Problem

**rednest** · November 25th 2008, 02:48 PM

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

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

Any suggestions will be appreciated

**repcvt** · November 26th 2008, 05:12 AM

Originally Posted by rednest

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

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

Any suggestions will be appreciated

$\Sigma n!x^n$ is biger

**Mathstud28** · November 26th 2008, 06:30 AM

Originally Posted by rednest

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

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

Any suggestions will be appreciated

$\sum_{n=0}^{\infty}n!x^n$

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

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

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

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