# Math Help - Another Series problem

1. ## Another Series problem

Hello, I have the problem:

I get:
$\lim_{n \to \infty} \left|\frac{(n+1)^9(x+2)^{n+1}}{9^{n+1}(n+1)^{29/3}} \cdot \frac{(9^n)(n^{29/3})}{n^9(x+2)^n} \right|$
then:
$\left |x+2 \right|\lim_{n \to \infty} \frac{(n+1)^9(n^{29/3})}{9(n+1)^{29/3}n^9}$
and I get stuck after that, is it correct so far?
Thanks,
Matt

2. $\left |x+2 \right|\lim_{n \to \infty} \frac{{\color{red}(n+1)^9}{\color{blue}(n^{29/3})}}{9{\color{red}(n+1)^{29/3}}{\color{blue}n^9}} = \frac{|x+2|}{9} \lim_{n \to \infty} \frac{n^{2/3}}{(n+1)^{2/3}}$ (Subtracted exponents)

So by the ratio test, the series absolutely converges and thus converges if $\frac{|x+2|}{9} < 1 \iff |x+2| < 9 \iff \hdots$

3. Thanks for the fast reply. Okay, I know that you took out the 9 and put it on the outside and I know when you have something like $\frac{x^7}{x^5}=x^2$ but could you explain a little more how you got the new fraction in the second part where you have (subtracted exponents)?

4. I factored out the 1/9 and simplified the exponents (it's colour-coded!): $\frac{29}{3} -3 = \frac{29}{3} - \frac{27}{3} = \frac{2}{3}$. The limit of the n-containing term is 1 so .. etc. etc.

5. Oh, I see... I should of saw that earlier...
Thank you!

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

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

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

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

10. 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.