Download Math 713 - hw 2.2 Solutions 2.16a Prove Proposition 2.6 on page 45

Math 713 - hw 2.2 Solutions
2.16a, 2.28a, 2.33, 2.36
2.16a Prove Proposition 2.6 on page 45.
Any monotone sequence of real numbers converges in R∗ . In fact, we have the following:
a) If {xn }∞
n=1 is nondecreasing, then
lim xn = sup{xn : n ∈ N}.
n→∞
In particular, the limit exists and is finite if {xn }∞
n=1 is bounded above and is ∞ otherwise.
Let {xn }∞
n=1 be a nondecreasing sequence of real numbers. First suppose the sequence is bounded above.
Then by the completness axiom x = sup{xn : n ∈ N} exists and is finite. We now show that x = limn→∞ xn .
Given ε > 0, since x − ε is not an upper bound for the set {xn : n ∈ N} (x is the least upper bound) there
is N ∈ N such that x − ε < xN ≤ x. Since {xn }∞
n=1 is nondecreasing we have x − ε < xN ≤ xn ≤ x for
all n ≥ N . It follows that |x − xn | < ε for all n ≥ N . Hence, x = limn→∞ xn ; moreover, the limit exists
and is finite. Suppose that {xn : n ∈ N} is not bounded above, then sup{xn : n ∈ N} = ∞. We now show
that limn→∞ xn = ∞. Let M ∈ R be given; then since {xn : n ∈ N} is not bounded above, M is not an
upper bound and therefore there is N ∈ N such that xN > M . Since {xn }∞
n=1 is nondecreasing we have
xn ≥ xN > M for all n ≥ N . Hence, limn→∞ xn = ∞. In either case, limn→∞ xn = sup{xn : n ∈ N}.
b) If {xn }∞
n=1 is nonincreasing, then
lim xn = inf{xn : n ∈ N}.
n→∞
In particular, the limit exists and is finite if {xn }∞
n=1 is bounded below and is −∞ otherwise.
Similar to the above.
∞
2.28a Let {xn }∞
n=1 and {xn }n=1 and assume that limn→∞ exists and is finite. Prove that
lim sup(xn + yn ) = lim sup xn + lim yn .
Let x = lim sup xn and y = lim yn . We first assume that x is a real number and use Proposition 2.8(a) to verify
that lim sup(xn + yn ) = x + y. Let ε > 0 be given. Since x = lim sup xn there is an N1 ∈ N such that xn ≤ x + ε/2
for all n ≥ N1 . Since yn → y, there is N2 ∈ N such that y − ε/2 < yn < y + ε/2 for all n ≥ N2 . Then for all
n ≥ N = max{N1 , N2 } both conditions hold and we have xn + yn ≤ x + y + ε. Now, let n ∈ N be given, then
since x = lim sup xn there is m ≥ max{n, N2 } such that xm > x − ε/2 and since m ≥ N2 , ym > y − ε/2. Hence,
xm + ym > x + y − ε for some m ≥ n. Hence, lim sup(xn + yn ) = x + y.
Now suppose that lim sup xn = ∞, then we use Proposition 2.8(b) to show that
lim sup(xn + yn ) = lim sup xn + lim yn = ∞ + y = ∞.
Since yn → y there is a K ∈ N such that yn > y − 1 for all n ≥ K. Now let M ∈ R and N ∈ N be given. Since
lim sup xn = ∞, there is n ≥ max{N, K} such that xn > M − y + 1. Since n ≥ K we have yn > y − 1. Hence,
xn + yn > M − y + 1 + y − 1 = M.
Hence, lim sup(xn + yn ) = ∞ as required.
Note that by Proposition 2.8(c) we have lim sup xn = −∞ iff limn→∞ xn = −∞. Now suppose that lim sup xn =
−∞ or equivalently that limn→∞ xn = −∞. Then to show that
lim sup(xn + yn ) = lim sup xn + lim yn = −∞ + y = −∞
it suffices by Proposition 2.8(c) to show that limn→∞ xn + yn = −∞. Now let M ∈ R. Since yn → y there is a
N1 ∈ N such that yn < y + 1 for all n ≥ N1 . Since xn → −∞, there is a N2 ∈ N such that xn < M − y − 1 for all
n ≥ N1 . Then for all n ≥ N = max{N1 , N2 } we have
xn + yn < M − y − 1 + y + 1 = M.
Hence, limn→∞ xn + yn = −∞.
1
2
2.33 Prove that an extended real number is a cluster point of a sequence if and only if the sequence has a subsequence
converging to that number. Conclude that the limit superior of a sequence is the limit of a subsequence of the
sequence and likewise for the limit inferior.
∗
Let {xn }∞
n=1 be a sequence of real numbers and let x ∈ R be an extended real number.
Suppose that x is a cluster point of the sequence. Then either x is finite or x = ±∞. Suppose first that x is
finite and construct a subsequence {xnk }∞
k=1 inductively such that |x − xnk | < 1/k as follows. Since x is a cluster
point of the sequence, there is n1 ∈ N such that |x − xn1 | < 1. Now suppose that there are n1 < · · · < nk in N such
1
.
that |x − xnj | < 1j for j = 1, . . . , k. Then since x is a cluster point, there is nk+1 > nk such that |x − xnk+1 | < k+1
∞
This is the desired subsequence, since xnk → x. Next suppose that x = ∞. We construct a subsequence {xnk }k=1
inductively such that xnk > k as follows. Since x is a cluster point of the sequence, there is n1 ∈ N such that
xn1 > 1. Now suppose that there are n1 < · · · < nk in N such that xnj > j for j = 1, . . . , k. Then since x is a
cluster point, there is nk+1 > nk such that xnk+1 > k + 1. This is the desired subsequence, since xnk → ∞. The
case x = −∞ is treated similarly.
Now suppose that there is a subsequence {xnk }∞
k=1 which converges to x. We show that x is a cluster point
of the sequence {xn }∞
n=1 . Suppose first that x is finite. Let ε > 0 and N ∈ N be given. Then since xnk → x as
k → ∞, there is K ∈ N such that |x − xnk | < ε for all k ≥ K. There is k ≥ K such that nk ≥ N , hence, there
is n ≥ N such that |x − xn | < ε (namely, n = nk ). Thus, x is a cluster point of the sequence. Next suppose that
x = ∞. Let M ∈ R and N ∈ N be given. Then since xnk → ∞ as k → ∞, there is K ∈ N such that xnk > M for
all k ≥ K. There is k ≥ K such that nk ≥ N , hence, there is n ≥ N such that xn > M (namely, n = nk ). Thus,
∞ is a cluster point of the sequence. The case x = −∞ is treated similarly.
It follows that an extended real number is a cluster point of a sequence if and only if the sequence has a subsequence converging to that number. Since the limit superior of a sequence is a cluster point of the sequence, it
follows that it is the limit of a subsequence of the sequence and likewise for the limit inferior.
∞
2.36 In this exercise, we will discuss infinite series. Let {xn }∞
n=1 be a sequence of real numbers. The sequence {sn }n=1
defined by
n
X
sn =
xk , n ∈ N,
k=1
is called the sequence of partial sums of {xn }∞
{sn }∞
n=1 . If the sequence P
n=1 converges to a real number, say, s, then
∞
we say that {xn }n=1 is summable to s or that the infinite series ∞
n=1 xn converges to s, and we write
s=
∞
X
xn .
n=1
∞
We also say that s is the sum of the infinite series. If the sequence {s
Pn∞}n=1 does not converge to a real number,
∞
then
n }n=1 is not summable or that the infinite series
n=1 xn diverges. For brevity we often write
P we say that {x
P∞
xn in place of n=1 xn .
P
a) Prove that if xn ≥ 0 for each n ∈ N, then either limn→∞ sn = ∞ or
xn converges.
Pn
∞
Suppose xn ≥ 0 for each n ∈ N. Then the sequence {sn }n=1 given by sn = k=1 xk is a nondecreasing
∞
sequence. Hence, by Proposition 2.6 either the sequence converges
P to ∞ if {sn }n=1 is not bounded above or
it is bounded above and thus has a finite limit, in which case
xn converges.
P
b) Show that if
xn converges, then limn→∞ xn = 0
P
∞
Suppose
xn converges. Then the sequences {sn }∞
n=1 and {sn+1 }n=1 share the same finite limit s. Hence,
lim xn+1 = lim sn+1 − sn = lim sn+1 − lim sn = s − s = 0.
n→∞
n→∞
n→∞
n→∞
It follows that limn→∞ xn = 0.
P
P∞
c) Show that if
xn converges, then limn→∞ k=n xk = 0.
P∞
P∞
P∞
Suppose that n=1 xn converges, say,Ps = n=1 xn . Then k=n xk also converges and if n ≥ 2 we have
P
∞
∞
k=n xk = s − sn−1 . Hence, limn→∞
k=n xk = limn→∞ s − sn−1 = 0.
3
d) Prove that if
P
|xn | converges, then so does
P
xn . Hint: Use the Cauchy criterion.
Pn
Suppose that
|xn | converges, then the sequence of partial sums {tn }∞
n=1 given by tn =
k=1
P|xk | converges
and therefore is a Cauchy sequence (by the Cauchy Criterion: Theorem 2.1).PTo show that
xn converges
n
it suffices to show that its sequence of partial sums {sn }∞
given
by
s
=
x
is
a
Cauchy
sequence.
n
n=1
k=1 k
Let ε > 0 be given. Then since {tn }∞
is
Cauchy,
there
is
N
∈
N
such
that
|t
−
t
|
<
ε
for
all
m,
n ≥ N.
m
n
n=1
Now let m, n ≥ N , we show that |sm − sn | < ε. If m = n there is nothing to prove; by m and n if necessary
we may assume that n < m. By the triangle inequality we have
m
n
m
m
X
X
X
X
|sm − sn | = xk −
xj = xk ≤
|xk | = |tm − tn | < ε.
P
k=1
Hence,
{sn }∞
n=1
j=1
k=n+1
is Cauchy and the series converges.
k=n+1