Very confused about the following:

Find the radius of convergence and interval of convergence of the series.

http://www.webassign.net/cgi-bin/sym...%2F%28n%2B6%29

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- Jul 1st 2010, 06:49 PMabel2Radius and Interval of Convergence
Very confused about the following:

Find the radius of convergence and interval of convergence of the series.

http://www.webassign.net/cgi-bin/sym...%2F%28n%2B6%29 - Jul 1st 2010, 08:09 PMJhevon
What have you tried? For this just use the ratio test. The interval of convergence is given by

$\displaystyle \displaystyle \lim_{n \to \infty} \left| \frac {a_{n + 1}}{a_n} \right| < 1$

where $\displaystyle \displaystyle a_n = \frac {x^{n + 8}}{n + 6}$

(Be sure to check the endpoints).

The radius of convergence, is 1/2 the length of the interval of convergence.

Can you continue? - Jul 2nd 2010, 06:35 AMabel2
Ok so using the ratio test where $\displaystyle \rho = lim_{n\rightarrow\infty}abs((a_{n+1})/(a_{n})) < 1 $

Then we have $\displaystyle lim_{n\rightarrow\infty} abs(\frac{\frac{(-1)^(n+1)*x^(n+9)}{(n+7)}}{\frac{(-1)^(n)*x^(n+8)}{(n+6)}}$

Simplifying, we get $\displaystyle \frac{(-1)^(n+1)*x^(n+9)*(n+6)}{(n+7)*(-1)^(n)*x^(n+8)}$

Then I get $\displaystyle =(-1)xlim_{n\rightarrow\infty}1/n$

$\displaystyle \rightarrow = lim_{n \rightarrow \infty} = 0 $

$\displaystyle \rho < 1$ and therefore converges.

Still not sure how to solve for the center point and find the two endpoints. I'm left with x = -1

I apologize in advance for my crappy latex. The higher parentheses are superscripts. And I'm not sure about the absolute value cmd. - Jul 2nd 2010, 08:12 AMabel2
Well that can't be right. 1/n = infinity. But I am sure this equation converges. Somehow I think I cancelled out the n's incorrectly. Should have ended up with a L'Hospital's method and gotten 1/1 is my suspicion. But if that is the case, wouldn't I just solve for x giving 0? But then I have no idea how to get the endpoints unless they are -1<x<1 where x=0. With radius of 1.

- Jul 2nd 2010, 08:57 AMskeeter
$\displaystyle \displaystyle \lim_{n \to \infty} \left|\frac{x^{n+9}}{n+7} \cdot \frac{n+6}{x^{n+8}}\right| < 1$

$\displaystyle \displaystyle |x| \cdot \lim_{n \to \infty} \frac{n+6}{n+7} < 1$

$\displaystyle \displaystyle |x| \cdot 1 < 1$

$\displaystyle -1 < x < 1$ ... radius of convergence is r = 1

checking the endpoints ...

when $\displaystyle x = 1$

$\displaystyle \displaystyle \sum_{n=2}^{\infty} \frac{(-1)^n \cdot 1}{n+6}$ ... alternating series that converges

when $\displaystyle x = -1$

$\displaystyle \displaystyle \sum_{n=2}^{\infty} \frac{(-1)^n \cdot (-1)^{n+8}}{n+6} = \sum_{n=2}^{\infty} \frac{(-1)^{2n+8}}{n+6} =\sum_{n=2}^{\infty} \frac{1}{n+6}$ ... series diverges

interval of convergence is $\displaystyle -1 < x \le 1$