# Radius and Interval of Convergence

• July 1st 2010, 06:49 PM
abel2

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

http://www.webassign.net/cgi-bin/sym...%2F%28n%2B6%29
• July 1st 2010, 08:09 PM
Jhevon
Quote:

Originally Posted by abel2

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

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

What have you tried? For this just use the ratio test. The interval of convergence is given by

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

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

(Be sure to check the endpoints).

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

Can you continue?
• July 2nd 2010, 06:35 AM
abel2
Ok so using the ratio test where $\rho = lim_{n\rightarrow\infty}abs((a_{n+1})/(a_{n})) < 1$

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

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

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

$\rightarrow = lim_{n \rightarrow \infty} = 0$
$\rho < 1$ 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.
• July 2nd 2010, 08:12 AM
abel2
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.
• July 2nd 2010, 08:57 AM
skeeter
$\displaystyle \lim_{n \to \infty} \left|\frac{x^{n+9}}{n+7} \cdot \frac{n+6}{x^{n+8}}\right| < 1$

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

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

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

checking the endpoints ...

when $x = 1$

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

when $x = -1$

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

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