absolute convergence of infinite series

**izzydoesit** · April 21st 2008, 06:46 AM

I need to show that if a series (an) is absolutely convergent then, the series (a^2 n) absolutely converges. Can anyone help with this. I thought of using the comparison test but wasnt sure if I was heading in the right direction. Any help would be appreciated.

**colby2152** · April 21st 2008, 07:00 AM

Originally Posted by izzydoesit

I need to show that if a series (an) is absolutely convergent then, the series (a^2 n) absolutely converges. Can anyone help with this. I thought of using the comparison test but wasnt sure if I was heading in the right direction. Any help would be appreciated.

What series are you talking about? You will find that the comparison test is your best weapon against proving the convergence/divergence of a series.

**mr fantastic** · April 21st 2008, 07:01 AM

Originally Posted by izzydoesit

I need to show that if a series (an) is absolutely convergent then, the series (a^2 n) absolutely converges. Can anyone help with this. I thought of using the comparison test but wasnt sure if I was heading in the right direction. Any help would be appreciated.

Posts #3 and #7 of this thread: http://www.mathhelpforum.com/math-he...questions.html

might get you started.

**izzydoesit** · April 21st 2008, 07:47 AM

The series is(a sub n) and it converges absolutely. It has to be shown that ( a sub n )^2 also is absolutely convergent. Im sorry. It is the summation of those series.

**Plato** · April 21st 2008, 08:57 AM

Because $\sum\limits_{n = 1}^\infty {\left| {a_n } \right|}$ converges, we know that $\left( {\exists N} \right)\left[ {n \geqslant N \Rightarrow \quad \left| {a_n } \right| < 1} \right]$ .
But $\left| {a_n } \right| < 1 \Rightarrow \quad \left( {a_n } \right)^2 = \left| {a_n } \right|^2 < \left| {a_n } \right|$ so by simple comparison we have it.

**mr fantastic** · April 21st 2008, 03:12 PM

Originally Posted by izzydoesit

The series is(a sub n) and it converges absolutely. It has to be shown that ( a sub n )^2 also is absolutely convergent. Im sorry. It is the summation of those series.

Originally Posted by mr fantastic

Posts #3 and #7 of this thread: http://www.mathhelpforum.com/math-he...questions.html

might get you started.

I never said the solution was there. I suggested it would get you started ...... A slight modification of the arguments was all that was required .....
Did you consider that?

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