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Math 320: Real Analysis
MWF 1pm, Campion Hall 302
Homework 3 Solutions
Please write neatly, and in complete sentences when possible.
Do the following problems from the book: 2.2.1, 2.2.2, 2.2.8, 2.3.2,
2.3.3, 2.3.8, 2.3.11
Solution 2.2.1. (a) Given > 0, let N be any integer larger than
s 1 1
−1 .
6 In this case, for any integer n ≥ N , we have
1 1
2
−1
n >
6 so that
1
6n2 + 1 > .
Equivalently,
1
1
=
−
0
6n2 + 1
6n2 + 1 < .
This implies that lim 1/(6n2 + 1) = 0, as desired.
(b) Given > 0, let N be any integer larger than
13 − 10
.
4
In this case, for any integer n ≥ N , we have
4n > 13 − 10,
in which case > 13/(4n + 10). For such n, we have
3 3n + 1 3(n + 25 ) 3n + 1 −
2 2n + 5 = 2n + 5 − 2n + 5 13 2 13 < ,
=
=
2n + 5 4n + 10 3n+1
so that we may conclude that lim 2n+5
= 3/2.
1
2
(c) Given > 0, let N be any integer larger than
for any integer n ≥ N , we have
4
n + 3 > 2,
√
so that n + 3 > 2/. For such n, we have
2
= √ 2
√
−
0
< ,
n+3
n+3
4
2
− 3. In this case,
2
so that we may conclude lim √n+3
= 0.
Solution 2.2.2. Given a sequence of real numbers (an ) and a real
number x, we claim that (an ) verconges to x if and only if (an ) is
bounded:
Suppose that (an ) is bounded, so that there exists an M with |an | < M
for all n. In this case, note that
|an − x| ≤ |an | + |x| < M + |x|
for all n. Let = M + |x|. Then it is true that for all N , if n ≥ N (in
fact, for any n) we have |an −x| < M +|x| = . Thus (an ) verconges to x.
On the other hand, if (an ) verconges to x, then there is an so that
(taking N = 1), for every n ∈ N we have |an − x| < . On the other
hand, we observe that
|an | = |an − x + x| ≤ |an − x| + |x|,
so that |an | < + |x| for all n ∈ N, and (an ) is bounded.
Finally, any divergent bounded sequence of real numbers will suffice
to see a vercongent divergent sequence. For example, an = (−1)n diverges, but since (an ) is bounded, it verconges to any real number x.
Solution 2.2.8. (a) The sequence ((−1)n ) is frequently in the set
{1}: For every N ∈ N, there is an even number n ≥ N so that
(−1)n = 1 ∈ {1}. On the other hand, ((−1)n ) is not eventually in
{1}: For every N ∈ N there is an odd number n ≥ N so that
(−1)n = −1 ∈
/ {1}.
(b) If (an ) is a sequence that is eventually in A, then (an ) is frequently
in A: By assumption, there is an N0 ∈ N so that for every n ≥ N0 ,
we have an ∈ A. This means that, given any N ∈ N, we may choose
3
n = max{N, N0 }. This n satisfies n ≥ N , and we have an ∈ A, since
n ≥ N0 .
On the other hand (an ) may be frequently in A, but not eventually in
A: The example in part (a) is one such example.
(c) Using the definition of ‘eventually’, we rephrase Definition 2.2.3B:
Definition. A sequence (an ) converges to a if, given any -neighborhood
V (a) of a, the sequence (an ) is eventually in V (a).
(d) If (xn ) is a sequence of real numbers so that infinitely many of
them are equal to 2, then (xn ) is frequently in the set (1.9, 2.1): For
every N ∈ N, there must exist n ≥ N so that xn = 2, otherwise the
number of terms xn equal to 2 would be finite. On the other hand, it
is not necessary that (xn ) is eventually in (1.9, 2.1). For example, if
xn = 1 + (−1)n , then the sequence (xn ) is frequently in the set {−2},
which is disjoint from (1.9, 2.1). It is thus not eventually in (1.9, 2.1),
as in part (a).
Solution 2.3.2. (a) Given > 0, since (xn ) → 0 there ispan N ∈ N so
√
that for every n ≥ N we have |xn | < 2 . Thus | xn | = |xn | < for
every n ≥ N , and we conclude that (xn ) → 0.
(b) By Theorem 2.3.4 part (i), xn ≥ 0 implies that x ≥ 0. We may
also assume that x 6= 0, since we’ve already proven the case x = 0 in
part (a). Thus we proceed with the assumption that x > 0.
Note that we have for any non-negative numbers a, b ≥ 0,
√ √
√
√
a − b = ( a + b)( a − b).
√
√
Thus, as long as a + b 6= 0, we have
√
√
a−b
√ .
a− b= √
a+ b
√
√
√
Since xn + x ≥ x > 0, we conclude that
xn − x √
√
|x − x|
1
√ = √ n √ ≤ √ |xn − x|
| xn − x| = √
xn + x
xn + x
x
√
for all n ∈ N. Given > 0, since x > 0 we have x > 0. As (xn ) →
√x,
there is an N ∈ N so that for every n ≥ N we have |xn − x| < x.
4
Thus, for any n ≥ N , we have
√
√
√
1
1
| xn − x| ≤ √ |xn − x| < √ · x = ,
x
x
√
√ and we conclude that
xn → x.
Solution 2.3.3. Since lim xn = lim zn = l, given > 0 there are
integers N1 and N2 so that, for any n satisfying n ≥ N1 and n ≥ N2
we have |xn − l| < and |zn − l| < . Let N = max{N1 , N2 }, so that
for any n ≥ N we have xn − x > − and zn − l < . The inequality
xn ≤ yn ≤ zn implies that xn − l ≤ yn − l ≤ zn − l, so that
− < xn − l ≤ yn − l ≤ zn − l < for n ≥ N . Equivalently,
|yn − l| < for n ≥ N . We conclude that lim yn = l.
Solution 2.3.8. (a) Let xn = (−1)n and yn = −(−1)n . In this case
both sequences (xn ) and (yn ) diverge. On the other hand, the sum
xn + yn = 0 for all n, so that (xn + yn ) converges to 0.
(b) By Theorem 2.3.3 part (i), as (xn ) converges, we know that (−xn )
converges. By Theorem 2.3.3 part (ii), as (xn + yn ) and (−xn ) each
converge, this implies that (xn + yn + (−xn )) = (yn ) converges, a contradiction.
(c) Let bn = 1/n, so that bn 6= 0 for all n ∈ N, and (bn ) is a convergent
sequence. On the other hand, 1/bn = n, so that (1/bn ) diverges.
(d) By Theorem 2.3.2, the convergent sequence (bn ) is bounded, so
that there exists a real number M1 so that |bn | < M1 for all n ∈ N. If
(an − bn ) is bounded, there exists a real number M2 so that |an − bn | <
M2 . Using the triangle inequality, this implies that
|an | = |an − bn + bn | ≤ |an − bn | + |bn | < M1 + M2
so that the sequence (an ) is bounded, a contradiction.
(e) Let (bn ) be any divergent sequence, and let an = 0 for all n ∈ N. In
this case, both sequences (an ) and (an bn ) are the sequence all of whose
terms are 0. Thus both are convergent, though (bn ) is divergent.
5
Solution 2.3.11. Since (xn ) is a convergent sequence, there is a real
number x so that, given any > 0, there is an N1 ∈ N such that, for
all n ≥ N1 we have |x − xn | < /2. Moreover, by Theorem 2.3.2 the
sequence (xn ) is bounded, so that there exists a real number M such
that |xn | < M for all n ∈ N. Note that by Theorem 2.3.4 part (iii) we
know as well that |x| < M . For any n ≥ N1 we have:
x1 + . . . + xn
(x1 − x) + . . . + (xn − x) − x = n
n
(x1 − x) + . . . + (xN1 − x) (xN1 +1 − x) + . . . + (xn − x) =
+
n
n
x1 + . . . + xN1 − N1 x (xN1 +1 − x) + . . . + (xn − x) +
≤ n
n
|x1 | + . . . + |xN1 | + N1 |x| |xN1 +1 − x| + . . . + |xn − x|
≤
+
n
n
M + . . . + M + N1 M
/2 + . . . + /2
<
+
n
n
(n − N1 ) 2N1 M
+
=
n
n
2
2N1 M
<
+ .
n
2
Let N2 ∈ N be any integer larger than 4N1 M/, so that for n ≥ N2 we
will have 2N1 M/n < /2.
Let N = max{N1 , N2 }. By the above calculations, if n ≥ N , then
n ≥ N1 and we have
x1 + . . . + xn
2N1 M
<
−
x
+ .
n
n
2
Moreover, n ≥ N2 , so that we have
x1 + . . . + x n
2N1 M
<
−
x
+ < + = .
n
n
2
2 2
We conclude that ((x1 + . . . + xn )/n) converges to x, as desired.
Finally, consider the sequence (xn ) given by xn = (−1)n . This sequence
is divergent (we showed in class that a very similar sequence doesn’t
converge, and that same proof will work here. Alternatively, we will
be developing a simple test soon to check divergence). On the other
hand,
x1 + . . . + xn
−1 + 1 . . . + (−1)n
=
.
n
n
These terms are given by yn = 0 for n even, and yn = 1/n for n odd.
This sequence converges to 0.
yn =