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April 26th, 2010, 08:35 AM   #1
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Abs. Conv. implies Abs. Conv. Squared

If is absolute convergent, then is absolute convergent.

How do I attempt to prove this? The only thing I know that is true is that if you multiply a abs. conv. series by another abs. conv. series that the product is another abs. conv. series.
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April 26th, 2010, 08:45 AM   #2
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Re: Abs. Conv. implies Abs. Conv. Squared

If converges (absolutely), then what can you say about ?

Then, which is bigger, or ?
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April 26th, 2010, 08:57 AM   #3
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Re: Abs. Conv. implies Abs. Conv. Squared

as . Then should be bigger. So since is bigger than this proves that is convergent? If what I just said is correct, why is it correct because I do not understand.
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April 26th, 2010, 09:36 AM   #4
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Re: Abs. Conv. implies Abs. Conv. Squared

Quote:
Originally Posted by HairOnABiscuit
as .
Correct.

Quote:
Then should be bigger.
Really? a_k*a_k > a_k, when a_k<1?
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April 26th, 2010, 09:50 AM   #5
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Re: Abs. Conv. implies Abs. Conv. Squared

Ok my bad is bigger <blushes>. So since and from the fact that converges (limit is equal to 0), must converge because must equal to less than infinity from , am I making any sense of this?
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April 26th, 2010, 10:31 AM   #6
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Re: Abs. Conv. implies Abs. Conv. Squared

That's almost it; but just because converges (to 0), doesn't necessarily mean converges.

But! Since (for big enough k) we know that .
So the sum of squares converges.
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April 26th, 2010, 11:56 AM   #7
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Re: Abs. Conv. implies Abs. Conv. Squared

Once again cknapp, you own. Thank you so much for all your help.
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