Question

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?

Answer 1

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?

Answer 2

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

Answer 3

and the sum is a/1-r

Answer 4

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

Answer 5

give me a little, I'm working on it

Answer 6

ok, thanks

Answer 7

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 \[\sum_{0}^{\infty}5(-2/3)^n\]

Answer 8

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

Answer 9

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

Answer 10

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? \[\sum_{n=1}^{\infty} 2\div (n ^{2} + 4n +3)\]

Answer 11

yeah, that's the first step. I think for convergence of the series, you have to calc the limit as n goes to infinity.

Answer 12

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

Answer 13

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

Answer 14

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