Math 317 HW #5 Solutions
1. Exercise 2.4.2.
(a) Prove that the sequence defined by x1 = 3 and
xn+1 =
1
4 − xn
converges.
Proof. I intend to use the Monotone Convergence Theorem, so my goal is to show that
(xn ) is decreasing and bounded. To do so, I will prove by induction that, for any
n ∈ {1, 2, 3, . . .},
0 < xn+1 < xn < 4.
Base Case When n = 1, we have that x1 = 3 and x2 =
1
4−3
= 1, so
0 < x2 < x1 .
Inductive Step Suppose 0 < xk+1 < xk < 4; our goal is to show that this implies that
0 < xk+2 < xk+1 < 4.
Since xk+1 < xk , we know that
4 − xk+1 > 4 − xk
or, equivalently,
xk+2 =
1
1
<
= xk+1 .
4 − xk+1
4 − xk
Moreover, since xk+1 < xk < 4, we have
xk+2 < xk+1 < 4.
The fact that xk+1 < 4 also implies that 4 − xk+1 > 0, so
0<
1
= xk+2 .
4 − xk+1
Putting this all together, then, we see that 0 < xk+1 < xk < 4 implies that 0 <
xk+2 < xk+1 < 4.
Therefore, by induction, we can conclude that
0 < xn+1 < xn < 4
for all n ∈ {1, 2, 3, . . .}. Thus, the sequence (xn ) is a bounded, decreasing sequence and
so, by the Monotone Convergence Theorem, it must converge.
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(b) Now that we know lim xn exists, explain why lim xn+1 must also exist and equal the
same value.
Answer. Intuitively, lim xn+1 converges because (xn+1 ) is identical to (xn ) except for
the very first term, which has no impact on the limiting behavior of the sequence.
If I want to be really rigorous, here’s how to do that: define the sequence (yn ) by
yn = xn+1 for all n ∈ {1, 2, 3, . . .}. Let > 0. Since (xn ) converges to, say, L ∈ R, there
exists an N ∈ N such that, if n ≥ N ,
|xn − L| < .
Therefore, if n ≥ N − 1,
|yn − L| = |xn+1 − L| < since n + 1 ≥ N . Therefore, since the choice of > 0 was arbitrary, we conclude that
(yn ) → L or, equivalently, (xn+1 ) → L.
(c) Take the limit of each side of the recursive equation in part (a) of this exercise to
implicitly compute lim xn .
Answer. On the one hand, if (xn ) → L, we know that (xn+1 ) → L. On the other hand,
we know
1
1
(xn+1 ) =
→
4 − xn
4−L
by the Algebraic Limit Theorem (we know L 6= 4 since 3 is an upper bound for (xn )).
Therefore, it must be the case that
L=
1
4−L
or, equivalently,
L(4 − L) = 1.
We can re-write this as
0 = L2 − 4L + 1,
so, by the quadratic formula,
p
√
√
4 ± (−4)2 − 4(1)(1)
4 ± 16 − 4
L=
=
= 2 ± 3.
2(1)
2
√
√
Since 2 + 3 > 3, which is an √
upper bound for (xn ), we know L 6= 2 + 3. Therefore,
we can conclude that L = 2 − 3 ≈ 0.268
2. Exercise 2.4.6. Let (an ) be a bounded sequence.
(a) Prove that the sequence defined by yn = sup{ak : k ≥ n} converges.
Proof. For any n ∈ {1, 2, 3, . . .}, define the set
An := {an , an+1 , an+2 , . . .}.
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In other words, yn = sup An . As defined, An+1 ⊂ An , so
yn+1 ≤ yn .
Since this is true for any n, we see that the sequence (yn ) is decreasing.
Also, since (an ) is bounded, there exists M > 0 such that |an | ≤ M for all n ∈
{1, 2, 3, . . .}. In particular, for any n,
−M ≤ an ≤ yn ≤ y1 ,
so we see that the sequence (yn ) is bounded.
Therefore, since (yn ) is a bounded, decreasing sequence, the Monotone Convergence
Theorem implies that it converges.
(b) The limit superior of (an ), or lim sup an , is defined by
lim sup an = lim yn ,
where yn is the sequence from part (a) of this exercise. Provide a reasonable definition
for lim inf an and briefly explain why it always exists for any bounded sequence.
Answer. Here’s the definition of lim inf an :
For a bounded sequence (an ), define the sequence (zn ) by
zn = inf{ak : k ≥ n}.
Then the limit inferior of (an ) is defined to be
lim inf an := lim zn .
The sequence (zn ) is increasing and bounded (it’s bounded below by z1 and above by
the upper bound for (an )), so the Monotone Convergence Theorem implies it converges,
so the above limit definitely exists.
(c) Prove that lim inf an ≤ lim sup an for every bounded sequence, and give an example of
a sequence for which the inequality is strict.
Proof. Note that, with (yn ) and (zn ) defined as above,
zn ≤ an ≤ yn
for all n ∈ {1, 2, 3, . . .}. In particular, zn ≤ yn . Therefore, by Theorem 2.3.4(ii),
lim inf an = lim zn ≤ lim yn = lim sup an .
Consider the sequence (an ) given by
an = 2 + (−1)n (1 + 1/n).
Then it’s straightforward to show that lim inf an = 1 and lim sup an = 3, so the inequality
is strict for this sequence.
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(d) Show that lim inf an = lim sup an if and only if lim an exists. In this case, all three share
the same value.
Proof. To show the forward implication, assume lim inf an = lim sup an . In other words,
with (yn ) and (zn ) defined as above, lim zn = lim yn . Now, we know that
zn ≤ an ≤ yn
for all n ∈ {1, 2, 3, . . .}, so the Squeeze Theorem implies that lim an exists and is equal
to the common limit of the yn ’s and zn ’s.
Turning to the reverse implication, assume lim an = L for some L ∈ R. Let > 0. Since
(an ) → L, there exists N ∈ N such that, whenever n ≥ N ,
|an − L| < /2
or, equivalently,
−/2 < an − L < /2.
Hence, L − /2 < an < L + for any n ≥ N , so L + is an upper bound for the set
An = {ak : k ≥ n}.
Since yn = sup An , this implies that yn ≤ L + /2, meaning that yn − L ≤ /2. On the
other hand, yn ≥ an > L − /2.
Putting this all together, we see that, for any n ≥ N ,
L − /2 ≤ yn ≤ L + /2,
which implies that
|yn − L| ≤ /2 < ,
Since our choice of > 0 was arbitrary, we conclude that (yn ) → L.
A totally equivalent argument shows that (zn ) → L as well, so we have that lim inf an =
lim sup an = lim an when (an ) converges.
Having proved both implications, we can conclude that lim inf an = lim sup an if an only
if lim an exists.
3. Exercise 2.5.3. Give an example of each of the following, or argue that such a request is
impossible.
(a) A sequence that does not contain 0 or 1 as a term but contains subsequences converging
to each of these values.
Example. Consider the sequence (an ) given by
(
1 + 1/n if n is odd
an =
1/n
if n is even.
Clearly, (1 + 1/n) → 1 and (1/n) → 0, so the subsequence of odd terms converges to 1
and the subsequence of even terms converges to 0, even though neither 1 nor 0 appear
as terms of the sequence.
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(b) A monotone sequence that diverges but has a convergent subsequence.
This is impossible. To see why, suppose (an ) is a monotone sequence with a convergent
subsequence (ank ). In fact, for simplicity, assume (an ) is increasing. Since (ank ) converges, it must be bounded, so there exists M ∈ R such that ank ≤ M for all of the ank .
Now, for any term an from the original sequence, there exists k ∈ N such that nk > n;
then
a1 ≤ an ≤ ank ≤ M.
Since our choice of n was arbitrary, we see that a1 ≤ an ≤ M for all n ∈ {1, 2, 3, . . .}, so
the sequence (an ) must be bounded. A similar argument works when (an ) is decreasing.
Therefore, any monotone sequence with a convergent subsequence must be bounded and,
therefore, convergent by the Monotone Convergence Theorem.
(c) A sequence that contains subsequences converging to every point in the infinite set
{1, 1/2, 1/3, 1/4, 1/5, . . .}.
Example. Consider the sequence
(1, 1, 1/2, 1, 1/2, 1/3, 1, 1/2, 1/3, 1/4, 1, 1/2, 1/3, 1/4, 1/5, 1, . . .).
In other words, I’m constructing the sequence by stringing together blocks of the form
1, 1/2, 1/3, . . . , 1/n,
starting with the block corresponding to n = 1, then moving on to the block corresponding to n = 2, etc.
Clearly, each number 1/k appears in the above sequence infinitely many times, which
gives a subsequence converging to 1/k.
(d) An unbounded sequence with a convergent subsequence.
Example. Consider the sequence (an ) given by
(
0 if n is odd
an =
n if n is even.
Then, for any M > 0, there exists N ∈ N such that N > M (by the Archimedean
Property); then either aN = N > M or aN +1 = N + 1 > M . Since this is true for any
M > 0, we see that (an ) is unbounded.
On the other hand, the subsequence of odd terms is just the sequence (0, 0, 0, 0, . . .),
which clearly converges to 0.
(e) A sequence that has a subsequence that is bounded but contains no subsequence that
converges.
This is impossible. To see why, suppose (an ) has a bounded subsequence (ank ). Then
(ank ) is a bounded sequence and so, by the Bolzano-Weierstrass theorem, contains a
convergent subsequence (ank` ). But, of course, the sub-subsequence (ank` ) is also a subsequence of the original sequence (an ), so (an ) must contain a convergent subsequence.
4. Exercise 2.5.4. Assume (an ) is a bounded sequence with the property that every convergent
subsequence of (an ) converges to the same limit a ∈ R. Show that (an ) must converge to a.
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Proof. I’ll prove the contrapositive. In other words, I want to show that, if (an ) is a bounded
sequence that does not converge to the number a ∈ R, then not every convergent subsequence
converges to a (or, more positively, there is some subsequence that converges to b 6= a).
To that end, suppose (an ) is a bounded sequence that does not converge to a. In other words,
it is not the case that, for every > 0, there is an N ∈ N such that, for all n ≥ N , |an − a| < .
Passing the negation through the various quantifiers, this means that there is some 0 > 0
such that, for all N ∈ N, there is an n ≥ N such that |an − a| ≥ 0 . Stated in plainer
language, we’re assuming that there is some 0 > 0 such that infinitely many of the an ’s are
at a distance at least 0 away from a.
Now, these infinitely many an ’s form a subsequence; call it (ank ). Since (an ) is bounded, the
subsequence (ank ) is also bounded and thus, by the Bolzano-Weierstrass Theorem, contains
a convergent subsequence (ank` ) → b. Since each of the ank` is at a distance at least 0 from
a, we definitely have that b 6= a, so we see that not every convergent subsequence of (an )
converges to a, which is what we wanted to show.
5. Exercise 2.6.2. Supply a proof for Theorem 2.6.2, which says that every convergent sequence
is a Cauchy sequence.
Proof. Suppose the sequence (an ) converges to a. Let > 0. Then, by the definition of
convergence, there exists N ∈ N such that, for any n ≥ N ,
|an − a| < /2.
Therefore, for any n, m ≥ N , we have that
|an − am | = |an − a + a − am | ≤ |an − a| + |a − am | < /2 + /2 = ,
where the first inequality is just an application of the triangle inequality and the second
follows from the assumption on n and m.
Therefore, since our choice of > 0 was arbitrary, we see that for all > 0 there exists N ∈ N
such that, if n, m ≥ N , |an − am | < , which means that (an ) is a Cauchy sequence.
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