SOLUTIONS TO HOMEWORK 6 In some cases, the statement of the

SOLUTIONS TO HOMEWORK 6
N. CHRISTOPHER PHILLIPS
In some cases, the statement of the problem has been rewritten to be more
precise.
Warning: Essentially no proofreading has been done.
1. Section 2.3
Problem 1.1 (Problem 2.3.10). Let (an )n∈Z>0 and (bn )n∈Z>0 be sequences in R,
and let b ∈ R. Suppose limn→∞ an = 0 and |bn − b| ≤ an for all n. Prove that
limn→∞ bn = b.
Solution. The hypotheses imply that an ≥ 0 for all n ∈ Z>0 . (Otherwise, we could
not possibly have |bn − b| ≤ an .)
Let ε > 0.
Since limn→∞ an = 0, there is N ∈ Z>0 such that for all n ∈ Z>0 with n ≥ N,
we have |an | < ε.
Let n ∈ Z>0 with n ≥ N. Then
|bn − b| ≤ an = |an | < ε.
This shows that limn→∞ bn = b.
2. Section 2.4
Problem 2.1 (Problem 2.4.3). Following the model of Exercise 2.4.2, show that
the sequence defined by y1 = 1 and yn+1 = 4 − 1/yn converges and find the limit.
Solution. We first prove by induction on n that 1 ≤ yn < yn+1 < 4 for all n ∈ Z>0 .
This is true for n = 1, since by definition y1 = 1, while the recursive relation
gives y2 = 3.
For the induction step, suppose it is known that 1 ≤ yn < yn+1 < 4. Then
1
1
1
1≥
>
> .
yn
yn+1
4
Therefore
1
1
1
3≤4−
<4−
<4− .
yn
yn+1
4
That is,
1
3 ≤ yn+1 < yn+2 < 4 − .
4
In particular,
1 ≤ yn+1 < yn+2 < 4,
as desired. The induction is complete.
It follows that (yn )n∈Z>0 is a bounded strictly increasing sequence. Therefore
y = limn→∞ yn exists. As before, we then also have limn→∞ yn+1 = y.
Date: 11 February 2015.
1
2
N. CHRISTOPHER PHILLIPS
Take limits as n → ∞ on both sides of the equation
yn+1 = 4 −
1
,
yn
and use the Algebraic Limit Theorem several times, to get
1
y =4− .
y
Rearrange to get
y 2 − 4y + 1 = 0.
This has the solutions
y=
4+
√
12
2
Simplify to get
√
12
2
or y = 2 −
.
√
3.
√
√
Since (yn )n∈Z>0 is nondecreasing, we have y ≥ y1 = 1 > 2 − 3, so y = 2 + 3. y =2+
3
or y =
√
4−
3. Section 2.5
Problem 3.1 (Problem 2.5.2).
(1) Prove that if an infinite series converges, then the associative property
holds. That is, assume a1 + a2 + a3 + a4 + · · · converges to a limit L (that
is, the sequence of partial sums (sn )n∈Z>0 converges to L). Show that any
regrouping of the terms
(a1 + a2 + · · · + an1 ) + (an1 +1 + · · · + an2 ) + (an2 +1 + · · · + an3 ) + · · ·
leads to a series which also converges to L.
(2) Compare this result to the example discussed at the end of Section 2.1
where infinite addition was shown not to be associative. Why doesn’t our
proof in part 1 apply to this example?
P∞
Solution. (1) The partial sums of the series n=1 are given by
sn = a1 + a2 + · · · + an
for n ∈ Z>0 .
Define n0 = 0, and define
bk = ank−1 +1 + · · · + an2
P∞
for k ∈ Z>0 . Thus the new series is k=1 bk . Its partial sums are given by
tk = b1 + b2 + · · · + bk .
It is immediate that
tk = a1 + a2 + · · · + ank = snk .
By hypothesis, limn→∞ sn = L. Therefore, by Theorem 2.5.2 of the book,
∞
X
(ank−1 +1 + · · · + an2 ) =
k=1
This completes the proof.
∞
X
k=1
bk = lim tk = lim snk = L.
k→∞
k→∞
SOLUTIONS TO HOMEWORK 6
3
(2) The series in the example discussed at the end of Section 2.1 does not converge. Indeed, the sequence of partial sums is
(−1, 0, −1, 0, −1, 0, −1, 0, . . .),
which has (constant) subsequences converging to the distinct limits −1 and 0.
4. Section 2.6
Problem 4.1 (Problem 2.6.1). Give an example of each of the following, or argue
that such a request is impossible.
(1)
(2)
(3)
(4)
A Cauchy sequence that is not monotone.
A monotone sequence that is not Cauchy.
A Cauchy sequence with a divergent subsequence.
An unbounded sequence containing a subsequence that is Cauchy.
Solution. (1) Define an = (−1)n /n for n ∈ Z>0 . Then limn→∞ an = 0. (Check: Let
ε > 0. Choose N ∈ Z>0 such that N1 < ε. Let n ≥ N. Then
(−1)n = 1 ≤ 1 < ε.
|an − 0| = n n
N
This proves limn→∞ an = 0.) Therefore (an )n∈Z>0 is Cauchy.
The sequence (an )n∈Z>0 is not monotone because a1 < a2 but a2 > a3 .
One can also easily prove directly that (an )n∈Z>0 is Cauchy.
There are many other examples. Here is a rather trivial one:
− 1, 1, −1, 0, 0, 0, 0, 0, 0, . . .).
(2) The sequence (an )n∈Z>0 defined by an = n for n ∈ Z>0 is monotone. It isn’t
Cauchy, since we have a theorem telling us that all Cauchy sequences are bounded.
(3) No such thing exists (at least not in R). Every Cauchy sequence converges,
so every subsequence converges to the same limit.
(4) Define
an =
n
0
n is odd
n is even.
Then (an )n∈Z>0 is clearly unbounded, but the subsequence (a2n )n∈Z>0 is constant,
therefore convergent, therefore Cauchy.
Problem 4.2 (Problem 2.6.5). If (xn )n∈Z>0 and (yn )n∈Z>0 are Cauchy sequences,
then one easy way to prove that (xn + yn )n∈Z>0 is Cauchy is to use the Cauchy
Criterion. By Theorem 2.6.4, (xn )n∈Z>0 and (yn )n∈Z>0 must be convergent, and
the Algebraic Limit Theorem then implies (xn + yn )n∈Z>0 is convergent and hence
Cauchy.
(1) Give a direct argument that (xn + yn )n∈Z>0 is a Cauchy sequence that does
not use the Cauchy Criterion or the Algebraic Limit Theorem.
(2) Do the same for the product (xn yn )n∈Z>0 .
Solution. (1) Let ε > 0. Choose N1 ∈ Z>0 such that for all m, n ∈ Z>0 with
m, n ≥ N1 , we have |xm − xn | < 21 ε. Choose N2 ∈ Z>0 such that for all m, n ∈ Z>0
4
N. CHRISTOPHER PHILLIPS
with m, n ≥ N2 , we have |ym − yn | < 12 ε. Define N = max(N1 , N2 ). Let m, n ∈ Z>0
with m, n ≥ N. Then
(xm +ym )−(xn +yn ) = (xm −xn )+(ym −yn ) ≤ |xm −xn |+|ym −yn | < 1 ε+ 1 ε = ε.
2
2
(2) We need a little preparation. By Lemma 2.6.3 of the book, (xn )n∈Z>0 and
(yn )n∈Z>0 are bounded. Thus, there are M1 , M2 ∈ [0, ∞) such that |xn | ≤ M1 and
|y2 ≤ M2 for all n ∈ Z>0 .
Let ε > 0. Choose N1 ∈ Z>0 such that for all m, n ∈ Z>0 with m, n ≥ N1 , we
have
ε
.
|xm − xn | <
2M2 + 1
Choose N2 ∈ Z>0 such that for all m, n ∈ Z>0 with m, n ≥ N2 , we have
ε
.
|ym − yn | <
2M1 + 1
Define N = max(N1 , N2 ). Let m, n ∈ Z>0 with m, n ≥ N. Then
xm ym − xn yn ) = (xm ym − xn ym ) + (xn ym − xn yn )
≤ |xm − xn | · |ym | + |xn | · |ym − yn |
ε
ε
≤
M2 + M1
2M2 + 1
2M1 + 1
< ε.
Remark 1: It is legitimate to use Lemma 2.6.3 of the book, since it doesn’t
depend on completeness. It isn’t legitimate to use the fact that Cauchy sequences
converge, and then the fact that convergent sequences are bounded, since this
requires completeness.
Remark 2: We divide by 2M1 + 1 and 2M2 + 1 in case M1 = 0 or M2 = 0. Problem 4.3 (Problem 2.6.6b). Use the Monotone Convergence Theorem to give
a proof of the Nested Interval Property.
Solution. For n ∈ Z>0 , let In = [an , bn ] be a closed bounded interval. Suppose that
I1 ⊃ I2 ⊃ I3 ⊃ I4 ⊃ · · · .
We have to prove that
∞
\
In 6= ∅.
n=1
Let n ∈ Z>0 . Since In+1 ⊆ In , we have an+1 ≥ an . This shows that (an )n∈Z>0 is
nondecreasing. Also, an ∈ I1 , so an ≤ b1 . This shows that (an )n∈Z>0 is bounded.
Therefore limn→∞ an exists. Call it a.
We claim that for all n ∈ Z>0 , we have a ∈ In . This will show that
∞
\
a∈
In ,
n=1
whence
∞
\
In 6= ∅.
n=1
So let n ∈ Z>0 . Then an , an+1 , an+2 , an+3 , . . . is a subsequence of (an )n∈Z>0 , so
it has the same limit, namely a. Since all the terms are greater than or equal to
SOLUTIONS TO HOMEWORK 6
5
an , so is the limit. Since all the terms are less than or equal to bn , so is the limit.
Thus a ∈ [an , bn ] = In .
Alternate solution. For n ∈ Z>0 , let In = [an , bn ] be a closed bounded interval.
Suppose that
I1 ⊃ I2 ⊃ I3 ⊃ I4 ⊃ · · · .
We have to prove that
∞
\
In 6= ∅.
n=1
Let n ∈ Z>0 . Since In+1 ⊆ In , we have an+1 ≥ an and bn+1 ≤ bn . This shows
that (an )n∈Z>0 is nondecreasing and (bn )n∈Z>0 is nonincreasing. Also, an ∈ I1 , so
an ≤ b1 . Similarly bn ≥ a1 . This shows that (an )n∈Z>0 and (bn )n∈Z>0 are bounded.
Therefore limn→∞ an and limn→∞ bn exist. Set a = limn→∞ an and b = limn→∞ bn .
We have
a = sup({an : n ∈ Z>0 })
and b = inf({bn : n ∈ Z>0 }).
For all n ∈ Z>0 , we have an ≤ bn . Therefore a ≤ b. In particular, [a, b] 6= ∅.
We claim that [a, b] ⊆ In for all n ∈ Z>0 . This will show that
[a, b] ⊆
∞
\
In ,
n=1
whence
∞
\
In 6= ∅.
n=1
So let x ∈ [a, b] and let n ∈ Z>0 . Then
an ≤ sup({am : m ∈ Z>0 }) = a ≤ x ≤ b = inf({bm : m ∈ Z>0 }) ≤ bn ,
so x ∈ [an , bn ] = In .
5. Section 2.7
Problem 5.1 (Problem 2.7.1). Do one of the three parts of this problem. That is:
Proving the Alternating Series Test (Theorem 2.7.7) amounts to showing that
the sequence of partial sums
sn = a1 + a2 + a3 + · · · ± an
converges. Different characterizations of completeness lead to different proofs. Use
one of the following methods:
(1) Prove the Alternating Series Test by showing that (sn )n∈Z>0 is a Cauchy
sequence.
(2) Supply another proof for this result using the Nested Interval Property
(Theorem 1.4.1).
(3) Consider the subsequences (s2n )n∈Z>0 and (s2n+1 )n∈Z>0 , and show how the
Monotone Convergence Theorem leads to a third proof for the Alternating
Series Test.
6
N. CHRISTOPHER PHILLIPS
Solution. We give all three proofs. All require the following notation and facts,
which we do first. Define (sn )n∈Z>0 to be the sequence of partial sums:
sn =
n
X
ak
k=1
for n ∈ Z>0 . Then we need:
(1) s1 ≥ s3 ≥ s5 ≥ · · · .
(2) s2 ≤ s4 ≤ s6 ≤ · · · .
(3) For all n ∈ Z>0 , we have s2n ≥ s2n+1 and s2n+1 ≤ s2n+2 .
For (1), observe that s2n+3 = s2n+1 − a2n+2 + a2n+3 and −a2n+2 + a2n+3 ≤ 0.
Part (2) is proved similarly. Part (3) is done as follows:
s2n+1 = s2n + a2n+1 ≤ s2n
and s2n+2 = s2n+1 − a2n+2 ≤ s2n+1 .
It now follows that for every n ∈ Z>0 , whenever m ≥ 2n we have sm ∈
[s2n+1 , s2n ].
(1) Let ε > 0. Choose an even number N ∈ Z>0 such that aN < ε. Then for all
m, n ∈ Z>0 with m, n ≥ N, we have sm , sn ∈ [sN , sN +1 ]. In particular,
|sm − sn | ≤ SN +1 − SN = aN +1 ≤ aN < ε.
(2) For n ∈ Z>0 , define In = [s2n+1 , s2n ]. Then, by the facts above,
I1 ⊃ I2 ⊃ I3 ⊃ I4 ⊃ · · · .
The Nested Interval Property implies that
∞
\
[s2n+1 , s2n ] 6= ∅.
n=1
Choose an element s in this intersection. We claim that limn→∞ sn = s.
Let ε > 0. Choose an even number N ∈ Z>0 such that aN < ε. Let n ≥ N. Then
s and sn are both in [sN +1 , sN ], so
|sn − s| ≤ SN +1 − SN = aN +1 ≤ aN < ε.
(3) We saw above that the subsequences (s2n )n∈Z>0 and (s2n+1 )n∈Z>0 are monotone and bounded. Therefor they converge. Call the limits x and y. It suffices to
prove that x = y.
Let ε > 0. Choose an even number N ∈ Z>0 such that aN < ε. Then for n ∈ Z>0
with n ≥ N, we have s2n , s2n+1 ∈ [sN +1 , sN ]. Therefore x, y ∈ [sN +1 , sN ]. It follows
that
|x − y| ≤ SN +1 − SN = aN +1 ≤ aN < ε.
Since ε > 0 is arbitrary, this implies that x = y.
Problem 5.2 (Problem 2.7.2b). Give another proof for the Comparison Test, using
the Monotone Convergence Theorem.
Solution. Let (ak )k∈Z>0 and (bk )k∈Z>0 be sequences such that 0 ≤ ak ≤ bk for all
k ∈ Z>0 . We have to prove two things:
P∞
P∞
(1) If Pk=1 bk converges, thenP k=1 ak converges.
∞
∞
(2) If k=1 ak diverges, then k=1 bk diverges.
SOLUTIONS TO HOMEWORK 6
7
Part (2) follows immediately from part (1), so we only prove part (1).
Define (sn )n∈Z>0 and (tn )n∈Z>0 to be the sequences of partial sums:
sn =
n
X
ak
and tn =
k=1
n
X
bk
k=1
for n ∈ Z>0 . Clearly (sn )n∈Z>0 and (tn )n∈Z>0 are nondecreasing. Also 0 ≤ sn ≤ tn
for all n ∈ Z>0 .
The hypotheses imply that limn→∞ tn exists. Since this limit is the same as
sup({tn : n ∈ Z>0 }), it follows that, for all n ∈ Z>0 , we have
0 ≤ sn ≤ tn ≤ lim tm .
m→∞
So (sn )n∈Z>0 is a bounded nondecreasing sequence, and therefore converges.
Problem 5.3 (Problem 2.7.3). Let (an )n∈Z>0 be a sequence of real numbers. Define
an
an ≥ 0
pn =
0
an < 0,
and define
an ≥ 0
an < 0,
P∞
P∞
P∞
(1) Suppose that n=1 an diverges. Prove that n=1 pn diverges or n=1 qn
diverges.
P∞
P∞
(2) Suppose that
n=1 pn diP∞ n=1 an converges conditionally. Prove that
verges and n=1 qn diverges.
P∞
P∞
Solution. (1) We have
an = pn +
n for all n ∈ Z>0 . So if
n=1 pn and
n=1 qn
P∞
Pq∞
both converge, then n=1 an = n=1 (pn + qn ) converges.
P∞
P∞
P∞
P∞
(2) If n=1 pnP
converges, then
If n=1 qn
P∞ n=1 qn = n=1 (an −pn ) converges.
P∞
∞
converges,
then
n=1 pn =
n=1 (an − qn ) converges. So either
n=1 pn and
P∞
both diverge.
n=1 qn both converge or theyP
P∞
P∞
∞
If they both converge, then P
(pn − qn ) converges, so n=1 an
n=1 |an | =
n=1
P∞
∞
converges absolutely. Thus both n=1 pn and n=1 qn must diverge.
qn =
0
an
6. Section 3.2
Problem 6.1 (Problem 3.2.1).
(1) Where in the proof of Theorem 3.2.3(ii) does the assumption that the collection of open sets be finite get used?
(2) Give an example of an infinite collection of nested open sets
U1 ⊃ U2 ⊃ U3 ⊃ U4 ⊃ · · ·
T∞
whose intersection n=1 Un is closed and nonempty.
Solution. (1) We need N to be finite so that we can say that inf({ε1 , ε2 , . . . , εN }) >
0. This is true because
inf({ε1 , ε2 , . . . , εN }) = min({ε1 , ε2 , . . . , εN }).
8
N. CHRISTOPHER PHILLIPS
(2) We have seen that [−1, 1] is closed, and that for n ∈ Z>0 , the interval
− 1 − n1 , 1 + n1 is open. Clearly
∞
\
− 1 − n1 , 1 +
1
n
= [−1, 1].
n=1
The following is also a correct, although trivial example (and one
T∞which misses
the point of the problem): Take Un = R for all n ∈ Z>0 . Then n=1 Un = R is
closed and nonempty.
Problem 6.2 (Problem 3.2.3). Decide whether the following sets are open, closed,
or neither. If a set is not open, find a point in the set for which there is no εneighborhood contained in the set. If a set is not closed, find a limit point that is
not contained in the set.
(1) Q.
(2) Z>0 .
(3) {x ∈ R : x > 0}.
(4) (0, 1] = {x ∈ R : 0 < x ≤ 1}.
(5) 1 + 41 + 19 + · · · + n12 : n ∈ Z>0 .
Solution. (1) The set Q is not open. Consider 0 ∈ Q. Let ε > 0. Choose n ∈ Z>0
such that n1 < ε. Set
√
2
.
a=
2n
√
We know 2 6∈ Q, so also a 6∈ Q. But 0 < a < n1 < ε, so a ∈ Vε (0).
√
√
The set Q is also not closed. We know 2 6∈ Q. However, we claim that 2 is a
limit point of Q. √
To see this,
of Q√implies that there is
√ let ε > 0. The order√density
r ∈ Q such that 2 < r < 2 + ε. Then r ∈ Vε 2 but r 6= 2, as desired.
We can also see that
√ Q is not closed by combining Theorem 3.2.10 and Theorem
3.2.5 to show that 2 is a limit point of Q.
(2) The set Z>0 is not open. Consider 1 ∈ Z>0 . Let ε > 0. Then a = 1 +
min 12 , 12 ε satisfies a 6∈ Z>0 but a ∈ Vε (0).
The set Z>0 is closed. It suffices to prove that R\Z>0 is open. First observe that
(−∞, 1) is open. One can prove this directly, or write it as the union of bounded
open intervals as follows:
∞
[
(−∞, 1) =
(−n, 1).
n=1
We can now write R \ Z>0 as a union of open sets:
R \ Z>0 = (−∞, 1) ∪
∞
[
(n, n + 1).
n=1
(3) The set {x ∈ R : x > 0} is open, since we can write it as a union of sets that
are known to be open:
∞
[
{x ∈ R : x > 0} =
(0, n).
n=1
However, this set is not closed. Clearly 0 6∈ {x ∈ R : x > 0}, but we claim that 0 is
a limit point of {x ∈ R : x > 0}. So let ε > 0. Then 21 ε ∈ Vε (0), proving the claim.
SOLUTIONS TO HOMEWORK 6
9
(4) The set (0, 1] is not open. Consider 1 ∈ (0, 1]. Let ε > 0. Then Vε (1) 6⊂ (0, 1],
because 1 + 21 ε ∈ Vε (1) but 1 + 12 ε 6∈ (0, 1].
This set is also not closed. Clearly 0 6∈ (0, 1], but we
claim that 0 is a limit point
of {x ∈ R : x > 0}. So let ε > 0. Then min 21 , 12 ε ∈ Vε (0) ∩ (0, 1], proving the
claim.
(5) We start with some preliminary work. Set
S = 1 + 41 + 19 + · · · + n12 : n ∈ Z>0 .
For n ∈ Z>0 , set
sn = 1 +
1
4
+
1
9
+ ··· +
1
n2 .
Thus
S = sn : n ∈ Z>0 .
P∞
Clearly s1 < s2 < s3 < · · · . We have already seen that n=1 n12 converges. Let s
be the sum of the series. Then limn→∞ sn = s. Since (sn )n∈Z>0 is nondecreasing, we
have s = sup(S). (See the proof of what the book calls the Monotone Convergence
Theorem.) Since (sn )n∈Z>0 is strictly increasing, for every n we have s > sn (since
s ≥ sn+1 > sn ). So s 6∈ S.
We claim that S is not open. Consider 1 = s1 ∈ S. Let ε > 0. Then a = 1 − 21 ε
satisfies a 6∈ S but a ∈ Vε (1). This proves the claim.
We claim that S is not closed. We saw above that s 6∈ S. However, Theorem
3.2.5, applied to the sequence (sn )n∈Z>0 in S, shows that s is a limit point of S. Problem 6.3 (Problem 3.2.4). Prove the converse of Theorem 3.2.5 by showing
that if x = limn→∞ an for some sequence (an )n∈Z>0 contained in A satisfying an 6= x
for all n ∈ Z>0 , then x is a limit point of A.
Solution. Let ε > 0. Choose N ∈ Z>0 such that for all n ∈ Z>0 with n ≥ N, we
have |an − x| < ε. Then aN ∈ Vε (x), and by the hypotheses on (an )n∈Z>0 , we have
aN ∈ A and aN 6= x.