Math 201-B
Introduction to Proofs
Instructor: Alex Roitershtein
Iowa State University
Department of Mathematics
Fall 2015
Exam #3 (solutions to the practice test)
December 18, 2015
Student name:
-
Student ID:
Duration of the exam 120 minutes.
The exam includes 5 questions.
The total mark is 100 points.
Please show all the work, not only the answers.
Calculators, textbooks, and help sheets are allowed.
1. [20 points]
(a) Let A and B be bounded subsets of R such that A ⊂ B. Show formally that sup A ≤ sup B
and inf B ≤ inf A.
(b) Let
A = {x ∈ R : x = n−s for some n ∈ N, n ≥ 2 and s ∈ R, s > 0}.
Show that sup A = 1 and inf A = 0.
Solution:
(a) Let a = sup A and b = sup B. By definition of b,
b ≥ x ∀ x ∈ B.
Since A ⊂ B, we obtain that b ≥ x for all x ∈ A. Thus b is an upper bound for A and
hence b ≥ a = sup A.
Similalrly, let c = inf A and d = inf B. By definition of d,
d ≤ x ∀ x ∈ B.
Since A ⊂ B, we obtain that d ≤ x for all x ∈ A. Thus d is a lower bound for A and
hence d ≤ c = inf A.
(b) For all n ∈ N, n ≥ 2 and s ∈ R, s > 0
n−s ≤ 2−s .
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Consider now a function f (s) = 2−s , s > 0. Then log f (s) = −s log 2 ≤ 0. Thus
sup A ≤ e0 = 1. On the other hand, if it would be the case that a := sup A ∈ (0, 1) we
couldn’t find s > 0 such that
−s log 2 > log a,
log a
or equivalently s < −log
. Since log a < 0 and s > 0 is arbitrary, the last assertion is
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clearly incorrect. This establishes that a = 1.
On the other hand, while n−s > 0 for the parameters n, s with the range under the
consideration, one can make n−1 (here we choose s = 1 for certainty) arbitrarily small
by picking n ∈ N large enough. Thus (apply the result of (a) for included sets)
0 ≤ inf A ≤ inf{x ∈ R : x = n−1 for some n ∈ N, n ≥ 2} = 0.
Thus sup A = 0.
2. [20 points]
(a) Exercise 2.1.6 from the textbook BA.
(b) Exercise 2.1.13 from the textbook BA.
(c) Exercise 2.1.15 from the textbook BA.
Solution:
(a) We have:
0≤
1
1
n
≤
≤
.
n2 + 1
n + n−1
n
Since limn→∞ n−1 = 0, we obtain from the monotonicity of the limit that limn→∞
n
n2 +1
= 0.
(b) Since xn is monotone, say monotone increasing 9decreasing sequences are treated in a
similar way),
lim xn ≥ xm
n→∞
for any m ∈ N. Therefore, if xm > xk for some m > k, then xk cannot be the limit of
the sequence. This shows that xm = xk for all m > k.
(c) The sequence is not bounded because the subsequence x2n+1 = 2n + 1 is not bounded.
An example of a converging subsequence is x2n . Any its further subsequence is also
converging.
3. [20 points]
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(a) Exercise 2.2.3 from the textbook BA.
(b) Exercise 2.2.8 from the textbook BA.
Solution:
(a) Use induction and the algebra of the limits. For k = 1 the result is trivia. On the
otehr hand, if it is true for all integer j ≤ k, then
lim xk+1
=
n
lim xkn · lim xn
n→∞
k
k+1
lim xn · lim xn = lim xn
,
n→∞
n→∞
=
n→∞
n→∞
n→∞
as desired.
(b) You are familiar with this kind of problems from Calculus II. Use, for instance, the
L’Hospital’s Rule twice:
n2
=
n→∞ 2n
x2
2x
= lim x
x
x→∞ 2
x→∞ 2 log 2
2
= lim x
= 0.
x→∞ 2 (log 2)2
lim
lim
4. [20 points] Exercise 2.1.8 from the textbook BA. Please argue using directly the definition
of the limit.
Solution: Let xn =
2n
n!
and Sn =
2n
n=0 n! .
Pn
Then
lim Sn = lim Sn−1 =
n→∞
n→∞
∞
X
2n
n=0
n!
= e2 .
Hence
lim xn =
n→∞
lim (Sn − Sn−1 ) = lim Sn − lim Sn−1
n→∞
2
n→∞
n→∞
= e − e2 = 0.
5. [20 points] Exercise 2.2.6 from the textbook BA.
Solution: limn→∞ zn = 0 while limn→∞ wn = +∞ (thus the latter doesn’t exist in reals).
Proposition 2.2.5 (iv) cannot be applied since limn→∞ xn = limn→∞ yn = 0 which rules out
the set of conditions of the proposition.
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