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hw7solutions - to 1 while s n does not converge 2 Prove...

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Math 521 - Advanced Calculus I Instructor: J. Metcalfe Due: February 5, 2010 Assignment 7 1. Prove that convergence of { s n } implies convergence of {| s n |} . Is the converse true? If s n s , then for any ε > 0 there is an N N so that | s n - s | < ε whenever n N . For any fixed ε , simply choose the same N as above. Then by the reverse triangle inequality, we have || s n | - | s || ≤ | s n - s | < ε which shows that | s n | → | s | . The converse is not true as if s n = ( - 1) n , then | s n | = 1 and {| s n |} converges trivially
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Unformatted text preview: to 1 while s n does not converge. 2. Prove that if lim x n = x and if x > 0, then there exists a natural number M such that x n > for all n ≥ M . Set ε = x/ 2. This is positive by assumption. Applying the definition of convergence, there is M ∈ N so that | x n-x | < x/ 2 , ∀ n ≥ M. or-x/ 2 < x n-x < x/ 2 . The left inequality implies that x n > x/ 2 > 0 for n ≥ M . 1...
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