim: (sn ) converges to 0 if and only if (|sn |) converges to 0.
Proof: First suppose that (sn ) converges to 0. Let > 0. Since lim sn = 0, there exists N such
that n > N implies that |sn | = |sn − 0| < . But, since | |sn | | = |sn |, this implies that | |sn | − 0| < .
Thus, n > N implies that | |sn | − 0| < . Therefore, since for all > 0 there exists N so that if
n > N , then | |sn | − 0| < , we see that (|sn |) converges to 0. So, if (sn ) converges to 0, then (|sn |)
converges to 0.
Now, suppose that (|sn |) converges to 0. Let > 0. Since lim |sn | = 0, there exists N such
that n > N implies that |sn | = | |sn | − 0| < . But, this implies that |sn | = |sn − 0| < . Thus,
n > N implies that |sn − 0| < . Therefore, since for all > 0 there exists N so that if n > N , then
|sn − 0| < , (sn ) converges to 0. So, if (|sn |) converges to 0, then (sn ) converges to 0.
8.9a) Claim: If (sn ) is a convergent sequence and sn ≥ a for all but finitely many values of n, then
lim sn ≥ a.
Proof: Suppose that lim sn = s < a. Let = a − s. By assumption, > 0. Since lim sn = s,
there exists N such that if n > N , then |sn − s| < . This implies that if n > N , then sn ∈
(s − , s + ) = (s − , a). Hence, if n > N , then sn < a, so there are infinitely many values of n
for which sn < a, which contradicts our initial assumption. Thus, it must actually be the case that
lim sn ≥ a as we originally claimed.
n
9.2) Claim: If lim xn = 3 and lim yn = 7 and and all yn are non-zero, then lim 3yny−x
= 18
2
49 .
n
Proof: Since lim yn = 7, Theorem 9.2 implies that (3yn ) converges to 21. Since (xn ) converges
to 3, Theorem 9.2 implies that (−1 · xn ) = (−xn ) converges to −1 · 3 = −3. Since lim 3yn = 21
and lim −xn = −3, Theorem 9.3 implies that (3yn + (−xn )) = (3yn − xn ) converges to 18. Since
lim yn = 7, Theorem 9.4 implies that (yn · yn ) = (yn2 ) converges to 49. Since
lim3yn − xn = 18 and
3yn −xn
2
2
lim yn = 49 6= 0 and yn is non-zero for all n, Theorem 9.6 implies that
converges to 18
49 .
y2
n
9.10a) Claim: If lim sn = +∞ and k > 0, then lim ksn = +∞.
Proof: Let M > 0. Since lim sn = +∞ and M
k > 0, there exists N such that n > N implies
that sn > M
,
which
in
turn
implies
that
ks
>
M
. So, n > N implies that ksn > M . Therefore,
n
k
for all M > 0, we have shown that there exists N so that n > n implies that ksn > M , so (ksn )
diverges to +∞.
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