1. ## Problem

Determine for which values of $\displaystyle x$ the following series converge and diverge.

$\displaystyle \Sigma n!x^n$ and $\displaystyle \Sigma \frac{(2x)^n}{n}$

Any suggestions will be appreciated

2. Originally Posted by rednest
Determine for which values of $\displaystyle x$ the following series converge and diverge.

$\displaystyle \Sigma n!x^n$ and $\displaystyle \Sigma \frac{(2x)^n}{n}$

Any suggestions will be appreciated
$\displaystyle \Sigma n!x^n$ is biger

3. Originally Posted by rednest
Determine for which values of $\displaystyle x$ the following series converge and diverge.

$\displaystyle \Sigma n!x^n$ and $\displaystyle \Sigma \frac{(2x)^n}{n}$

Any suggestions will be appreciated
$\displaystyle \sum_{n=0}^{\infty}n!x^n$

You can either apply the Root or Ratio test to find that the interval of covnergence is $\displaystyle \{0\}$, or you can use the fact that $\displaystyle \forall{x}\in\mathbb{R}-\{0\}~\lim_{n\to\infty}n!x^n=\infty$

$\displaystyle \sum_{n=1}^{\infty}\frac{2^nx^n}{n}$

Once again the Root, Ratio, or integral tests would work. Or you could note that $\displaystyle \sum\frac{2^n}{n}$ is divergent and the only thing that will make it converge in this case is a "dominant" geometric series provided by $\displaystyle x^n$, so $\displaystyle \lim_{n\to\infty}2^nx^n=0\implies{x\in\left(\frac{-1}{2},\frac{1}{2}\right)}$