My Math Forum Abs. Conv. implies Abs. Conv. Squared

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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 $\sum_{k=1}^\infty a_k$ is absolute convergent, then $\sum_{k=1}^{\infty} a^2_k$ 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 $\sum{a_k}$ converges (absolutely), then what can you say about $\lim a_k$? Then, which is bigger, $|a_k|$ or $a_k^2$?
 April 26th, 2010, 08:57 AM #3 Member   Joined: Oct 2009 Posts: 85 Thanks: 0 Re: Abs. Conv. implies Abs. Conv. Squared $\lim a_k= 0$ as $k \to \infty$. Then $a_k^2$ should be bigger. So since $a_k^2$ is bigger than $|a_k|$ this proves that $a^2_k$ 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
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Re: Abs. Conv. implies Abs. Conv. Squared

Quote:
 Originally Posted by HairOnABiscuit $\lim a_k= 0$ as $k \to \infty$.
Correct.

Quote:
 Then $a_k^2$ should be bigger.
Really? a_k*a_k > a_k, when a_k<1?

 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 $|a_k|$ is bigger . So since $|a_k| > a^2_k$ and from the fact that $|a_k|$ converges (limit is equal to 0), $a^2_k$ must converge because $\lim_{k\to \infty} a^2_k$ must equal to less than infinity from $a_k^2 < |a_k|$, 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 $a_k^2$ converges (to 0), doesn't necessarily mean $\sum a_k^2$ converges. But! Since $a_k^2<|a_k|$ (for big enough k) we know that $\sum a_k^2 \leq \sum |a_k|$. 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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