JhalyssaWilliams.Analysis.HW5 - Homework 5 Section 4.4 2a...

Homework 5
Section 4.4
2a) False; By theorem 4.4.8, “ Every unbounded sequence contains a monotone subsequence that
has either +∞ ?? − ∞ as a limit.” Every sequence does not have a convergent subsequence.
2b) True; By theorem 4.4.7, “ Every bounded sequence has a convergent subsequence .” So by
definitions 4.4.9, the subsequential limit must not be empty.
2c) False; By definition 4.4.9, if S
n
converges to s, then the lim inf S
n
= lim sup S
n
. This is true
only if s ∈ ℝ
2d) True
2e)False: S
n
=n for every natural number n. It is not bounded about but the lim Inf S
n
= 1.
3a) Since S
1
= 0, S
2
=2, S
3
=0, S
4
=2, and S
5
=0, then the subsequential limit is {0,2}. The Lim Sup
S
n
=2 and the Lim Inf S
n
=0
3b) t
n
= ( 0, ½, 2/3, ¼, 4/5, 1/6, 6/7, …)  The subsequential limit is {0,1},thus the Lim Sup t
n
=1
and the Lim Inf t
n
=0
3c) Since u
1
=-2, u
2
=0, u
3
=-18, u
4
=0, u
5
= -50, u
6
=0  The subsequential limit is { −∞ , 0}. The
lim Sup u
n
=0 and the Lim Inf u
n
= −∞ .
3d) Let V
n
=
𝑛𝑠𝑖𝑛(𝜋𝑛)
2
. Then V
1
= 1, V
2
=0, V
3
= -3, V
4
=0, V
5
=5, V
6
=0, and V
8
= -8. When n is even,
the sequence is 0, for all n ∈ ℕ . When n is odd, the sequence diverges to +∞ ?? − ∞  The
subsequential limit is { +∞, 0, −∞ } with +∞ as is the Lim Sup V
n
and −∞ is the limit inf V
n.
4a) Since w
1
= -1, w
2
=1/2, w
3
= -1/3, w
4
= ¼, then w
n
converges to 0 ∀? ∈ ℕ  lim sup w
n
= lim
inf w
n
= 0
4b) The subsequence can be written as X
3n-2
=0 ∀?
,
X
3n-1
=1 ∀? , and X
3n
converges to +∞
 The subsequential limit is {0, 1, +∞} and the Lim Sup X
n
= +∞ ; Lim Inf X
n
= 0
4c) The subsequence can be written as Y
2n
=6n ∀?, and Y
2n-1
=n. Both subsequences converge to
+∞