SOLUTION OF HW2
MINGFENG ZHAO
September 19, 2012
1. [20 Points] a. If xn is strictly decreasing and 0 ≤ xn ≤ 21 , then xn → 0 as n → ∞.
b. If xn =
(n − 1) cos(n2 + n + 1))
, then xn has a convergent subsequence.
2n − 1
c. If xn is a strictly increasing sequence and |xn | < 1 + n1 , then xn → 1 as n → ∞.
d. If xn has a convergent subsequence, then xn is bounded.
Proof. a. False. For example, let xn =
1
1
+
for all n ≥ 1, then xn > xn+1 for all n ≥ 1 and
4 4n
0 ≤ xn =
1
1 1
1
1
+
≤ + = .
4 4n
4 4
2
But we know that
lim xn = lim
n→∞
n→∞
1
1
+
4 4n
=
1
6= 0.
4
b. True. For any n ≥ 1, we know that
(n − 1) cos(n2 + n + 1)) |xn | = 2n − 1
=
n−1
cot | cos(n2 + n + 1)|
2n − 1
≤ 2.
Hence {xn } is a bounded sequence. By the Weierstrass’s Theorem, we know that {xn } has a
convergent subsequence.
c. False. For example, let xn =
1
1
−
for all n ≥ 1, then xn < xn+1 for all n ≥ 1 and
2 2n
1
1 1
1
1
|xn | = −
≤ +
<1− .
2 2n
2 2n
n
1
2
MINGFENG ZHAO
But we know that
lim xn = lim
n→∞
n→∞
1
1
−
2 2n
=
1
6= 0.
2
d. False. For example, let xn = n [1 + (−1)n ] for all n ≥ 1. Consider the subsequence {x2k−1 } = {0}
is convergent, but {xn } is not bounded, since x2k = 4k → ∞ as k → ∞.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu