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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 doesn’t change for the second question for nonnegative series as our counterexample was for nonnegative series. The answer does change for the ﬁrst 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 suﬃciently large, we have  a n  ≤ 1 and  b n  ≤ 1 (deﬁnition of limit with ε = 1). Thus a 2 n ≤ a n and b 2 n ≤ b n for n suﬃciently 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, Mathematical Series, J. Metcalfe

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