
Another Series problem
Hello, I have the problem:
http://webwork.csufresno.edu/webwork...ced109bcf1.png
I get:
$\displaystyle \lim_{n \to \infty} \left\frac{(n+1)^9(x+2)^{n+1}}{9^{n+1}(n+1)^{29/3}} \cdot \frac{(9^n)(n^{29/3})}{n^9(x+2)^n} \right$
then:
$\displaystyle \left x+2 \right\lim_{n \to \infty} \frac{(n+1)^9(n^{29/3})}{9(n+1)^{29/3}n^9}$
and I get stuck after that, is it correct so far?
Thanks,
Matt

$\displaystyle \left x+2 \right\lim_{n \to \infty} \frac{{\color{red}(n+1)^9}{\color{blue}(n^{29/3})}}{9{\color{red}(n+1)^{29/3}}{\color{blue}n^9}} = \frac{x+2}{9} \lim_{n \to \infty} \frac{n^{2/3}}{(n+1)^{2/3}}$ (Subtracted exponents)
So by the ratio test, the series absolutely converges and thus converges if $\displaystyle \frac{x+2}{9} < 1 \iff x+2 < 9 \iff \hdots$

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 $\displaystyle \frac{x^7}{x^5}=x^2$ but could you explain a little more how you got the new fraction in the second part where you have (subtracted exponents)?

I factored out the 1/9 and simplified the exponents (it's colourcoded!): $\displaystyle \frac{29}{3} 3 = \frac{29}{3}  \frac{27}{3} = \frac{2}{3}$. The limit of the ncontaining term is 1 so .. etc. etc.

Oh, I see... I should of saw that earlier...(Shake)
Thank you!

Hmm, webwork is saying that my answer isn't correct, I have $\displaystyle 7 \leq x<11$ Is that correct?

$\displaystyle x+2 < 9 \ \iff \  9 < x + 2 < 9$. You're subtracting 2 from all 3 sides of the inequality, not adding.

So what should the interval of convergence be? I don't understand exactly what you are saying.

$\displaystyle 9 < x + 2 < 9 \ \iff \ 11 < x < 7$ is what I'm saying. Test for convergence for $\displaystyle x = 11, 7$ and you'll find your interval of convergence.

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)