Math 341 Section 2 — Fall 2009 — Midterm Exam 1 — Solutions
1. State the definitions of each of the following words or phrases:
(a) the sequence (an ) is bounded
Answer 1: There exists a real number M such that |an | ≤ M for all n ∈ N.
Answer 2: There exist real numbers A and B such that A ≤ an ≤ B for all n ∈ N.
(b) the supremum of a set A
Answer 1: The supremum of s of A is a real number s such that
i. s is an upper bound for A, and
ii. if b is any upper bound for A, then s ≤ b.
Answer 2: The supremum of A is a real number x such that
i. a ≤ x for every a ∈ A, and
ii. for every B ∈ R such that a ≤ B for every a ∈ A, then x ≤ B.
(c) the set A is countable
The set A is countable if there exists a bijection f : N → A.
(d) the set A is uncountable
Answer 1: A is uncountable if there does not exist a surjection f : N → A
Answer 2: A is uncountable if A is an infinite set such that there does not exist a
bijection f : N → A.
(e) the sequence (an ) is a Cauchy sequence
The sequence (an ) is Cauchy if, for every > 0, there exists N ∈ N, such that
|an − am | < whenever m, n ≥ N .
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2. Classify each of the sets as finite, countable, or uncountable by circling the appropriate response. No justification is necessary.
{x ∈ Q : |x| <
√
2}
finite
countable
uncountable
2 or a < −3}
finite
countable
uncountable
{x ∈ (0, 1) : x is irrational}
finite
countable
uncountable
{a/b ∈ R : a ∈ N, b ∈ N, a + b = 10}
finite
countable
uncountable
{a/b ∈ R : a ∈ R, b ∈ R, b 6= 0, a + b = 10}
finite
countable
uncountable
{a ∈ Z : a2 <
√
3. Give a simple example of each of the following, or argue that such a request is impossible.
(a) Sequences (xn ) and (yn ), which both diverge, but whose sum (xn + yn ) converges.
(xn ) = (n),
(yn ) = (−n),
(xn + yn ) = (0, 0, 0, . . .).
(b) A convergent sequence (bn ) with bn 6= 0 for all n such that (1/bn ) diverges.
(bn ) = (1/n),
(1/bn ) = (n).
(c) A sequence that does not contain 0 or 1 as a term but contains subsequences converging
to each of these values.
Define an by
(
1
if n is odd,
an = n+1
n
if n is even.
n+1
That is,
(an ) =
1 2 1 4 1 6
, , , , , ,...
2 3 4 5 6 7
Then, (a2n ) → 1 and (a2n−1 ) → 0.
(d) A monotone sequence that diverges but has a convergent subsequence.
Answer 1: Impossible. Let (an ) be a monotone sequence that diverges. By the Monotone
Convergence Theorem, (an ) is unbounded. Any subsequence of an unbounded monotone
sequence is also unbounded and therefore does not converge.
Answer 2: Impossible. Let (an ) be a monotone sequence that diverges. If (ank ) is a
convergent subsequence, let s = lim ank . Then, for every n, an belongs to the closed
interval with endpoints a1 and s. But then (an ) is bounded, and by the Monotone
Convergence Theorem, (an ) converges, a contradiction.
Part (d) was worth 3 points.
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4. By directly using the definition of the limit of a sequence, show that if lim(an ) = 5, then
2an
5
= .
n→∞ an − 1
2
lim
[For this question you should not use the Algebraic Limit Theorem.]
Proof. Let > 0 be given. We must show that there exists some N such that
2an
5
−
an − 1 2 < whenever n ≥ N .
Since lim(an ) = 5, there exists N1 ∈ N such that
|an − 5| < 1
whenever n ≥ N1 and so
−1 < an − 5 < 1
3 < an − 1 < 5
1
1
1
<
< .
5
an − 1
3
Since lim(an ) = 5, there also exists N2 ∈ N such that for n ≥ N2
|an − 5| < 6.
Then, for n ≥ N = max{N1 , N2 },
2an
5 2(2an ) − 5(an − 1) |an − 5|
|an − 5|
6
=
<
<
= .
an − 1 − 2 = 2(an − 1)
2|an − 1|
6
6
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5. Let a1 = 1 and an+1 =
√
1 + an for n ≥ 1.
(a) Prove that (an ) is a positive increasing sequence.
[Hint: Is the sequence positive? Is a1 < a2 ? If an < an+1 , show that an+1 < an+2 .]
The first√term of the sequence a1 is positive. If an > 0, then from the recurrence formula
an+1 = 1 + an is also positive. So, all the terms of the sequence are positive.
√
The first two terms are related by 0 < a1 = 1 < a2 = 2. Now suppose 0 < an < an+1
for some n. Then
0 < an < an+1
1 < 1 + an < 1 + an+1
p
√
1 < 1 + an < 1 + an+1
1 < an+1 < an+2 .
This shows that (an ) is a positive increasing sequence.
(b) Prove that (an ) has an upper bound.
0 < an < an+1 =
√
1 + an
⇒
since
√
1± 5
2
a2n < 1 + an
⇒
√
√
1− 5
1+ 5
< an <
2
2
⇒
a2n − an − 1 < 0
are the two root of x2 − x − 1 = 0. So,
√
1+ 5
an <
2
for all n ∈ N.
(c) Find the exact value of A = lim(an ).
[If your reasoning relies on a major theorem, don’t forget to mention it as part of the
explanation.]
By parts (a) and (b) and the Monotone Convergence Theorem, A = lim(an ) exists. Then
by the Algebraic Limit Theorem,
a2n+1 = 1 + an
lim a2n+1 = lim (1 + an )
n→∞
n→∞
2
lim (an+1 ) = lim (1) + lim (an )
n→∞
n→∞
n→∞
2
A =1+A
Thus, A =
√
1± 5
.
2
Since A is positive, the correct choice is A =
4
√
1+ 5
2
.
6. Let S = (0, 1) ∩ Q. That is, S is the set of all rational numbers contained in the open
interval (0, 1). Assume f : N → S is a function that is both one-to-one and onto, and consider
the sequence (xn ) where xn = f (n). If c is a given irrational number
in the interval (0, 1),
explain how to construct a strictly increasing subsequence of xn that converges to c.
There are, of course, many ways to solve this problem. Here are two ways:
Solution 1. Consider the strictly increasing sequence (bk ) of real numbers bk defined by
1
bk = c 1 − k
2
Note that limk→∞ bk = c. Define Ik to be the interval
c c
Ik = (bk , bk+1 ) = c − k , c − k+1 .
2
2
Choose n1 such that
xn1 ∈ I1 .
Choose n2 > n1 such that
xn2 ∈ I2 .
In general, choose nk > nk−1 such that
xnk ∈ Ik .
Since xnk ∈ Ik ,
xn1 < xn2 < xn3 < · · ·
and |xnk − c| <
c
.
2k
Thus
lim xnk = c.
k→∞
Here is a picture illustrating the choice of xnk :
I1
0
b1
xn1
I2
I3
b2 xn2 b3 xn2
c
1
Solution 2. Write the irrational number c as an infinite decimal expansion as
c = 0.c1 c2 c3 . . .
where ci ∈ {0, 1, . . . , 9} is the ith decimal digit. Let rn = 0.c1 c2 c3 . . . cn . Then the sequence
(rn ) is an increasing sequence of rational numbers that converges to c.
Now choose xn1 to be the first term of the sequence (xn ) that equals rn for some n. Choose
xn2 to be the first term of the sequence (xn ) after xn1 such that xn2 = rn for some n and
xn1 < xn2 . In general, give that xnk has been chosen, let xnk+1 be the first term of the sequence
(xn ) after xnk such that xnk+1 = rn for some n and xnk < xnk+1 . Then the subsequence (xnk )
is an increasing subsequence of (xn ). Because (xnk ) is also a subsequence of (rn ), we know
that (xnk ) converges to c.
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7. (Squeeze Theorem) By directly using the definition of the limit, show that, if xn ≤ yn ≤ zn
for all n ∈ N and if lim xn = lim zn = L, then lim yn = L as well.
Solution. Let > 0 be given. Since lim xn = lim zn = L, there exist N1 , N2 ∈ N such that
|xn − L| < for all n ≥ N1 ,
|zn − L| < for all n ≥ N2 .
Let N = max{N1 , N2 }. If n ≥ N , then
L − < xn ≤ yn ≤ zn < L + .
Hence, |yn − L| < . This shows that lim yn = L.
8. Prove that the set of real numbers R is uncountable by using an argument that relies on the
Nested Interval Property.
Solution. Since N ⊂ R, we know that R is an infinite set. So, R is either countable or
uncountable. Suppose, by way of contradiction, that R is countable. Let x1 , x2 , x3 , . . . be an
enumeration of R. Let I1 be a closed interval of nonzero length not containing x1 . Let I2 be a
closed interval of nonzero length contained in I1 not containing x2 . Let I3 be a closed interval
of nonzero length contained in I2 not containing x3 . Continuing in this fashion we obtain a
nested sequence of closed intervals
I1 ⊃ I2 ⊃ I3 ⊃ · · ·
such that xk 6∈ Ik . Therefore,
xn 6∈
∞
\
Ik
k=1
for any n. In other words,
∞
\
Ik = ∅.
k=1
By the Nested Interval Theorem, ∩∞
k=1 Ik 6= ∅, a contradiction. Thus, R is uncountable.
Note: Omitting the word ‘closed’ would prevent us from using the Nested Interval Property
of the reals.
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9. Let (an ) and (bn ) be convergent sequences such that lim an = A and lim bn = B. Also assume
that A 6= 0 and B 6= 0. By directly using the definition of the limit, show that
lim(an bn ) = AB.
Hint: |an bn − AB| = |an bn − Abn + Abn − AB|.
Solution. Let > 0 be given. Since (bn ) converges, it it bounded. Thus, there exists positive
M such that |bn | < M for all n. Since lim an = A and lim bn = B, there exist N1 , N2 ∈ N
such that
2M
|bn − B| <
2|A|
|an − A| <
for all n ≥ N1 ,
for all n ≥ N2 .
Let N = max{N1 , N2 }. If n ≥ N , then
|an bn − AB| = |an bn − Abn + Abn − AB|
≤ |an bn − Abn | + |Abn − AB|
= |bn | |an − A| + |A| |bn − B|
+ |A|
= .
<M
2M
2|A|
This shows that lim an bn = AB.
10. Assume (xn ) is a Cauchy sequence and that 0 < A ≤ xn ≤ B for all n ∈ N. Give a direct
argument that ( x1n ) is also Cauchy that does not use the Cauchy Criterion or the Algebraic
Limit Theorem.
Solution. Let > 0 be given. Since (xn ) is a Cauchy sequence, there exists N ∈ N such that
|xn − xm | < A2 whenever m, n ≥ N . Then for m, n ≥ N ,
1
1 xn − xm |xn − xm |
A2 −
=
≤
<
= .
xm xn x n xm A2
A2
Thus ( x1n ) is also a Cauchy sequence.
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