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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 nonnegative series as our counterexample was for nonnegative 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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 Spring '10
 MetCalfe
 Calculus

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