I need to know if this happens Always, Sometimes, or Never GoodwiForanyreal sequence, limXnYn=Tim (x₂)tim (yn

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Answered: •oodwi For any real sequence, lim Xn Yn…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section: Chapter Questions

Problem 2RE

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Lemma 4. Let {yn} be a positive solution of equation (1.1) with the corresponding sequence {z„} e So for n > N. Then: (i) (1– p)zn N; (ii) {zn¢n} is increasing for all n > N. Proof. Assume that {yn} is a positive solution of equation (1.1) with the corres- ponding sequence {zn} E So. Then z, is positive, Zn 2 yn, and Yn = Zn – Pnyo(n) 2 (1– p)zn, n>N> no, so (i) is proved. It easy to see that z, E So implies lim b,(Az,)“ = 0; n00 otherwise we would eventually have Az, > 0 contradicting z, E So. Similarly, lim a„A(b,(Azn)“) = 0. A summation of equation (1.1) then yields 9szs+1 Zn+1 LIs. s=n s=n s=n Summing once more, we obtain as t=s s=n or Azn 2-Zn+1Qn. Hence, A(zn&n) = PnAzn + Zn+1Ao, > Zn+1(A0n – OnQn) =0 since {0,} is a solution of the difference equation (Ao, – Qnºn) = 0. Therefore, {z,9n} is increasing and this completes the proof of the lemma.
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Using basic real analysis 1, solve the question
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•oodwi For any real sequence, lim Xn Yn = Tim (x₂) tim(yn)