Another Series problem

**matt3D** · October 22nd 2008, 07:56 PM

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

**o_O** · October 22nd 2008, 08:20 PM

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

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

**matt3D** · October 22nd 2008, 08:29 PM

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)?

**o_O** · October 22nd 2008, 08:31 PM

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.

**matt3D** · October 22nd 2008, 08:34 PM

Oh, I see... I should of saw that earlier...

Thank you!

**matt3D** · October 22nd 2008, 09:22 PM

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

**o_O** · October 22nd 2008, 09:30 PM

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

**matt3D** · October 22nd 2008, 10:22 PM

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

**o_O** · October 22nd 2008, 10:51 PM

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

**matt3D** · October 22nd 2008, 11:30 PM

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.

Thread: Another Series problem

LinkBack

Thread Tools

Display

Another Series problem

Similar Math Help Forum Discussions

Exploiting geometric series with power series to find Taylors series

series problem

question about power series, taylor series, maclaurin series

help /w series problem

Series problem. Cant seem to get this one.

Search Tags