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anonymous
 5 years ago
So the question is "find a convergent geometric series with a first term of 5 and sum of 3." Can anyone explain how to do this?
anonymous
 5 years ago
So the question is "find a convergent geometric series with a first term of 5 and sum of 3." Can anyone explain how to do this?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cool question, so I think that the form for a geometric series is \[\sum_{0}^{\infty}ar^(n1)\], that is a times r to the n1 right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually it is\[\sum_{0}^{\infty}ar^n\], my bad

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, thats the right equation. what do i do with it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0give me a little, I'm working on it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no sweat, okay so we have those two relations right, and we know we want the first term to be = 5, and that n= 0 there. So \[5=a(r^0)\], a = 5.Now, we want our sum to be equal to 3 so \[a/(1r) =3 \]\[5/(1r)=3; r =( 2/3)\] The series is then \[\sum_{0}^{\infty}5(2/3)^n\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh that makes a lot of sense actually. that helps me a lot in understanding this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cool, it's one of the few things I actually like to mess with, with infinite series stuff.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so another question is to determine whether the series is convergent or divergent. The first step is to write this as the sum of a series right? \[\sum_{n=1}^{\infty} 2\div (n ^{2} + 4n +3)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, that's the first step. I think for convergence of the series, you have to calc the limit as n goes to infinity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can't solve it by having 0 in the denominator, so factor out an n squared \[\sum_{0}^{\infty}[(n^2)(1/n^2)]/[n^2(1+(4/n)+(3/n^2)]\], then the n^2 cancels out and you have 0/1, and the series conv to 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh. I guess this stuff is pretty easy once you understand it. Thanks again

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it gets harder though, so keep your head up going into it.
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