anonymous
  • anonymous
what is a p-series?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • 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
anonymous
  • anonymous
could you help me find p on a p-series
anonymous
  • anonymous
let me take a look at it.

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anonymous
  • anonymous
hold on let me write the problem
anonymous
  • anonymous
\[\sum_{n=1}^{\infty} \sqrt[3]{(6n^5-6n^3+7n)} / (6n^5+2n^4-4)\]
anonymous
  • anonymous
gee whiz I wan't expecting that there, give me a few minutes
anonymous
  • anonymous
lol thank you
anonymous
  • anonymous
no problem, is that the 3rd root over the numerator?
anonymous
  • anonymous
yes sorry it was hard to make it look like it was only over the numerator ha
anonymous
  • anonymous
so they want you to resolve this thing down to some ort of series all taken up to the same power then?
anonymous
  • anonymous
heres an example.... 1/n and p=1
anonymous
  • anonymous
and if it was \[7/\sqrt{n}\] p=1/2
anonymous
  • 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.
anonymous
  • 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.
anonymous
  • anonymous
*expecting
anonymous
  • anonymous
lol idk either
anonymous
  • 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.
anonymous
  • anonymous
you did help me, thank you :)
anonymous
  • anonymous
no problem

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