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Unformatted text preview: limSn=+ . (a) Counterexample An=n^2, Bn= 4n (b) Counterexample An=n5, Bn=5n 2.36 Remark Since Sn is bounded by the completeness axiom s=sup{Sn,n N}=:S exists, such that Sn <= s for all n. Let eps>0 then Sn <= s +eps. By thm 2.9 Sns<eps. So limSn=sup(Sn). Simular for infSn....
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This note was uploaded on 05/11/2008 for the course MAT 371 taught by Professor Thieme during the Fall '07 term at ASU.
 Fall '07
 thieme

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