Problem session solutions
1. The sequence an = (−1)n is not convergent, but some of its subsequences are. Which ones?
Solution The ones with only finitely many odd n converge to 1. The ones with finitely many
even n converge to −1. None of the others converge.
2. Find
∞
X
n=2
n3
1
.
−n
Solution After computing the first few partial sums one can guess that
n
X
j=2
j3
1
1
1
= − 2
.
−j
4 2n + 2n
This is then easy to prove by induction. Taking limits as n → ∞,
∞
X
n=2
1
1
=
.
n3 − n
4
3. Which of the following sets of real numbers contains a least element?
[Recall that if S ⊆ R, then S contains a least element if there is an element x ∈ S such that
x ≤ y for all y ∈ S].
(a) N0
(b) {−1, −2, −3, . . . }
(c) Z
(d) [0, ∞)
Solution (a) has a least element by the Least Integer Principle, since it is a non-empty
subsets of N0 .
(b), (c) have no least element, since in those cases x ∈ S =⇒ x > x − 1 and x − 1 ∈ S, so x
is not a least element of S.
(d) 0 is the least element of [0, ∞), since 0 ∈ [0, ∞) and 0 ≤ x for all x ∈ [0, ∞).
4. Use the Euclidean algorithm to compute:
(a) gcd(2325, 2316)
(b) gcd(1000, −2369) [Hint: gcd(a, b) = gcd(b, a) = gcd(a, −b) = gcd(−a, b)]
(c) gcd(−1104, −1679)
Solution (a)
2325 = 2136 × 1 + 189
2136 = 189 × 11 + 57
189 = 57 × 3 + 18
57 = 18 × 3 + 3
18 = 3 × 6 + 0
Hence gcd(2325, 2136) = 3.
(b)
2369 = 1000 × 2 + 369
1000 = 369 × 2 + 262
369 = 262 × 1 + 107
262 = 107 × 2 + 48
107 = 48 × 2 + 11
48 = 11 × 4 + 4
11 = 4 × 2 + 3
4=3×1+1
3=1×3+0
Hence gcd(1000, −2369) = gcd(2369, 1000) = 1.
(c)
1679 = 1104 × 1 + 575
1104 = 575 × 1 + 529
575 = 529 × 1 + 46
529 = 46 × 11 + 23
46 = 23 × 2 + 0
Hence gcd(−1104, −1679) = gcd(1679, 1104) = 23.
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