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onk+1 k + 1 By induction, we found a subsequence (Sn) such that sne - tk < . It is easy to show that t = lim Sn. (left as an exercise.)

onk+1 k + 1 By induction, we found a subsequence (Sn) such that sne - tk < . It is easy to show that t = lim Sn. (left as an exercise.)

Algebra & Trigonometry with Analytic Geometry
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 72E
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please see the blue line， show t= limSnk.

You should use the definition of convergence, namely, using the ϵ language, and find the corresponding N.

Theorem 9. Let S be the set of subsequential limits of (sn). If (tn) is a sequence in SnR and lim tnt, then t E S. Proof. We use "construction by induction". (1) Since t₁ is the limit of some subsequence of (sn), there exists n₁, such that |Sn₁ − t₁| < 1; (2) Assume n₁ < < nk have been selected (they may come from different subsequences of (sn)) with |8n, − tj| < —, for j = 1, ···, k. Since tk+1 is the limit for some subsequence of (sn), there exists nk+1 | Snx+1 tk+1|< 1 k+1 By induction, we found a subsequence (Sn) such that |Snx −tk| < lim Sn. (left as an exercise.) = nk, such that It is easy to show that
Transcribed Image Text:Theorem 9. Let S be the set of subsequential limits of (sn). If (tn) is a sequence in SnR and lim tnt, then t E S. Proof. We use "construction by induction". (1) Since t₁ is the limit of some subsequence of (sn), there exists n₁, such that |Sn₁ − t₁| < 1; (2) Assume n₁ < < nk have been selected (they may come from different subsequences of (sn)) with |8n, − tj| < —, for j = 1, ···, k. Since tk+1 is the limit for some subsequence of (sn), there exists nk+1 | Snx+1 tk+1|< 1 k+1 By induction, we found a subsequence (Sn) such that |Snx −tk| < lim Sn. (left as an exercise.) = nk, such that It is easy to show that
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