## Requiem Group Title 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)? one year ago one year ago

1. Requiem Group Title

Or do i need to use a comparison test?

2. Requiem Group Title

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

3. satellite73 Group Title

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

4. Requiem Group Title

yes thats it

5. satellite73 Group Title

it not only converges, but it converges absolutely

6. satellite73 Group Title

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

7. Requiem Group Title

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 Group Title

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 Group Title

and i wasnt sure if the former was true

10. satellite73 Group Title

yeah only need the terms go to zero

11. Requiem Group Title

ok, thanks satellite

12. satellite73 Group Title

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

13. satellite73 Group Title

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

14. Requiem Group Title

its converges conditionally right?

15. satellite73 Group Title

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

16. Requiem Group Title

ok thanks you for your help! much appreciated

17. satellite73 Group Title

yw