• Let (xn) and (Yn) be bounded sequences and xn < Yn. Show that lim inf xn < lim inf Yn . n00

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.2: Arithmetic Sequences
Problem 68E
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How do you solve the second bullet point?

Let (xn) be a bounded sequence. Recall that
lim inf xn = sup inf xm =
sup inf{xn, &n+1,..}, lim sup an = inf sup am = infsup{rn, &n+1,• .}.
m>n
n m>n
n00
Show that L = lim inf xn is the smallest limit point of (xn), i.e., any subsequential limit of (xn) is not smaller
n-00
than L, and there is a subsequence that has L as its limit.
Remark: Similarly, lim sup xn is the largest limit point of (xn). This also shows that a sequence (xn) converges if
n00
and only if its lim sup and lim inf are equal.
• Let (xn) and (Yn) be bounded sequences and xn < Yn. Show that
lim inf xn <lim inf yn .
n00
n00
• Let (xn) and (Yn) be bounded sequences. Show that
lim inf xn + lim inf yn < lim inf(xn + Yn) < lim inf xn + lim sup Yn,
n00
n00
n00
n00
n00
Hint: for parts 2) and 3), you can take some proper subsequences and use part 1).}
Transcribed Image Text:Let (xn) be a bounded sequence. Recall that lim inf xn = sup inf xm = sup inf{xn, &n+1,..}, lim sup an = inf sup am = infsup{rn, &n+1,• .}. m>n n m>n n00 Show that L = lim inf xn is the smallest limit point of (xn), i.e., any subsequential limit of (xn) is not smaller n-00 than L, and there is a subsequence that has L as its limit. Remark: Similarly, lim sup xn is the largest limit point of (xn). This also shows that a sequence (xn) converges if n00 and only if its lim sup and lim inf are equal. • Let (xn) and (Yn) be bounded sequences and xn < Yn. Show that lim inf xn <lim inf yn . n00 n00 • Let (xn) and (Yn) be bounded sequences. Show that lim inf xn + lim inf yn < lim inf(xn + Yn) < lim inf xn + lim sup Yn, n00 n00 n00 n00 n00 Hint: for parts 2) and 3), you can take some proper subsequences and use part 1).}
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