# Thread: conditional convergence v.s. converges absolutely

1. ## [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!

2. ## Ok

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 $\displaystyle |a_n|$ converges, then it conditionally converges...but if $\displaystyle |a_n|$ and $\displaystyle a_n$ converges it is absolutely convergent

3. Originally Posted by Mathstud28
If a series $\displaystyle |a_n|$ converges, then it conditionally converges...but if $\displaystyle |a_n|$ and $\displaystyle a_n$ converges it is absolutely convergent
ok
How to say a series is conditional converges

4. ## Ok

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 $\displaystyle |a_n|$ converges...for example since $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{n}$ converges but $\displaystyle \sum_{n=0}^{\infty}\bigg|\frac{(-1)^{n+1}}{n}\bigg|=\frac{1}{n}$ diverges it is only conditionally convergent

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

6. 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 $\displaystyle u_k$ and $\displaystyle a_k$ is but if you could prove that a series converges but the absolute value of that series diverges it is conditionally convergent

7. Originally Posted by Mathstud28
I have no idea what $\displaystyle u_k$ and $\displaystyle a_k$ 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

8. Originally Posted by soleilion
ok
I understood
Thank you very much
No problem...on the ones I didnt mistype