## 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?

1. 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?

2. anonymous

actually it is$\sum_{0}^{\infty}ar^n$, my bad

3. anonymous

and the sum is a/1-r

4. anonymous

yeah, thats the right equation. what do i do with it?

5. anonymous

give me a little, I'm working on it

6. anonymous

ok, thanks

7. 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 $\sum_{0}^{\infty}5(-2/3)^n$

8. anonymous

oh that makes a lot of sense actually. that helps me a lot in understanding this

9. anonymous

cool, it's one of the few things I actually like to mess with, with infinite series stuff.

10. 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? $\sum_{n=1}^{\infty} 2\div (n ^{2} + 4n +3)$

11. 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.

12. 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

13. anonymous

Oh. I guess this stuff is pretty easy once you understand it. Thanks again

14. anonymous

it gets harder though, so keep your head up going into it.