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**Unformatted text preview: **vextendsingle vextendsingle < Ç« since 1 n + 1 > = â‡’ vextendsingle vextendsingle vextendsingle vextendsingle âˆ’ 1 n + 1 vextendsingle vextendsingle vextendsingle vextendsingle < Ç« = â‡’ vextendsingle vextendsingle vextendsingle vextendsingle n âˆ’ n âˆ’ 1 n + 1 vextendsingle vextendsingle vextendsingle vextendsingle < Ç« = â‡’ vextendsingle vextendsingle vextendsingle vextendsingle n âˆ’ ( n + 1) n + 1 vextendsingle vextendsingle vextendsingle vextendsingle < Ç« = â‡’ vextendsingle vextendsingle vextendsingle vextendsingle n n + 1 âˆ’ 1 vextendsingle vextendsingle vextendsingle vextendsingle < Ç« = â‡’ vextendsingle vextendsingle vextendsingle vextendsingle s n +1 s n âˆ’ 1 vextendsingle vextendsingle vextendsingle vextendsingle < Ç« 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 âˆˆ 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 Ç« > 0, there exists N âˆˆ R such that for n âˆˆ N , n > N = â‡’ vextendsingle vextendsingle vextendsingle vextendsingle t n +1 t n âˆ’ 1 vextendsingle vextendsingle vextendsingle vextendsingle < Ç« 2 Real Analysis II: Section 17 and 18 David Joseph Stith Suppose Ç« > 0. Then let N = 1 Ç« . Now, n > N = â‡’ n > 1 Ç« and 0 < n = â‡’ 1 n < Ç« and 0 < 1 n = â‡’ 1 + 1 n < 1 + Ç« and 1 < 1 + 1 n = â‡’ n + 1 n < 1 + Ç« and 1 < n + 1 n = â‡’ n + 1 n âˆ’ 1 < Ç« and 0 < n + 1 n âˆ’ 1 = â‡’ vextendsingle vextendsingle vextendsingle vextendsingle n + 1 n âˆ’ 1 vextendsingle vextendsingle vextendsingle vextendsingle < Ç« = â‡’ vextendsingle vextendsingle vextendsingle vextendsingle t n +1 t n âˆ’ 1 vextendsingle vextendsingle vextendsingle vextendsingle < Ç« Therefore lim( t n +1 /t n ) = 1. Q.E.D. 17.15 Exercise. Prove the following. (a) lim( âˆš n + 1 âˆ’ âˆš n ) = 0 . Suppose Ç« > 0. We need to show that there exists N âˆˆ R such that for n âˆˆ N , n > N = â‡’ vextendsingle vextendsingle ( âˆš n + 1 âˆ’ âˆš n ) âˆ’ vextendsingle vextendsingle < Ç« . Let N = 1 Ç« 2 . Then, n > N = â‡’ n > 1 Ç« 2 = â‡’ vextendsingle vextendsingle âˆš n vextendsingle vextendsingle > 1 Ç« = â‡’ vextendsingle vextendsingle âˆš n + 1 + âˆš n vextendsingle vextendsingle > 1 Ç« = â‡’ 1 vextendsingle vextendsingle âˆš n + 1 + âˆš n vextendsingle vextendsingle < Ç« = â‡’ vextendsingle vextendsingle vextendsingle vextendsingle âˆš n + 1 âˆ’ âˆš n ( âˆš n + 1 + âˆš n )( âˆš n + 1 âˆ’ âˆš n ) vextendsingle vextendsingle vextendsingle vextendsingle < Ç« = â‡’ vextendsingle vextendsingle âˆš n + 1 âˆ’ âˆš n vextendsingle vextendsingle < Ç« = â‡’ vextendsingle vextendsingle ( âˆš n + 1 âˆ’ âˆš n ) âˆ’ vextendsingle vextendsingle < Ç« Therefore lim( âˆš n + 1 âˆ’ âˆš n ) = 0. Q.E.D....

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- Fall '08
- Akhmedov,A
- Mathematical analysis, Tn, Dominated convergence theorem, David Joseph Stith