## anonymous 5 years ago what is a p-series?

1. anonymous

it is a summation the is set up (An)^n$\sum_{?}^{?}(A _{n})^n$, it's a series that is made to a power

2. anonymous

could you help me find p on a p-series

3. anonymous

let me take a look at it.

4. anonymous

hold on let me write the problem

5. anonymous

$\sum_{n=1}^{\infty} \sqrt[3]{(6n^5-6n^3+7n)} / (6n^5+2n^4-4)$

6. anonymous

gee whiz I wan't expecting that there, give me a few minutes

7. anonymous

lol thank you

8. anonymous

no problem, is that the 3rd root over the numerator?

9. anonymous

yes sorry it was hard to make it look like it was only over the numerator ha

10. anonymous

so they want you to resolve this thing down to some ort of series all taken up to the same power then?

11. anonymous

heres an example.... 1/n and p=1

12. anonymous

and if it was $7/\sqrt{n}$ p=1/2

13. anonymous

I could see$\sum_{?}^{?}(1/(6n^5+2n^4-4))*(6n^5-6n^3+7n)^(1/3)$ that is to the (1/3) toward the end there. In that case the power it is taken to is (1/3). I just don't see anyway of reducing the n's down so it is more clear. I guess if I were answering it, I would say (1/3) is the only determinable power. It's not as if the numerator and denominater have like bases to see it being anything else. Neither one of them are factorable from wht I can see.

14. anonymous

unless in some sick way they are expection 1-1/3 = 2/3, but I really can't see that, because like I said the bases would have to be the same.

15. anonymous

*expecting

16. anonymous

lol idk either

17. anonymous

I ws thinking you had something vageuly reducible that could be taken up to a single power. I've had stuff like that thrown at me, but jsut to say what the power is alone, is different. Sorry I could not be more help.

18. anonymous

you did help me, thank you :)

19. anonymous

no problem