# Math Help - Radius of Convergence of Power Series

1. ## Radius of Convergence of Power Series

Hello,

I need a little bit help with the following power series. The exercise is to calculate the radius of convergence.

$\sum\limits_{n=1}^{\infty} (\mathrm{ln}(n))^{n} x^{n}$

I know the criterion of Euler and the stronger criterion of Hadamard.

Euler:

$r= \frac{\mid a_{n} \mid}{ \mid a_{n+1} \mid }$

$r= \frac{1}{\mathrm{limsup} ~ \sqrt[n]{\mid a_{n} \mid} }$

and of course $a_{n}$ is

$a_{n} = (\mathrm{ln(n)})^{n}$

My problem now, is that I don't know how to calculate this expressions with this logarithm

Thanks for help

2. Originally Posted by Besserwisser
Hello,

I need a little bit help with the following power series. The exercise is to calculate the radius of convergence.

$\sum\limits_{n=1}^{\infty} (\mathrm{ln})^{n} x^{n}$

I know the criterion of Euler and the stronger criterion of Hadamard.

Euler:

$r= \frac{\mid a_{n} \mid}{ \mid a_{n+1} \mid }$

$r= \frac{1}{\mathrm{limsup} ~ \sqrt[n]{\mid a_{n} \mid} }$

and of course $a_{n}$ is

$a_{n} = \mathrm{ln}^{n}$

My problem now, is that I don't know how to calculate this expressions with this logarithm

Thanks for help
Is it $\sum_{n = 1}^{\infty}(\ln{x})^nx^n$?

3. Originally Posted by Prove It
Is it $\sum_{n = 1}^{\infty}(\ln{x})^nx^n$?
I made typing errors. Sorry. The Power Seris is

$\sum_{n = 1}^{\infty}(\mathrm{ln}(n))^{n} x^{n}$

and

$a_{n} = (\mathrm{ln}(n))^{n}$

4. What is the neccesary condition for the convergence of an 'infinite sum'?...

Kind regards

$\chi$ $\sigma$

5. Originally Posted by Prove It
Is it $\sum_{n = 1}^{\infty}(\ln{x})^nx^n$?
But that would not be a power series.

6. Originally Posted by Besserwisser
Hello,

I need a little bit help with the following power series. The exercise is to calculate the radius of convergence.

$\sum\limits_{n=1}^{\infty} (\mathrm{ln}(n))^{n} x^{n}$

I know the criterion of Euler and the stronger criterion of Hadamard.

Euler:

$r= \frac{\mid a_{n} \mid}{ \mid a_{n+1} \mid }$

$r= \frac{1}{\mathrm{limsup} ~ \sqrt[n]{\mid a_{n} \mid} }$
and of course $a_{n}$ is
$a_{n} = (\mathrm{ln(n)})^{n}$
For it to converge you clearly need $\ln^n(n)\cdot x^n=\left(x\ln(n)\right)^n\to0$. For this to happen you need only prove that $x\ln(n)$ goes to zero for some $x$..but...there is a problem with that.