## Requiem 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. Requiem

Or do i need to use a comparison test?

2. Requiem

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

3. satellite73

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

4. Requiem

yes thats it

5. satellite73

it not only converges, but it converges absolutely

6. satellite73

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

7. Requiem

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. satellite73

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. Requiem

and i wasnt sure if the former was true

10. satellite73

yeah only need the terms go to zero

11. Requiem

ok, thanks satellite

12. satellite73

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

13. satellite73

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

14. Requiem

its converges conditionally right?

15. satellite73

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

16. Requiem

ok thanks you for your help! much appreciated

17. satellite73

yw