Radius of Convergence and Absolute Convergence of a Series

[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

 

[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.

 

[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\} ??

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.

 

[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?