Theorem 2. Let (sn) be a sequence with sn #0. Then lim inf Sa+15 Sn ≤ lim inf |sn|¹/¹ ≤ lim sup |sn|¹/" ≤ lim sup Proof. The middle inequality is obvious. The first and third inequalities have similar proofs. We will prove the third inequality and leave the first inequality as an exercise. Let a = lim sup|sn|¹/n, L = lim sup Sn+1 and we want to prove a ≤ L. If L = +∞, then we are done. Sn Assume L L. (in this case a is a lower bound of the set {L₁ L₁ > L}, and L is the infimum of this set.) Since L = lim N→∞ then there exists N > 0, such that Sn+1 (sup { $ +1| :n > N}) < Sn Sn+1 Sn Sn+1 Sup{|||2x}<4₂ v} :n> Sn L₁. < L₁
Theorem 2. Let (sn) be a sequence with sn #0. Then lim inf Sa+15 Sn ≤ lim inf |sn|¹/¹ ≤ lim sup |sn|¹/" ≤ lim sup Proof. The middle inequality is obvious. The first and third inequalities have similar proofs. We will prove the third inequality and leave the first inequality as an exercise. Let a = lim sup|sn|¹/n, L = lim sup Sn+1 and we want to prove a ≤ L. If L = +∞, then we are done. Sn Assume L L. (in this case a is a lower bound of the set {L₁ L₁ > L}, and L is the infimum of this set.) Since L = lim N→∞ then there exists N > 0, such that Sn+1 (sup { $ +1| :n > N}) < Sn Sn+1 Sn Sn+1 Sup{|||2x}<4₂ v} :n> Sn L₁. < L₁
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
show the first inequality， The third inequality solution has been given. Please refer that， Thank you
Transcribed Image Text:Theorem 2. Let (sn) be a sequence with sn ‡0. Then
lim inf
Sn+1
<lim inf|sn|¹/" <lim sup |sn|¹/" <lim sup
Proof. The middle inequality is obvious. The first and third inequalities have similar proofs. We
will prove the third inequality and leave the first inequality as an exercise.
Let a = lim sup|sn|¹/¹, L lim sup 4/₂
=
Sn+1
Sn
done.
and we want to prove a ≤ L. If L
=
Sn+1
L- Jim (up {|-|
=
Sn
then there exists N > 0, such that
Assume L<H+∞. It suffices to show a ≤ L₁ for any L₁ > L. (in this case a is a lower bound of
the set {L₁ L₁ > L}, and L is the infimum of this set.) Since
:
n > N})
:n
Sn+1
Sn
Sn+1
sup{">N}<L₁.
Sn
<I
< L₁
+∞, then we are
Transcribed Image Text:Then by iteration, for n > N,
N
Let a = L₁ |sn| > 0, then
As lim a¹/n
=
1, we have
|sn|
=
'n
Sn-1
Sn-1
Sn-2
< Ln-~ |SN| = L₁.
SN+1
SN
|SN|
L
•|SN|
|sn| <a · L₁ ⇒ |sn|¹/n < L₁a¹/n.
a = lim sup |sn|¹/n <L₁.
'n
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