Question

what is a p-series?

Answer 1

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

Answer 2

could you help me find p on a p-series

Answer 3

let me take a look at it.

Answer 4

hold on let me write the problem

Answer 5

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

Answer 6

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

Answer 7

lol thank you

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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.

Answer 14

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.

Answer 15

*expecting

Answer 16

lol idk either

Answer 17

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.

Answer 18

you did help me, thank you :)

Answer 19

no problem