# Radius of Convergence of Power Series

• Jan 26th 2010, 12:43 AM
Besserwisser
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.

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

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

Euler:

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

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

and of course $\displaystyle a_{n}$ is

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

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

Thanks for help
• Jan 26th 2010, 01:10 AM
Prove It
Quote:

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.

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

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

Euler:

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

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

and of course $\displaystyle a_{n}$ is

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

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

Thanks for help

Is it $\displaystyle \sum_{n = 1}^{\infty}(\ln{x})^nx^n$?
• Jan 26th 2010, 01:28 AM
Besserwisser
Quote:

Originally Posted by Prove It
Is it $\displaystyle \sum_{n = 1}^{\infty}(\ln{x})^nx^n$?

I made typing errors. Sorry. The Power Seris is

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

and

$\displaystyle a_{n} = (\mathrm{ln}(n))^{n}$
• Jan 27th 2010, 03:55 AM
chisigma
What is the neccesary condition for the convergence of an 'infinite sum'?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jan 27th 2010, 04:43 AM
HallsofIvy
Quote:

Originally Posted by Prove It
Is it $\displaystyle \sum_{n = 1}^{\infty}(\ln{x})^nx^n$?

But that would not be a power series.
• Jan 27th 2010, 09:14 AM
Drexel28
Quote:

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.

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

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

Euler:

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

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