# Thread: Convergence at limit points

1. ## Convergence at limit points

Here's another question a student hit me with tonight. (I'm not having a good night!)
What is the radius of convergence of the sum
$\displaystyle \sum_{n = 0}^{\infty}\frac{(-1)^n}{ln(n)}x^n$
The answer, if I am not mistaken is $\displaystyle |x| > 1$.

So then the student asks me, what about $\displaystyle x = \pm 1$?

I have no answer to that. Could someone please explain how to find that out? Thanks again!

-Dan

2. Originally Posted by topsquark
Here's another question a student hit me with tonight. (I'm not having a good night!)

The answer, if I am not mistaken is $\displaystyle |x| > 1$.

So then the student asks me, what about $\displaystyle x = \pm 1$?

I have no answer to that. Could someone please explain how to find that out? Thanks again!

-Dan
First, I think you want to start from n = 2, not 0 (or 1).

x = 1 is easy. you have an alternating series which will converge (by the Cauchy alternating series test) since $\displaystyle a_n > a_{n+1}$ and $\displaystyle a_n \rightarrow 0$.

x = -1. All terms are positive. And since $\displaystyle \frac{1}{\ln n} > \frac{1}{n}$ a basic comparison test will show that this series diverges.

3. A more interesting question is if $\displaystyle x$ is a complex variable, and to find all points on the unit disk where it converges.