# Absolute converge of a series

1. May 28, 2006

### Mathman23

Hello

I have this problem here:

Given the series

$$\sum_{n=1} ^{\infty} \frac{x^{n+1}}{n(n+1)}$$

show that this converges for every $$x \in \{ w \in \mathbb{C} \| \|w \| \leq 1 \}$$

Solution:

Since

$$\sum_{n=1} ^{\infty} \left| \frac{x^{n+1}}{n(n+1)} \right| = \sum_{n=1} ^{\infty} \frac{1}{n(n+1)} \right|$$

Since $$\sum_{n=1} ^{\infty} \frac{1}{n(n+1)} \right|$$ convergence then according the definition then

$$\sum_{n=1} ^{\infty} \frac{x^{n+1}}{n(n+1)}$$ is absolute convergent.

Thus

$$x \in \{ w \in \mathbb{C} \| \|w \| \leq 1 \}$$. Right?

Best Regards
Fred

Last edited: May 28, 2006
2. May 28, 2006

### shmoe

This:

$$\sum_{n=1} ^{\infty} \left| \frac{x^{n+1}}{n(n+1)} \right| = \sum_{n=1} ^{\infty} \frac{1}{n(n+1)} \right|$$

shouldn't be an equality. You hopefully meant an inequality here which is valid for x on the unit disc. Otherwise it looks good.

3. May 28, 2006

### Hummingbird25

Then it should have say ?

$$\sum_{n=1} ^{\infty} \left| \frac{x^{n+1}}{n(n+1)} \right| \leq \sum_{n=1} ^{\infty} \frac{1}{n(n+1)} \right|$$

Regards Humminbird

p.s. Do I use uniform continuety to show that

$$g(x) = \sum_{n=1} ^{\infty} \frac{x^{n+1}}{n(n+1)}$$

is continious on $$x \in \{ w \in \mathbb{C} \| \|w \| \leq 1\}$$ ??

Last edited: May 28, 2006
4. May 28, 2006

### HallsofIvy

No, if
$$\sum_{n=1} ^{\infty} \frac{x^{n+1}}{n(n+1)}$$
then all you can say is "if x= 1 this is
$$\sum_{n=1} ^{\infty} \frac{1}{n(n+1)}$$
which converges absolutely" (since all terms are positive anyway).

Of course, I'm not sure why you chose to make x= 1. Have you determined what the radius of convergence is? And if the radius of convergence is 1, what happens when x= -1?