Question

I have this problem: It is a series from n=0 to infinity. The series is ((3^n)-1)/2^n.
I know that it diverges. My question is, how do you prove that? Should I break it up into two pieces, one of which is infinity, and the other goes to zero?

Answer 1

\[\sum_{n=0}^{\infty}\frac{ 3^{n} -1 }{ 2^{n} }\] correct?

Answer 2

yes, thats it. Sorry, I don't have a math symbol plug in

Answer 3

No worries.

\[\sum_{n=0}^{\infty}\frac{ 3^{n} -1 }{ 2^{n} }= \sum_{n=0}^{\infty}\frac{ 3^{n} }{ 2^{n} }-\sum_{n=0}^{\infty}\frac{ 1 }{ 2^{n} }=\sum_{n=0}^{\infty}(\frac{ 3 }{ 2 })^{n}-\sum_{n=0}^{\infty}(\frac{ 1 }{ 2 })^{n}\]

Do those two series look like something familiar to you? :)

Answer 4

Yes, thats what I was thinking of doing. A similar problem in my book suggests doing that with the difference rule. I just don't understand if that is necessary to prove that there is no convergence?

Answer 5

Well, what kind of series are those?

Answer 6

Geometric I suppose, arent they? I don't really know. I am new with these series

Answer 7

Yes, they are geometric. I understand that you're new, but you'll have to be able to identify them as such eventually :) So these are both geometric series and there are conditions for convergence or divergence for a geometric series. Remember what they are by chance?

Answer 8

Yes, if they are above or below one. Above goes to infinity and diverges and below one goes to zero.

Answer 9

When common ratio is less than 1, the term eventually goes to zero. Not the sum.

Answer 10

Not quite. To be more specific: A geometric series of the form \[\sum_{n=0}^{\infty}ar^{n}\]converges if |r| <1. Have to be careful, though, those are absolute value bars. So we converge if -1< r < 1. And diverge if this is not true. For your first series, you can see r = 3/2 and for the second series, r = 1/2. Separately, the first series diverges since -1 < r < 1 is not true while the second one would converge. If any one of the series you have diverges, though, the whole thing diverges. So this is why your series diverges.

Answer 11

oh, ok, that makes sense. I though about the first part being 3^n/2^n, and that starting at 3/2 when n=1but I figured that would be breaking a rule somehow. I can see how this works though, if one goes to infinity, it doesn't matter if the other part does have some sum, as it's clearly not bigger than infinity. Thanks a lot for your help! This was my first question here.

Answer 12

Well, given that you have groups of multiplication and division with a common exponent, that exponent can be factored out. As long as I'm not trying to factor out an exponent through +'s and -'s, I'm okay. For the 2nd series, it's knowing that even though I only have a 1, I can treat it as if it's 1^n, which allows me to get it to match the form of a geo series.

Answer 13

Awesome answer, thank you so much

Answer 14

No problem. Good luck :)