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MATH 355 HOMEWORK # 5 SOLUTIONS
Hi everyone, here are the solutions to Homework # 5. Enjoy!
√
3.2.1 (a) Let > 0. Find N according to the Archimedean Property so that N1 < 3 .
Then for n ≥ N , we have | n13 − 0| = n13 ≤ N13 < . (b) Let > 0 be given. Find
√
N according to the Archimedean Property so that N1 < 5 . Then for n ≥ N ,
n
1
1
1
− 0| = 1+n
we have | (−1)
5 ≤ n5 ≤ N 5 < . (c) Let > 0 be given. Choose
1+n5
N according to the Archimedean Property so that N1 < 2 . Then for n ≥ N , we
−2
2
1
have | n−1
n+1 − 1| = | n+1 | ≤ N < . (d) Notice that n − 4 ≥ 2 n when n ≥ 8. Let
. Then for
> 0 be given. Choose N by the Archimedean Principle so that Ñ1 < 14
−7
7
7
14
14
n ≥ N = max{8, Ñ }, we have | n−4 − 0| = n−4 ≤ 1 n = n ≤ N < .
√2
√
3.2.2 (a) Note first that 12 n3 ≤ n3 − 15 exactly when 3 30 ≤ n, so, since 4 ≥ 3 30, this
1 2
3
2 n ≤ n − 15 for n ≥ 4. Now we claim that the sequence converges to 1. For any
n ≥ 4, we have
3
n −7
8
8
16
n3 − 15 − 1 = |n3 − 15| ≤ 1 n3 = n3 .
2
p
1
Let > 0 be given. Now choose N so that N < 3 16
, so that N163 < . Now for
n ≥ N , we have n163 ≤ N163 , and so |an − 1| < .
n2
n2
(b) This sequence diverges, since it is not bounded. Notice that n+1
≥ n+n
= n2 ,
which has no finite bound.
(f) Let > 0 be given. If we choose N so that N1 < , then for n ≥ N we have
|an − 0| is either 0 (for even n) or n1 (for odd n), both of which are less than or
equal to N1 , which is less than .
(g) This sequence diverges. Choose = 0.1, and consider the neighborhood
U (x) about some real number x. Notice that an = 1 for even n, so if an → x, then
1 ∈ U (x). But if this is the case, x cannot be any smaller than 1 − 0.1 = .9, and
thus the numbers n1 for n ≥ 2 are too small to be in U (x) for any x ≥ 0.9. So,
the sequence is not eventually in U (x) for any x, and so, the sequence does not
converge to x for any x, and so it diverges.
3.2.3 Let > 0 be given. Choose N so that N1 < 2 , so that √1N < . Then for n ≥ N ,
we have | √1n − 0 =
√1
n
≤
√1
N
< .
3.2.4 (a) Suppose an → a, but that a < 0. Then choose = |a|
2 . The neighborhood
U (a) = (a − |a|/2, a + |a|/2) = (3a/2, a/2), and all of the numbers in this set are
strictly negative. Since an → a, the sequence is eventually in this neighborhood,
which is impossible, since all of the numbers an ≥ 0.
1
2
MATH 355 HOMEWORK # 5 SOLUTIONS
(b) Let > 0. Suppose that a = 0. Then since an → 0, there exists an N so that if
√
√
n ≥ N , we have |an −0| = an < 2 . Then, for these n, we have | an −0| = an < .
Now suppose a > 0 (according to √
Part √(a), this is the only alternative). Then
a + a
by rationalizing (i.e., multiplying by √ann +√a ), we have
√
√
|an − a|
|a − a|
√ ≤ n√
| an − a| = √
.
an + a
a
3.2.9
3.2.10
3.2.11
3.2.14
√
Since an → a, we may find a N so that if n ≥ N , then |an − a| < a. For these
√
√
n, we have | an − a| ≤ |an√−a|
< .
a
Let > 0. Since an → L, we may choose N so that if n ≥ N , then |an − L| < .
But ||an | − |L|| ≤ |an − L| < for these n, proving that |an | → |L|.
(a) Suppose |an | → 0. Let > 0 be given. Choose N so that ||an | − 0| = |an | < .
For these n ≥ N , we have |an − 0| = |an | < as well, so an → 0.
(b) Let an = (−1)n . We have |an | = 1 → 1, but (an ) diverges.
Choose = |L|
2 > 0. Notice that for every x ∈ R, the neighborhood U (x) does not
contain both L and −L. So, suppose there is a number x for which (−1)n an → x.
Therefore the sequence must eventually be in U (x). But the numbers L and −L
are not both in this neighborhood, so since (−1)n an is either L or −L, it is not
possible for this sequence to eventually be in U (x), a contradiction.
(d) (i) It is so noted. (ii) Since x > 1, there is a positive number h for which
x = 1 + h. Using Bernoulli’s inequality, xn = (1 + h)n ≥ 1 + nh. (iii) Notice that
1
→ 0 for any h > 0. Let > 0 be given, and choose N
by a previous exercise, 1+nh
1
so that 1+nh < for all n ≥ N . Now, for n ≥ N , we have
|rn − 0| = rn =
3.3.1 (a) By factoring out an n, we have
1
1
≤
< .
n
x
1 + nh
n−1
n+1
=
1
1− n
1
1+ n
→
1−0
1+0
= 1.
(b) Again factoring out an n in the fraction, we have
1
1/n
0
=
→ = 0.
2n − 1
2 − 1/n
2
1
Since 7 → 7, we have 7 + 2n−1
→ 7 + 0 = 7.
e
1
n
e−π
= nπ−e → 0, by a previous Theorem (since p = π−e > 0).
(e) We have nπ = n
3.3.10 Recall that the irrational numbers are dense in R. For every n ∈ N, we may
then choose an irrational number ζn so that α − n1 < ζn < α. Since α → α, and
α − n1 → α − 0 = α as well, the Squeeze Theorem forces ζn → α as well.
3.3.21 (a) Notice that
(1 − r)(1 + r + r2 + · · · + rn ) = 1(1 + r + r2 + · · · + rn ) − r(1 + r + r2 + · · · + rn )
= 1 + r + r2 + · · · + rn − r − r2 − · · · − rn − rn+1
= 1 − rn+1 .
MATH 355 HOMEWORK # 5 SOLUTIONS
3
So, by dividing by 1 − r, we arrive at the result.
(b) The sequence we are taking the limit of, according to the previous part, is
1−rn+1
1−r . Notice that the denominator 1 − r → 1 − r. Notice also that if |r| < 1,
according to a previous exercise, rn → 0, so rn+1 = r · rn+1 → r · 0 = 0. So,
n+1
1
1 − rn+1 → 1 − 0 = 1. Putting this together, we have 1−r
1−r → 1−r .
3.3.23 Suppose |bn | ≤ M . Let > 0 be given. Since an → 0, we may choose N so
that if n ≥ N , |an − 0| = |an | < M+1 . Then, for these same n ≥ N , we have
|an bn − 0| = |an ||bn | < M+1 M < .