## anonymous 2 years ago Determine convergence or divergence. (-1)^n/(n^3) I know that I want to use the alternating series test. I used (bsubn) as 1/n^3 and (bsubn+1) as 1/(n+1)^3. I know that (bsubn) as n ----> infinity equals 0, but is (bsubn+1) less than or equal to (bsubn)?

1. anonymous

Or do i need to use a comparison test?

2. anonymous

I know that those two conditions have to be met for it to be convergent...

3. anonymous

is it $\sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}$

4. anonymous

yes thats it

5. anonymous

it not only converges, but it converges absolutely

6. anonymous

all you need for the alternating series to converge is that the terms go to zero, which thsee certainly do

7. anonymous

oh ok..in my book it says that (bsubn+1) has to be less than or equal to (bsubn) but also that (bsubn) has to equal 0 when the limit is taken...

8. anonymous

but $\sum_{n=1}^{\infty}\frac{1}{n^3}$ also converges, so this series is not just convergent, it is absolutely convergent as well

9. anonymous

and i wasnt sure if the former was true

10. anonymous

yeah only need the terms go to zero

11. anonymous

ok, thanks satellite

12. anonymous

$\sum\frac{(-1)^n}{n}$ for example converges, although not absolutely

13. anonymous

hell, even $\sum\frac{(-1)^n}{\ln(n)}$ converges

14. anonymous

its converges conditionally right?

15. anonymous

the last two i wrote are only conditionally convergent, the one you posted is absolutely convergent, a stronger condition

16. anonymous

ok thanks you for your help! much appreciated

17. anonymous

yw