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319HW6Correction

# 319HW6Correction - MAT 319/320 Correction of HW6 Exercise 1...

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MAT 319/320 Correction of HW6 Exercise 1. Section 3.6, #1. Proof. Pick α 1 =1 . Since the sequence is unbounded, one can find x n 1 > α 1 . Pick α 2 = max (2 , x n 1 +1) . For the same reason, one can find x n 2 > α 2 . Continue like this: by construction, the subsequence is increasing and satisfies x n k > k , there- fore it is properly divergent. square Exercise 2. Section 3.6, #8d. Proof. Read Example 3.4.6(c).In it, they construct two sequences ( n k ) k greaterorequalslant 1 and ( m k ) k greaterorequalslant 1 of nat- ural numbers, such that sin ( n k ) [1/2 , 1] and sin ( m k ) [ - 1 , - 1/2] . By taking subsequences and using Bolzano-Weierstrass, one can even assume that sin ( n k ) converges to c 1 [1/2 , 1] and that sin ( m k ) c 2 [ - 1 , - 1/2] . Now if you consider the subsequences ( n k ) = ( n k 2 ) and ( m k ) = ( m k 2 ) , they are such that sin ( n k radicalbig ) and sin ( m k radicalbig ) converge towards two distinct numbers, there- fore the sequence sin ( n ) cannot converge.

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