2/1/2017
Solutions to Math 425 Homework 3
1. (Exercise 2.2.2)
(a) Let > 0 be given. By the Archimedean Principle, choose N ∈ N such that N >
n ≥ N . Then
−3 2n + 1 2 5n + 4 − 5 = 25n + 20 |−3|
=
|25n + 20
3
=
25n + 20
3
≤
25n
3
≤
25N
< .
3
25 .
Now assume
(b) Let > 0 be given. By the Archimedean Principle, choose N ∈ N such that N > 2 . Now assume
n ≥ N . Then
2n2 2n2
n3 + 3 − 0 = n3 + 3 2n2
=
n3 + 3
2n2
≤
n3
2
=
n
2
≤
N
< .
(c) (Discussed in class)
2. (2.2.3)
(a) To disprove the statement, we’d have to find a college in the U.S., where every student is less than
seven feet tall.
(b) We’d have to find a college where every professor has given at least one student a grade other than
A or B.
(c) We’d have to show that every college in the US has a student who is under six feet tall.
∞
3. (2.2.1) Let an = 0 for all n, and let bn = (−1)n for all n. Then the sequences (an )∞
n=1 and (bn )n=1 are
both vercongent to 0. The sequence an is also convergent to 0, but bn is divergent.
1
To show that bn is vercongent to 0: we need to show the following:
(∃ > 0)(∀N ∈ N)(n ≥ N =⇒ |bn − 0| < )
Let N ∈ N and assume n ≥ N . Then bn = (−1)n = ±1, so |bn − 0| = |±1| = 1 < 2.
Thus we have shown that there exists > 0 such that for all N ∈ N, if n ≥ N then |bn − 0| < .
(Specifically, = 2 works.)
Indeed, the same = 2 will also show that bn verconges to the value 21 . In fact, bn also verconges to 100
(take = 102 to demonstrate this). Indeed, an also verconges to all of these values, so a sequence may
indeed verconge to multiple values.
Theorem: A sequence an verconges to a real number x if and only if an is bounded.
Proof: Suppose an verconges to x, so there exists > 0 such that for all N ∈ N, if n ≥ N then
|an − x| < . In particular, setting N = 1, we get that if n ≥ 1 then |an − x| < . That is, |an − x| < for all n ∈ N. In other words, for all n ∈ N, we have x − < an < x + , so the sequence (an ) is bounded.
Conversely, suppose (an ) is bounded, so there exists M ∈ R such that −M ≤ an ≤ M for all n ∈ N.
Then let x ∈ R. Now −M − x ≤ an − x ≤ M − x for all n ∈ N, so |an − x| ≤ max({|−M − x|, |−M + x|}),
for all n ∈ N.
Now let N ∈ N, and assume n ∈ N and n ≥ N . Then from above, |an −x| ≤ max({|−M −x|, |−M +x|}).
Hence, we have shown
(∀N ∈ N)(n ≥ N =⇒ |an − 0| < 1 + max({|−M − x|, |−M + x|}))
Setting = 1 + max({|−M − x|, |−M + x|}), we have shown that
(∃ > 0)(∀N ∈ N)(n ≥ N =⇒ |an − 0| < ),
as desired. 4. (2.2.4)
(a) The sequence (0, 1, 0, 1, 0, 1, 0, 1, . . .) has an infinite number of 1’s but does not converge to 1.
(b) This is impossible. Suppose an → a and (an ) has an infinite number of 1’s. That is, (an ) has an
infinite number of 1’s, and for every > 0, there exists N ∈ N such that if n ≥ N then |an − a| < .
Assume for contradiction that a 6= 1. Then |a − 1| > 0, so there exists N ∈ N such that if n ≥ N
then |an − a| < |a − 1|. But (an ) has an infinite number of terms equal to 1, and only finitely many
of its terms before the N ’th term. So (an ) must have a term equal to 1, that is found somewhere
later than its N ’th term. Hence there exists n ≥ N such that an = 1. But then n ≥ N and
|an − a| = |1 − a| = , which is a contradiction.
(c) Construct a sequence like (0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, . . .).
5. (2.2.6)
Assume an → a, and also an → b. Now assume for contradiction that a 6= b. Then |b − a| > 0.
> 0. Then there exists N1 ∈ N such that n ≥ N implies |an − a| <
Let = |b−a|
2
N2 ∈ N such that n ≥ N2 implies |an − b| < |b−a|
2 .
Now let n = max({N1 , N2 }), so n ≥ N1 and n ≥ N2 . Therefore |an −a| <
Finally, we get
|a − b| = |a − an + an − b| ≤ |a − an | + |an − b| = |an − a| + |an − b| <
so |a − b| < |a − b|, which is impossible. Hence we must have a = b.
6. (2.3.1) Assume xn ≥ 0 for all n ∈ N.
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|b−a|
2 ,
|b−a|
2 ,
and there exists
and also |an −b| <
|b−a|
2 .
|b − a| |b − a|
+
= |b − a| = |a − b|,
2
2
(a) Suppose xn → 0. Let > 0 be given. Then 2 > 0, so (since xn → x) there exists N ∈ N such that
if n ≥ N then |xn − 0| < 2 . Equivalently, xn < 2 for all n ≥ N . Now assume n ≥ N , so
√
√
√
| xn − 0| = xn < 2 = .
√
Hence xn → 0.
(b) Suppose xn → x. If x = 0 then we are done by part (a), so assume x 6= 0. Since xn → x and xn ≥ 0
for all n, it follows that x > 0. (It is not possible for x to be negative; if x < 0 and xn ≥ 0 for all n,
then |xn − x| ≥ |0 − x| = |x|, for all n, which would contradict the convergence of xn to x.) Thus
√
x > 0.
√
√
Let > 0 be given. Then x > 0, so there exists N ∈ N such that n ≥ N implies |xn − x| < x.
Now assume n ≥ N . Then
√
√
√
|xn − x|
|xn − x|
x
|xn − x|
√ ≤ √
| xn − x| = √
= √
< √ = .
| xn + x|
| x|
x
x
7. (2.3.2)
Assume xn → 2.
(a) Let > 0 be given. Then 32 > 0, so there exists N ∈ N such that if n ≥ N then |xn − 2| < 23 .
Now assume n ≥ N . Then
2xn − 4 2xn − 1
2
2
3
− 1 = = .
= 3 |xn − 2| < 3
3
3
2
(b) Let > 0. Then 2 > 0, so there exists N1 ∈ N such that n ≥ N1 implies |xn − 2| < 2.
Also, there exists N2 ∈ N such that n ≥ N2 implies |xn − 2| < 1. Thus for all n ∈ N, if n ≥ N2 then
1 < xn < 3. Choose N = max({N1 , N2 }). Now assume n ≥ N , so n ≥ N1 and n ≥ N2 . Then
1
1 2 − xn |2 − xn |
2
2
xn − 2 = 2xn = 2xn < 2xn < 2 = .
8. (a)
lim
1 + 2an
1 + 3an − 4a2n
=
lim(1) + lim(2an )
1 + 2 lim(an )
lim(1 + 2an )
=
=
2
2
lim(1 + 3an − 4an )
lim(1 + 3an − 4an )
1 + 3 lim(an ) − 4 lim(a2n )
=
1 + 2(0)
1
=
=1
1 + 3(0) − 4 lim(an ) lim(an )
1 − 4(0)2
(b)
lim
(an + 2)2 − 4
an
= lim
a2n + 4an + 4 − 4
an
= lim
(an )(an + 4)
an
= lim(an + 4) = lim(an ) + lim(4) = 0 + 4 = 4.
(c)
lim
2
an
1
an
+3
+5
!
= lim
2 + 3an
1 + 5an
=
lim(2 + 3an )
lim(2) + 3 lim(an )
2+0
=
=
= 2.
lim(1 + 5an )
lim(1) + 5 lim(an )
1+0
9. (2.3.7)
(a) Let xn = n and let yn = −n. Then (xn ) = (1, 2, 3, . . .) and (yn ) = (−1, −2, −3, . . .) both diverge,
but (xn + yn ) = (0)∞
n=1 = (0, 0, 0, ...) converges (to 0).
(b) Impossible. If (xn ) converges and (xn + yn ) converges, then (yn ) = (xn + yn ) − (xn ) must also
converge.
(c) Let bn =
1
n
for all n. Then bn 6= 0 for all n, and bn → 0, but ( b1n ) = (n) diverges.
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(d) Impossible. Suppose (bn ) is convergent and (an − bn ) is bounded. Then (bn ) must also be bounded,
by Theorem 2.3.2. Therefore (an ) = (an − bn ) + (bn ) must also be bounded, as the sum of bounded
sequences is bounded. (If (xn ) is bounded by M and (yn ) is bounded by N , then (xn + yn ) is
bounded by M + N .)
(e) Let an = 0 and bn = n, for all n.
10. (2.3.9)
(a) Suppose (an ) is bounded, so there exists M > 0 such that |an | ≤ M for all n. Assume also that
bn → 0.
> 0 so there exists N ∈ N such that if n ≥ N then |bn − 0| < M
. Now assume
Let > 0. Then M
n ≥ N . Then
|an bn − 0| = |an bn | = |an ||bn | ≤ M |bn | < M
= .
M
The Algebraic Limit Theorem doesn’t apply, since one of its assumptions is that an converges.
(b) Suppose that bn → b, and b 6= 0. Then (an bn ) converges if and only if (an ) converges.
Proof: Suppose (an ) converges. Then (an bn ) converges, by the Algebraic Limit Theorem (part iii).
On the other hand, suppose (an bn ) converges. Then (an ) = (an bn /bn ), which converges by the ALT
(part iv), since bn converges to b 6= 0.
(c) Assume an → a = 0, and bn → b. Then (bn ) is bounded, by Theorem 2.3.2. So by part (a) (with
the roles of an and bn transposed), the sequence (an bn ) is convergent and lim(an bn ) = 0 = 0b = ab,
as desired.
11. (2.3.12) Assume an → a.
(a) True. Suppose that an is an upper bound for B, for all n ∈ N. Show that a is an upper bound for
B.
Let b ∈ B. Then for all n ∈ N, an ≥ b. By Theorem 2.3.4 (iii), a ≥ b. But b was an arbitrary
element of B, so a is an upper bound for B.
(b) True. Suppose that an ∈ (−∞, 0] ∪ [1, ∞), for all n ∈ N. Assume for contradiction that a ∈ (0, 1).
Then let = min({|a − 0|, |a − 1|}) > 0. Then for all n ∈ N, either an ≤ 0 < a in which case
|an − a| ≤ |0 − a|, or else an ≥ 1 > a, in which case |an − a| ≥ |1 − a|. In either case, |an − a| ≥ ,
so for all n ∈ N, |an − a| ≥ . This contradicts the assumption that a = lim(an ).
(c) False. Let
decimal representa√ (an ) = (1.4, 1.41, 1.414, 1.4142, . . .) be the sequence of (truncated)
√
tions of 2. Thus an is a 1. followed by the first n decimal digits of 2. Each an may be written
as a fraction
√ of integers whose denominator is a power of 10, so each an is rational. However,
lim(an ) = 2 ∈
/ Q.
12. (2.4.2)
(a) The proof is invalid. The problem is that lim(yn ) does not exist. Hence, setting y = lim(yn ) makes
no sense.
(b) We define y1 = 1 and yn+1 = 3 − y1n for each n ∈ N. We propose to show that yn converges, by
showing that yn is monotone and bounded.
√
√
3− 5
≤ yn ≤ 3+2 5 , for all n ∈ N.
2
√
√
that 3−2 5 ≤ 1 ≤ 3+2 5 . The right-hand
√
√
First let us show that (yn ) is bounded. In fact, let us show that
Proof (by induction). First, y1 = 1, so we need to show
√
3+ 5
inequality holds, since 1 = 1+1
2 ≤
2 . Next, we know that
√
5
>
2
>
1
>
as desired.
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3−2
√
3− 5
√
3− 5
,
2
5>
4 = 2 > 1 = 3 − 2, so
Now assume
√
3− 5
2
≤ yn ≤
√
3+ 5
2 .
Then
2
1
2
√ ≥
√
≥
yn
3− 5
3+ 5
√ !
√ !
2
3+ 5
1
2
3− 5
√
√
√
√
≥
≥
yn
3− 5
3+ 5
3+ 5
3− 5
√
√
2(3 + 5)
1
2(3 − 5)
≥
≥
9−5
yn
9−5
√
√
1
3− 5
3+ 5
≥
≥
2
yn
2
√ !
√ !
3+ 5
1
3− 5
−
≤−
≤−
2
yn
2
√ !
√ !
6
3+ 5
1
6
3− 5
−
≤3−
≤ −
2
2
yn
2
2
Therefore:
√
√
3− 5
3+ 5
≤ yn+1 ≤
,
2
2
as desired.
√
Note that this lower bound also shows that yn > 0 for all n (since 3 > 5).
Next, we wish to show that (yn ) is increasing; in other words, we wish to show that yn+1 − yn ≥ 0
for all n. Let n ∈ N. Then
1
1
yn+1 − yn = 3 −
− yn =
(3yn − 1 − yn2 )
yn
yn
Since yn > 0, we conclude that yn+1 −yn > 0 if and only if 3yn −1−yn2 > 0. Let f (x) = −x2 +3x−1,
so we wish to show that f (yn ) > 0 for all n.
The graph of f is a concave-down parabola, so f (x) is positive exactly when x is between the roots
of f . The quadratic formula shows that the roots of f are precisely the lower and upper bounds we
found before, so f (yn ) > 0 for all n, as desired.
Thus (yn ) is increasing and bounded. Therefore (yn ) is convergent, by the Monotone Convergence
Theorem, so lim(yn ) exists. Set y = lim(yn ) = lim(yn+1 ). Then by the Algebraic Limit Theorem,
y =3−
√
1
y
√
Therefore either y = 3+2 5 , or y = 3−2 5 . But we know yn is increasing, so y ≥ yn for all n. In
√
√
particular, y ≥ y1 = 1 > 3−2 5 . Therefore y = 3+2 5 . 13. (2.4.5)
Let x1 = 2, and for all n ∈ N define
xn+1 =
1
2
2
xn +
xn
(a) First show that x2n ≥ 2, for all n ∈ N:
(By induction.) For the base case n = 1: we said x1 = 2, so x21 = 4 ≥ 2.
Now assume x2n ≥ 2. Then
(xn+1 )2 =
2 1 2
1
4
1 2
x2n + 4 + 2 ≥
xn + 4 ≥
2 + 4 = 2.
2
xn
4
4
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So x2n ≥ 2 for all n ∈ N.
Next
xn − xn+1 = xn −
1
1
xn +
2
xn
=
1
1
1
xn −
=
2
xn
2
xn −
2
xn
.
We wish to show that xn − xn+1 ≥ 0; it is equivalent to show that xn − x2n ≥ 0 for all n. If we knew
that xn > 0 for all n, then it would suffice to show that x2n − 2 ≥ 0 for all n. But we have already
proven that inequality! So we need only show that xn > 0 for all n, and then we can conclude that
(xn ) is a decreasing sequence.
Show that xn > 0 for all n, by induction: First, x1 = 2 > 0. Next, assume xn > 0. Then x2n > 0,
and indeed 21 (xn + x2n ) > 0, so xn+1 > 0.
Thus (xn ) is decreasing, and bounded below by 0. Therefore (xn ) is a convergent sequence. Let
x = lim(xn ) = lim(xn+1 ). Then by the Algebraic Limit Theorem,
1
2
x=
x+
,
2
x
so x = x2 and therefore x2 = 2. Since xn ≥ 0 for all n, we conclude that x ≥ 0 as well and therefore
√
x = 2.
(b) Let x1 = 2, and xn+1 = 12 xn + xcn .
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