Hello, I have the problem:

http://webwork.csufresno.edu/webwork...ced109bcf1.png

I get:

then:

and I get stuck after that, is it correct so far?

Thanks,

Matt

Printable View

- Oct 22nd 2008, 08:56 PMmatt3DAnother Series problem
Hello, I have the problem:

http://webwork.csufresno.edu/webwork...ced109bcf1.png

I get:

then:

and I get stuck after that, is it correct so far?

Thanks,

Matt - Oct 22nd 2008, 09:20 PMo_O
(Subtracted exponents)

So by the ratio test, the series absolutely converges and thus converges if - Oct 22nd 2008, 09:29 PMmatt3D
Thanks for the fast reply.(Hi) Okay, I know that you took out the 9 and put it on the outside and I know when you have something like but could you explain a little more how you got the new fraction in the second part where you have (subtracted exponents)?

- Oct 22nd 2008, 09:31 PMo_O
I factored out the 1/9 and simplified the exponents (it's colour-coded!): . The limit of the n-containing term is 1 so .. etc. etc.

- Oct 22nd 2008, 09:34 PMmatt3D
Oh, I see... I should of saw that earlier...(Shake)

Thank you! - Oct 22nd 2008, 10:22 PMmatt3D
Hmm, webwork is saying that my answer isn't correct, I have Is that correct?

- Oct 22nd 2008, 10:30 PMo_O
. You're

**subtracting**2 from all 3 sides of the inequality, not adding. - Oct 22nd 2008, 11:22 PMmatt3D
So what should the interval of convergence be? I don't understand exactly what you are saying.

- Oct 22nd 2008, 11:51 PMo_O
is what I'm saying. Test for convergence for and you'll find your interval of convergence.

- Oct 23rd 2008, 12:30 AMmatt3D
For -11 I got infinity and 7 came out to negative infinity when put in the series. So both do not converge. But I still can't get the interval of convergence. (Headbang)