anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
cool question, so I think that the form for a geometric series is \[\sum_{0}^{\infty}ar^(n-1)\], that is a times r to the n-1 right?
anonymous
  • anonymous
actually it is\[\sum_{0}^{\infty}ar^n\], my bad
anonymous
  • anonymous
and the sum is a/1-r

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anonymous
  • anonymous
yeah, thats the right equation. what do i do with it?
anonymous
  • anonymous
give me a little, I'm working on it
anonymous
  • anonymous
ok, thanks
anonymous
  • anonymous
no 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 \[a/(1-r) =3 \]\[5/(1-r)=3; r =( -2/3)\] The series is then \[\sum_{0}^{\infty}5(-2/3)^n\]
anonymous
  • anonymous
oh that makes a lot of sense actually. that helps me a lot in understanding this
anonymous
  • anonymous
cool, it's one of the few things I actually like to mess with, with infinite series stuff.
anonymous
  • anonymous
so 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
  • anonymous
yeah, that's the first step. I think for convergence of the series, you have to calc the limit as n goes to infinity.
anonymous
  • anonymous
you 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
  • anonymous
Oh. I guess this stuff is pretty easy once you understand it. Thanks again
anonymous
  • anonymous
it gets harder though, so keep your head up going into it.

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