# Math Help - Radius and Interval of Convergence

1. ## Radius and Interval of Convergence

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

2. Originally Posted by abel2

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

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?

3. 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.

4. 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.

5. $\displaystyle \lim_{n \to \infty} \left|\frac{x^{n+9}}{n+7} \cdot \frac{n+6}{x^{n+8}}\right| < 1$

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

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

$-1 < x < 1$ ... 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 $-1 < x \le 1$