Prove that if ∑ a_n converges absolutely, then ∑ (a_n)^2 converges
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Prove that if ∑ a_n converges absolutely, then ∑ (a_n)^2 converges

[From: ] [author: ] [Date: 12-05-01] [Hit: ]
-a) For n sufficiently large,So,Since ∑ |a_n| converges (as ∑ a_n converges absolutely), we conclude that ∑ (a_n)^2 also converges by the Comparison Test.b) This is false; consider ∑ (-1)^n / n^(1/4), which only converges conditionally,......
hi
i need help !

a) prove that if ∑ a_n (n=1 to infinity) converges absolutely, then ∑ (a_n)^2 (n=1 to infinity) converges.
b) if ∑ a_n (n=1 to infinity) converges, does ∑ (a_n)^2 (n=1 to infinity) necessarily converge?

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a) For n sufficiently large, we have |a_n| < 1.
So, (a_n)^2 < |a_n| for sufficiently large n.

Since ∑ |a_n| converges (as ∑ a_n converges absolutely), we conclude that ∑ (a_n)^2 also converges by the Comparison Test.

b) This is false; consider ∑ (-1)^n / n^(1/4), which only converges conditionally, but
∑ [(-1)^n / n^(1/4)]^2 = ∑ 1/n^(1/2) is a divergent p-series.

I hope this helps!

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X = 7
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