Download Math 347 - Fall 2015 Midterm 3 practice solutions 1. Prove from the

Math 347 - Fall 2015
Midterm 3 practice solutions
1. Prove from the definition of a convergent sequence that if an → L and bn → M , then
an + b n → L + M .
Solution. This is theorem 14.5(a) in the book.
2. (a) Prove that every infinite subset of a countable set is countable.
(b) Prove that every set that contains an uncountable set is uncountable.
Solution. This is problem 13.7 on homework 6.
3. (a) State the definition of a Cauchy sequence and the Cauchy convergence criterion.
(b) State the definition of a subsequence and the Bolzano–Weierstrass Theorem.
Solution. These are Definition 14.12, Theorem 14.19, and Theorem 14.17 in the book.
4. Let a, b ∈ N.
(a) State the definition of gcd(a, b).
(b) Prove that the set of integer combinations of a and b equals the set of integer multiples
of gcd(a, b).
Solution. a) gcd(a, b) is the largest natural number that divides both a and b.
b) This is Theorem 6.12 in the book.
5. Prove that a bounded sequence need not be Cauchy by providing a counterexample and a
formal -style argument showing that your sequence is not a Cauchy sequence.
Solution. Consider an = (−1)n and take = 1/2 and set m = n + 1. Then for all N , if
n, m ≥ N we have
|an − am | = |an − an+1 | = |2| ≥ 1/2 = ,
so the sequence is not Cauchy.
6. The set of irrational numbers is R \ Q. Prove that the irrational numbers are uncountable.
Solution. We argue by contradiction, so assume R \ Q is countable, and let f : N → R \ Q be
a bijection. Since Q is countable, there is also a bijection g : N → Q. We use these bijections
to define a new bijection, h : N → R, by
f (n/2)
n even
h(n) =
.
g((n + 1)/2) n odd
Since h has an inverse h−1 : R → N defined by
2f −1 (x)
x∈R\Q
−1
h (x) =
−1
2g (x) − 1
x∈Q
h is indeed a bijection as claimed. But since R is uncountable, there cannot exist a bijection
h : N → R, which gives a contradiction to R \ Q being countable. Hence, R \ Q is uncountable.
1
7. (a) State the comparison test for series.
Solution. See Proposition 14.29 in the book.
(b) Prove by induction that 3n − 1 ≥ 2n when n ≥ 1.
Solution. For a base case, we have 31 − 1 = 2 ≥ 21 . For the inductive step, assume that
3n − 1 ≥ 2n up to some n. Then
3n+1 − 1 = 3(3n − 1) + 2 > 3(3n − 1) > 2(3n − 1) ≥ 2 · 2n = 2n+1
where the last greater than or equal to sign uses the inductive hypothesis. This completes
the induction.
∞
X
1
(c) Prove that
converges.
n
3
−
1
n=1
Solution. By the comparison test, we have
0≤
1
1
≤
3n − 1
2n
P∞
P
1
and since the series ∞
n=1
n=1 2n converges (since it is a geometric series) the series
also converges by the comparison test.
1
3n −1
8. For the following questions, decide whether the statement is true or false and provide a brief
proof or counterexample justifying your choice.
P
(a) If lim an does not exist, then ∞
n=1 an is divergent.
n→∞
P
Solution. TRUE. The contrapositive of the statement reads “If ∞
n=1 an is convergent,
then limn→∞ an exists” is true, since this limit must exist and equal zero.
(b) If lim an = 1 and lim bn does not exist, then lim an bn does not exist.
n→∞
n→∞
n→∞
Solution. TRUE. We argue by contradiction. Suppose limn→∞ an bn = L. Then since
1/an → 1/1 = 1, we have that
L = lim an bn lim (1/an ) = lim (an bn /an ) = lim bn ,
n→∞
n→∞
n→∞
n→∞
but since limn→∞ bn does not exist, this is a contradiction.
Note: It would be wrong to start out by writing
lim an bn = lim an lim bn
n→∞
n→∞
n→∞
since this formula only holds with both limits on the right exist.
(c) If lim (an+k − an ) = 0 for every k ∈ N, then {an } converges.
n→∞
√
Solution. FALSE. The sequence an = n provides a counterexample since
√
√
lim ( n + k − n) = lim √
n→∞
n→∞
for every k ∈ N.
2
k
√ =0
n+k+ n
(d) If
∞
X
a2k
diverges, then
∞
X
ak diverges.
k=1
k=1
√
P
2
Solution. FALSE. A counterexample isPak = (−1)k / k. Then the series ∞
k=1 ak is the
harmonic series (which diverges) while ∞
k=1 ak converges.
9. (a) Compute the gcd of 78 and 90 using the Euclidean algorithm.
Solution. We did this one in review: the Euclidean algorithm for this pair is
gcd(78, 90) =
=
=
=
gcd(90, 78) = gcd(90 − 78, 78) = gcd(12, 78)
gcd(78, 12) = gcd(78 − 6 · 12, 12) = gcd(6, 12)
gcd(12, 6) = gcd(12 − 2 · 6, 6) = gcd(0, 6)
6.
(b) Compute the gcd of 78 and 90 using prime factorization. (HINT: 78 = 6 · 13.)
Solution. We have 90 = 9 · 10 = 2 · 32 · 5 = 6 · 15 and 78 = 6 · 13. Since 13 and 15 are
relatively prime, we have gcd(90, 78) = 6.
10. (a) Carefully explain how an infinite decimal expansion of a real number defines a Cauchy
sequence whose limit is the real number given by the decimal expansion.
Solution. An infinite decimal expansion of a real number x ∈ [0, 1] defines a sequence
k
X
an
xk =
10n
n=1
with xk → x. This sequence is Cauchy since for all m ≥ n ≥ N we have
|xm − xn | <
1
1
≤
10n−1
10N −1
(b) Give a careful proof (using the definition of a convergent sequence) that .999 · · · = 1.
Solution. The infinite decimal .9̄ is the limit of the sequence
k
X
9
.
xk =
10n
n=1
We have 1 − xk =
n ≥ N,
1
.
10k
Hence, for any given > 0, there exists an N such that for all
1
1
≤ N <
n
10
10
and hence the limit of the sequence xk is 1. This shows .9̄ = 1.
|1 − xn | =
3