# Radius of convergence / endpoints

• Apr 26th 2009, 04:25 PM
VkL
I am really having trouble with this problem. The n'th term is given in the problem,

http://www.webassign.net/www22/symIm...fc23e2e01e.gif
(a) Determine the radius of convergence of the series.
(b) What are the endpoints of the interval of convergence?
left endpoint = x =
right endpoint = x =
Does the series converge at the left endpoint?
Does the series converge at the right endpoint?

I did the ratio test and got [(-1^n)*(x^n+1) / (n+1)] * [n / -1^(n-1) * x^n]

I get infinity when i get an x in the numerator when i take the limit. Usually the x's cancel out. is this a special case? maybe i am doing something wrong, or don't know something. And I have no clue on how to find the end points even.

• Apr 26th 2009, 05:21 PM
skeeter
Quote:

Originally Posted by VkL
I am really having trouble with this problem. The n'th term is given in the problem,

http://www.webassign.net/www22/symIm...fc23e2e01e.gif
(a) Determine the radius of convergence of the series.
(b) What are the endpoints of the interval of convergence?
left endpoint = x =
right endpoint = x =
Does the series converge at the left endpoint?
Does the series converge at the right endpoint?

I did the ratio test and got [(-1^n)*(x^n+1) / (n+1)] * [n / -1^(n-1) * x^n]

I get infinity when i get an x in the numerator when i take the limit. Usually the x's cancel out. is this a special case? maybe i am doing something wrong, or don't know something. And I have no clue on how to find the end points even.

$\displaystyle \lim_{n \to \infty} \left|\frac{(-1)^n x^{n+1}}{n+1} \cdot \frac{n}{(-1)^{n-1} x^n}\right| < 1$

since we're working with the absolute value of the ratio, the (-1) factors can be ignored. simplifying ...

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

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

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

$\displaystyle -1 < x < 1$

I leave you to check the endpoints for possible convergence.
• Apr 26th 2009, 09:52 PM
VkL
on step 3, why was the absolute value of x taken out of the limit?
• Apr 26th 2009, 10:04 PM
redsoxfan325
Quote:

Originally Posted by VkL
on step 3, why was the absolute value of x taken out of the limit?

Because $\displaystyle n>0$, so it is irrelevant.

For the endpoints, plugging in $\displaystyle 1$ gives you $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{n}$. This is the alternating harmonic series, so what can we say about it?

Plugging in $\displaystyle -1$ gives you $\displaystyle \sum_{n=1}^{\infty}\frac{-1}{n}$. This is the negative harmonic series, so what can we say about it?

Spoiler:
The first one converges (to $\displaystyle \ln 2$) and the second one diverges. So this power series converges on $\displaystyle x=(-1,1]$.