# conditional convergence v.s. converges absolutely

• Apr 15th 2008, 05:55 PM
soleilion
[SOLVED]conditional convergence v.s. converges absolutely
1.After I use Alternating Series Test and result is converges
Which one do I say? converges or converges absolutely
Is it only way to prove series is converges absolutely by Ratio Test for Absolute Convergence???

2.I know how to prove the series is converges absolutely by Ratio Test for Absolute Convergence

How to prove a series is conditional convergence? Which method do I need to use?

Thank you!
• Apr 15th 2008, 06:26 PM
Mathstud28
Ok
Quote:

Originally Posted by soleilion
1.After I use Alternating Series Test and result is converges
Which one do I say? converges or converges absolutely
Is it only way to prove series is converges absolutely by Ratio Test for Absolute Convergence???

2.I know how to prove the series is converges absolutely by Ratio Test for Absolute Convergence

How to prove a series is conditional convergence? Which method do I need to use?

Thank you!

If a series $|a_n|$ converges, then it conditionally converges...but if $|a_n|$ and $a_n$ converges it is absolutely convergent
• Apr 15th 2008, 06:28 PM
soleilion
Quote:

Originally Posted by Mathstud28
If a series $|a_n|$ converges, then it conditionally converges...but if $|a_n|$ and $a_n$ converges it is absolutely convergent

ok
How to say a series is conditional converges
• Apr 15th 2008, 06:56 PM
Mathstud28
Ok
Quote:

Originally Posted by soleilion
ok
How to say a series is conditional converges

You would say a series is...o wait I am sorry I mistyped it is actually absolutely convergent if $|a_n|$ converges...for example since $\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{n}$ converges but $\sum_{n=0}^{\infty}\bigg|\frac{(-1)^{n+1}}{n}\bigg|=\frac{1}{n}$ diverges it is only conditionally convergent
• Apr 15th 2008, 07:00 PM
soleilion
Quote:

Originally Posted by Mathstud28
You would say a series is...o wait I am sorry I mistyped it is actually absolutely convergent if $|a_n|$ converges...for example since $\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{n}$ converges but $\sum_{n=0}^{\infty}\bigg|\frac{(-1)^{n+1}}{n}\bigg|=\frac{1}{n}$ diverges it is only conditionally convergent

that means if I can prove uk is converges and ak is diverges
then, I can say the series is conditional converges
right?
• Apr 15th 2008, 07:08 PM
Mathstud28
Quote:

Originally Posted by soleilion
that means if I can prove uk is converges and ak is diverges
then, I can say the series is conditional converges
right?

I have no idea what $u_k$ and $a_k$ is but if you could prove that a series converges but the absolute value of that series diverges it is conditionally convergent
• Apr 15th 2008, 07:09 PM
soleilion
Quote:

Originally Posted by Mathstud28
I have no idea what $u_k$ and $a_k$ is but if you could prove that a series converges but the absolute value of that series diverges it is conditionally convergent

ok
I understood
Thank you very much
• Apr 15th 2008, 07:10 PM
Mathstud28
Quote:

Originally Posted by soleilion
ok
I understood
Thank you very much

No problem...on the ones I didnt mistype(Doh)