show the first inequality， The third inequality solution has been given. Please refer that， Thank you Theorem 2. Let (sn) be a sequence with sn ‡0. Thenlim infSn+1 0, such thatAssume L L. (in this case a is a lower bound ofthe set {L₁ L₁ > L}, and L is the infimum of this set.) Since:n > N}):nSn+1SnSn+1sup{">N} Then by iteration, for n > N,NLet a = L₁ |sn| > 0, thenAs lim a¹/n=1, we have|sn|='nSn-1Sn-1Sn-2< Ln-~ |SN| = L₁.SN+1SN|SN|L•|SN||sn|

Answered: Image /qna-images/answer/45d40725-1f68-49ad-9ce1-0f93800e3ab1.jpg

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

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.1: Infinite Sequences And Summation Notation

Problem 32E

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