vextendsingle vextendsingle � since 1 n 1 vextendsingle vextendsingle

Vextendsingle vextendsingle ? since 1 n 1

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Unformatted text preview: vextendsingle vextendsingle < ǫ since 1 n + 1 > = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle − 1 n + 1 vextendsingle vextendsingle vextendsingle vextendsingle < ǫ = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle n − n − 1 n + 1 vextendsingle vextendsingle vextendsingle vextendsingle < ǫ = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle n − ( n + 1) n + 1 vextendsingle vextendsingle vextendsingle vextendsingle < ǫ = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle n n + 1 − 1 vextendsingle vextendsingle vextendsingle vextendsingle < ǫ = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle s n +1 s n − 1 vextendsingle vextendsingle vextendsingle vextendsingle < ǫ Therefore lim( s n +1 /s n ) = 1. Q.E.D. (But note that ( s n ) defined by s n = 1 would suffice, and is a much simpler example). (b) Give an example of a divergent sequence ( t n ) of positive numbers such that lim( t n +1 /t n ) = 1 . Consider the sequence ( t n ) defined by t n = n . Now t n is positive for all n ∈ N and ( t n ) is divergent since ( t n ) is unbounded. We will show that lim( t n +1 /t n ) = 1 by showing that for any ǫ > 0, there exists N ∈ R such that for n ∈ N , n > N = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle t n +1 t n − 1 vextendsingle vextendsingle vextendsingle vextendsingle < ǫ 2 Real Analysis II: Section 17 and 18 David Joseph Stith Suppose ǫ > 0. Then let N = 1 ǫ . Now, n > N = ⇒ n > 1 ǫ and 0 < n = ⇒ 1 n < ǫ and 0 < 1 n = ⇒ 1 + 1 n < 1 + ǫ and 1 < 1 + 1 n = ⇒ n + 1 n < 1 + ǫ and 1 < n + 1 n = ⇒ n + 1 n − 1 < ǫ and 0 < n + 1 n − 1 = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle n + 1 n − 1 vextendsingle vextendsingle vextendsingle vextendsingle < ǫ = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle t n +1 t n − 1 vextendsingle vextendsingle vextendsingle vextendsingle < ǫ Therefore lim( t n +1 /t n ) = 1. Q.E.D. 17.15 Exercise. Prove the following. (a) lim( √ n + 1 − √ n ) = 0 . Suppose ǫ > 0. We need to show that there exists N ∈ R such that for n ∈ N , n > N = ⇒ vextendsingle vextendsingle ( √ n + 1 − √ n ) − vextendsingle vextendsingle < ǫ . Let N = 1 ǫ 2 . Then, n > N = ⇒ n > 1 ǫ 2 = ⇒ vextendsingle vextendsingle √ n vextendsingle vextendsingle > 1 ǫ = ⇒ vextendsingle vextendsingle √ n + 1 + √ n vextendsingle vextendsingle > 1 ǫ = ⇒ 1 vextendsingle vextendsingle √ n + 1 + √ n vextendsingle vextendsingle < ǫ = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle √ n + 1 − √ n ( √ n + 1 + √ n )( √ n + 1 − √ n ) vextendsingle vextendsingle vextendsingle vextendsingle < ǫ = ⇒ vextendsingle vextendsingle √ n + 1 − √ n vextendsingle vextendsingle < ǫ = ⇒ vextendsingle vextendsingle ( √ n + 1 − √ n ) − vextendsingle vextendsingle < ǫ Therefore lim( √ n + 1 − √ n ) = 0. Q.E.D....
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  • Fall '08
  • Akhmedov,A
  • Mathematical analysis, Tn, Dominated convergence theorem, David Joseph Stith

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