Pre-Calc Homework Solutions 371

# Pre-Calc Homework Solutions 371 - Section 9.2 55 Given an 0...

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55. Given an e . 0, by definition of convergence there corresponds an N such that for all n , N , ) L 1 2 a n ) , e and ) L 2 2 a n ) , e . (There is one such number for each series, and we may let N be the larger of the two numbers.) Now ) L 2 2 L 1 ) 5 ) L 2 2 a n 1 a n 2 L 1 ) # ) L 2 2 a n ) 1 ) a n 2 L 1 ) , e 1 e 5 2 e . ) L 2 2 L 1 ) , 2 e says that the difference between two fixed values is smaller than any positive number 2 e . The only nonnegative number smaller than every positive number is 0, so ) L 2 2 L 1 ) 5 0 or L 1 5 L 2 . 56. Consider the two subsequences a k ( n ) and a i ( n ) , where lim n a k ( n ) 5 L 1 ,lim n a i ( n ) 5 L 2 , and L 1 ± L 2 . Given an e . 0 there corresponds an N 1 such that for k ( n ) . N 1 , ) a k ( n ) 2 L 1 ) , e , and an N 2 such that for i ( n ) . N 2 , ) a i ( n ) 2 L 2 ) , e . Assume a n converges. Let N 5 max { N 1 , N 2 }. Then for n . N , we have that ) a n 2 L 1 ) , e and ) a n 2 L 2 ) , e for infinitely many n . This implies that lim
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