MATH 104 HW #3
James Ni
Problem 1. Ross 10.6.
(a) Let (s
n
) be a sequence such that
|s
n+1
− s
n
| < 2
−n
for all n ∈ N.
Prove (s
n
) is a Cauchy sequence and hence a convergent sequence.
(b) Is the result in (a) true if we only assume |s
n+1
− s
n
| <
1
n
for all n ∈ N?
We first prove a lemma which will assist us in proving both (a) and (b).
Lemma 1.1. Let (s
n
) be a sequence and (a
n
) be a strictly positive sequence
such that |s
n+1
− s
n
| < a
n
. Define a sequence (A
n
) such that A
n
=
P
n
i=0
a
n
. If
(A
n
) is convergent then (s
n
) is Cauchy.
Proof. We claim that |s
n+k
− s
n
| < A
n+k−1
− A
n−1
. We can prove this using
induction on k. The base case trivially holds from our assumption. Now assume
that |s
n+k
− s
n
| < A
n+k−1
− A
n−1
. Then, we have
|s
n+k+1
− s
n
| = |(s
n+k+1
− s
n+k
) + (s
n+k
− s
n
)|
≤ |s
n+k+1
− s
n+k
| + |s
n+k
− s
n
|
< a
n+k
+ A
n+k−1
− A
n−1
= A
n+k
− A
n−1
.
Because (A
n
) converges, let A = lim A
n
. Because (a
n
) is strictly positive,
we know that (A
n
) is strictly increasing. This implies that (A
n
) is bounded
from above by A. Without loss of generality, suppose that for two m, n ∈ N, we
have m > n. By extension, we have |s
m
− s
n
| < A
m−1
− A
n−1
< A − A
n−1
.
For ∀ε > 0, let N be the smallest integer such that A
N−1
≥ A − ε. Observe
that because A−A
n
is strictly decreasing and lim A−A
n
= 0, such an N always
exists for positive ε. Then we have ∀m, n > N, |s
m
− s
n
| < ε. Thus, (s
n
) is
Cauchy.
Proof. Endowed with this lemma, proving (a) and (b) becomes trivial, as in the
case a
n
= 2
−n
, we have lim
P
n
i=0
2
−i
= 2 which implies that (s
n
) is a Cauchy
sequence, and hence, is also convergent. However, in the case a
n
=
1
n
, A
n
diverges which implies that (s
n
) is not necessarily Cauchy.
Problem 2. Ross 11.2. Consider the sequences defined as follows:
a
n
= (−1)
n
, b
n
=
1
n
, c
n
= n
2
, d
n
=
6n + 4
7n − 3
.
1
(a) For each sequence, give an example of a monotone subsequence.
(b) For each sequence, give its set of subsequential limits.
(c) For each sequence, give its lim sup and lim inf.
(d) Which of the sequences converges? diverges to +∞? diverges to −∞?
(e) Which of the sequences is bounded?
Proof. Consider the subsequence a
k
n
= a
2n
= (−1)
2n
= 1. Because a
k
n
is
constant, it is monotone, and lim a
k
n
= 1. Also consider the subsequence a
k
n
=
a
2n+1
= (−1) · (−1)
2n
= −1. Because a
k
n
is constant, lim a
k
n
= −1.
We claim the set of subsequential limits S of a
n
is {−1, 1}. Suppose there
exists another subsequential limit a. It is easy to see that −1 ≤ a
n
≤ 1.
Because a
n
is bounded, it is impossible for a = ±∞. We can then conclude that
there exists a increasing integer sequence (k
n
) such that ∀ε > 0∃N > 0∀n >
N, |a
k
n
− a| < ε. Becuase {a
n
} = {−1, 1}, we can split a
k
n
into 3 cases.
If a
k
n
strictly consists of 1, then a = 1; if a
k
n
strictly consists of −1, then
a = −1. If a
k
n
consists of ±1, then a + ε > 1 and a − ε > −1. This implies that
ε > 1, which implies that convergence fails for ε ≤ 1. Thus, such an a cannot
exist, so therefore, S = {−1, 1}.
We can then conclude lim sup a
n
= sup S = 1, lim inf a
n
= inf S = −1.
Because lim sup a
n
6= lim inf a
n
, lim a
n
is divergent.
Proof. Because b
n+1
=
1
n+1
< b
n
=
1
n
, b
n
is monotone decreasing. Hence, any
subsequence of b
n
is also monotone decreasing. It is easy to see that lim b
n
=
0. Thus, its set of subsequential limits is simply S = {0} and lim sup b
n
=
lim inf b
n
= 0. Because b
n
is monotone decreasing and converges to 0, it is
bounded from below by 0 and bounded from above by b
1
= 1.
Proof. Because c
n+1
= (n + 1)
2
> c
n
= n
2
, c
n
is monotone increasing. Hence,
any subsequence of c
n
is also monotone increasing. It is easy to see that lim c
n
=
+∞. Thus, its set of subsequential limits is simply S = {+∞} and lim sup c
n
=
lim inf c
n
= +∞. Because c
n
diverges to +∞, it is not bounded.
Proof. We first show that d
n
is monotone decreasing.
d
n+1
≤ d
n
⇔
6(n + 1) + 4
7(n + 1) − 3
=
6n + 10
7n + 4
≤
6n + 4
7n − 3
⇔ (6n + 10)(7n − 3) ≤ (6n + 4)(7n + 4) (n ≥ 1)
⇔ 42n
2
+ 52n − 30 ≤ 42n
2
+ 52n + 16
⇔ −30 ≤ 16 ⇔ n ≥ 1.
Because of this, any subsequence of cd
n
is also monotone decreasing. We now
evaluate lim d
n
.
lim d
n
= lim
6n + 4
7n − 3
= lim
6 + 4/n
7 − 3/n
=
lim 6 + 4/n
lim 7 − 3/n
=
6
7
.
2
Because d
n
converges, its set of subsequential limits is simply S = {
6
7
} and
lim sup d
n
= lim inf d
n
=
6
7
. Because d
n
is monotone decreasing and converges
to
6
7
, it is bounded from below by
6
7
and bounded from above by d
1
=
5
2
.
Problem 3. Ross 11.3. Repeat Problem 2 for the sequences:
s
n
= cos(
nπ
3
), t
n
=
3
4n + 1
, u
n
= (−
1
2
)
n
, v
n
= (−1)
n
+
1
n
.
Proof. Observe that we can utilize the periodicity of cos to rewrite s
n
as
s
n
=
1 n = 6k
1/2 n = 6k + 1 or n = 6k + 5
−1/2 n = 6k + 2 or n = 6k + 4
−1 n = 6k + 3
,
where k is an integer. Thus, a simple example of a monotone subsequence is
s
k
n
= s
6n
= 0. We claim the set of subsequential limits S = {−1, −1/2, 1/2, 1}.
The proof of this claim is a simple proof of variations of a
n
of Problem 2 via
the additive and multiplicative properties of convergent sequences. We quickly
find that lim sup s
n
= sup S = 1, lim inf s
n
= inf S = −1. Because lim sup s
n
6=
lim inf s
n
, lim s
n
is divergent. However, −1 ≤ s
n
≤ 1 which implies s
n
is
bounded.
Proof. Because t
n+1
=
3
4(n+1)+1
=
3
4n+5
< t
n
=
3
4n+5
, t
n
is monotone decreas-
ing. Hence, any subsequence of t
n
is also monotone decreasing. Evaluating
lim t
n
, we find
lim t
n
= lim
3
4n + 1
= lim
3/n
4 + 1/n
=
lim 3/n
4 + 1/n
= 0.
Thus, its set of subsequential limits is simply S = {0} and lim sup t
n
= lim inf t
n
=
0. Because t
n
is monotone decreasing and converges to 0, it is bounded from
below by 0 and bounded from above by t
0
= 3.
Proof. Consider the subsequence u
k
n
= u
2n
= (
1
4
)
n
= 4
−n
. Clearly, 4
−(n+1)
=
4
−n
4
< 4
−n
, so this subsequence is monotone decreasing. We demonstrate that
lim u
n
= 0. First, observe that |u
n
| is monotone decreasing since |u
n+1
| =
(
1
2
)
n+1
< |u
n
| = (
1
2
)
n
. Thus, for any ε > 0, let N = max(0, log
1/2
(ε)). Then
we have for any n > N , |u
n
| < |u
N
| < ε. Thus, u
n
converges to 0, its set of
subsequential limits is simply S = {0}, and lim sup u
n
= lim inf u
n
= 0. Because
|u
n
| is monotone decreasing, it is bounded from above by the first positive term
u
0
= 1 and bounded from below by the first negative term u
1
= −1/2.
Proof. Consider the subsequence v
k
n
= v
2n
= 1 +
1
n
. Clearly, 1 +
1
n+1
< 1 +
1
n
,
so this subsequence is monotone decreasing.
We claim the set of subsequential limits S = {−1, 1}. Suppose there exists
another subsequential limit v. We first check that v
n
is bounded by observing
3
that it is the sum of two bounded sequences. Because v
n
is bounded, it is
impossible for v = ±∞.
We can then conclude that there exists a increasing integer sequence (k
n
)
such that ∀ε > 0∃N > 0∀n > N, |v
k
n
− v| < ε. This implies that v − ε −
1
k
n
<
(−1)
k
n
< v + ε −
1
k
n
. Because (−1)
k
n
can only take values in {−1, 1}, we have
v − ε −
1
k
n
< −1, v + ε −
1
k
n
> 1. This implies that ε > 1, which implies
that convergence fails for ε ≤ 1. Thus, such an v cannot exist, so therefore,
S = {−1, 1}.
We can then conclude lim sup v
n
= sup S = 1, lim inf v
n
= inf S = −1.
Because lim sup v
n
6= lim inf v
n
, lim v
n
is divergent.
Problem 4. Ross 11.5. Let (q
n
) be an enumeration of all the rationals in the
interval (0, 1].
(a) Give the set of subsequential limits for (q
n
).
(b) Give the values of lim sup q
n
and lim inf q
n
.
Proof. We claim that the set of subsequential limits for (q
n
) is [0, 1]. We first
prove that if q ∈ [0, 1], then q ∈ S. In order to prove this, we define an inductive
algorithm which defines a subsequence a
n
= q
k
n
which converges to q.
For our base case, we let p
0
be any arbitrary rational in {q
n
} with finite
index i
p
such that p
0
= q
i
p
6= q and k
0
= i
p
.
Now suppose that we have defined k
n
up to k
j
. Let p = q
k
j
and i
p
= k
j
.
Here, it may seem natural to utilize the denseness of Q to construct a new
member of the subsequence. However, this condition alone is not sufficient to
create a subsequence which converges. Instead, we let r = (p + q)/2 with finite
index i
r
such that if q < p, we have q < r < p and if p < q, we have p < r < q.
It is easy to show that r is rational as
r =
p + q
2
=
c
1
/d
1
+ c
2
/d
2
2
=
c
1
d
2
+ c
2
d
1
2d
1
d
2
.
If i
r
> i
p
, then k
j+1
= i
r
, and the inductive step advances forward to k
j+1
.
If i
r
< i
p
, then we set r
0
= (r + q)/2 with finite index i
0
r
. We then repeat the
current inductive step for r
0
instead of r. Because there are only a finite number
of indicies smaller than i
p
, we must eventually find an r
0
such that i
0
r
> i
p
, which
completes the inductive algorithm.
If p
0
< q, our algorithm gives us a monotone increasing sequence with q >
a
n
> p
0
, and if p
0
> q, our algorithm gives us a monotone decreasing sequence
with q < a
n
< p
0
. This implies that in both cases, our sequences are convergent.
An essential observation to make is that because our inductive step may repeat,
the next term in the sequence may be a successive average of the previous
term with q. This implies that if p
0
< q, we have the additional condition that
a
n+1
>
a
n
+q
2
, and if p
0
> q, we have the additional condition that a
n+1
<
a
n
+q
2
.
In the first case, our additional condition implies a
n+1
> q − (a
n+1
− a
n
).
For any ε > 0, let a
n
− a
n−1
< ε. Because (a
n
) is monotone increasing and
convergent, we know that (a
n
) is also Cauchy. This implies that it is always
possible to find such an n which satisfies the ε condition. This implies ∀ε >
4
0∃n > 0 such that a
n
> q − ε, which implies that q is a least upper bound. This
implies that lim q
k
n
= lim a
n
= q.
The proof for the second case follows similarly, utilizing the additional con-
dition and Cauchy condition to show that q is the greatest lower bound which
implies lim q
k
n
= lim a
n
= q.
Now, we show that if q /∈ [0, 1], then q /∈ S. Suppose that q ∈ S. This
implies that there exists a positive, increasing sequence of integers (k
n
) such
that ∀ε > 0∃N > 0∀n > N, |q
k
n
− q| < ε. We can split the proof into two cases:
1) q < 0 and 2) q > 1.
If q < 0, then using the fact that q
k
n
− q < ε and q
k
n
> 0 we have −q < ε
which implies that our condition fails for ε < −q, which is a contradiction. If
q > 1, then using the fact that −ε < q
k
n
− q and q
k
n
≤ 1 we have −ε < 1 − q
which implies that our condition fails for ε < q − 1, which is a contradiction.
Thus, q /∈ S.
Because S = [0, 1], we have lim sup q
n
= 1 and lim inf q
n
= 0.
Discussion 1. What exactly is lim sup?
For some sequence (a
n
), lim sup a
n
is formally defined as lim
N→∞
sup{a
n
:
n > N }. However, the meaning of lim sup is lost in the formal definition.
Instead, it is easier to think of lim sup and lim inf as the upper and lower bounds
of a sequence for sufficiently large n. In other words, it characterizes the long-
run behavior of a sequence by giving us a range of possible values of a
n
.
One important note is that lim sup a
n
6= sup{a
n
} and lim inf a
n
6= inf{a
n
}.
Consider the sequence (a
n
) = 1, −1, 0, 0, .... Clearly, lim sup a
n
= lim inf a
n
= 0
because the only possible value of a
n
for sufficiently large n (in this case, simply
n ≥ 2) is 0. However, sup{a
n
} = 1 and inf{a
n
} = −1.
Instead, it is more appropriate to think of lim sup a
n
as inf sup a
n
, or the
smallest supremum of the successive elements of the sequence, and lim inf a
n
as sup inf a
n
, or the largest infimum of the sucessive elements of the sequence.
Formally, this is written as lim sup a
n
= inf{sup{a
n
: n > N } : N ≥ 0} and
lim inf a
n
= sup{inf{a
n
: n > N} : N ≥ 0}.
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