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April 26th, 2010, 08:35 AM  #1 
Member Joined: Oct 2009 Posts: 85 Thanks: 0  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. 
April 26th, 2010, 08:45 AM  #2 
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Re: Abs. Conv. implies Abs. Conv. Squared
If converges (absolutely), then what can you say about ? Then, which is bigger, or ? 
April 26th, 2010, 08:57 AM  #3 
Member Joined: Oct 2009 Posts: 85 Thanks: 0  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.

April 26th, 2010, 09:36 AM  #4  
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Re: Abs. Conv. implies Abs. Conv. Squared Quote:
Quote:
 
April 26th, 2010, 09:50 AM  #5 
Member Joined: Oct 2009 Posts: 85 Thanks: 0  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?

April 26th, 2010, 10:31 AM  #6 
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  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. 
April 26th, 2010, 11:56 AM  #7 
Member Joined: Oct 2009 Posts: 85 Thanks: 0  Re: Abs. Conv. implies Abs. Conv. Squared
Once again cknapp, you own. Thank you so much for all your help.


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