hw15solutions - 1 /n which diverges. For the second...

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Math 521 - Advanced Calculus I Instructor: J. Metcalfe Due: March 3, 2010 Assignment 15 1. Prove that the convergence of a n implies the convergence of X a n n if a n 0. Note that a n n a n , if a n 1 n 2 . And if a n 1 n 2 , then a n n 1 n 2 . Thus, 0 a n n max( a n , 1 /n 2 ) a n + 1 n 2 . Since a n converges and 1 /n 2 converges, we have that X ( a n + 1 /n 2 ) converges. By the basic comparison test, it follows that a n n converges. This could also easily be proved using the Cauchy-Schwarz inequality, if you recall that from multivariable calculus. Recall there that x · y ≤ k x kk y k . It then follows that k X n =1 a n n ± k X n =1 a n ² 1 / 2 ± k X n =1 1 n 2 ² 1 / 2 . And the Cauchy criterion for a n /n follows easily from those for a n and 1 /n 2 . 2. If a n and b n are both convergent series, is a n b n convergent? If a n and b n are both divergent series, is a n b n divergent? Do your answers change if it is assumed that a n and b n are non-negative for each n ? Justify your answers completely. The answer to the first question is no. By the alternating series test, ( - 1) n / n converges. Setting a n = b n = ( - 1) n / n , we see that a n b n =
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Unformatted text preview: 1 /n which diverges. For the second question, we can take a n = b n = 1 /n . Then a n = b n diverges (it is the harmonic series), yet a n b n = 1 /n 2 converges. 1 The answer clearly doesnt change for the second question for non-negative series as our counterexample was for non-negative series. The answer does change for the rst question. In particular, if a n ,b n 0, then a n b n 1 2 ( a 2 n + b 2 n ) . And if a n and b n converge, then a n 0 and b n 0. Thus, for n suciently large, we have | a n | 1 and | b n | 1 (denition of limit with = 1). Thus a 2 n a n and b 2 n b n for n suciently large. That is, a n b n 1 2 ( a n + b n ) but the sum of the right side converges. Hence by the basic comparison test, we have that a n b n converges. 2...
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hw15solutions - 1 /n which diverges. For the second...

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