Homework_Set_4_Solution

Homework_Set_4_Solution - Hw 4 solutions 10/19/2005...

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Unformatted text preview: Hw 4 solutions 10/19/2005 Solution 1 Suppose n =1 x n converges. We shall show that n =1 y n converges as well. From the definition of the limit, N such that | x n y n c | < c/ 2 n N . This implies, from the the fact that all terms involved are positive, that x n y n > c/ 2. Hence y n < 2 x n c n N and by the comparison test it follows that n = N y n converges. Clearly adding the finitely many terms y n , 1 n < N cannot affect convergence. Conversely suppose n =1 y n converges. We shall show that n =1 x n con- verges as well. From the definition of the limit, N such that | x n y n c | < c n N . This implies that x n y n < 2 c . Hence x n < 2 cy n n N and by the comparison test it follows that n = N x n converges. Clearly adding the finitely many terms x n , 1 n < N cannot affect convergence....
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Homework_Set_4_Solution - Hw 4 solutions 10/19/2005...

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