CalcS6

From Exampleproblems

Determine the interval of convergence for the power series $\sum_{n=0}^\infty\frac{n(3x-4)^n}{\sqrt[3]{n^4}(2x)^{n-1}}.$

Note $\sum_{n=0}^\infty\frac{n(3x-4)^n}{\sqrt[3]{n^4}(2x)^{n-1}}=2x\sum_{n=0}^\infty\left(\frac{3x-4}{2x}\right)^nn^{-\frac{1}{3}}$ . Use the Ratio Test to check for absolute convergence:

$\lim_{n\rightarrow\infty}\left\|\frac{a_{n+1}}{a_n}\right\|$	$=\lim_{n\rightarrow\infty}\left\|\frac{3x-4}{2x}\cdot\sqrt[3]{\frac{n}{n+1}}\right\|$
	$=\left\|\frac{3x-4}{2x}\right\|=\left\|\frac{3}{2}-\frac{2}{x}\right\|<1$

$\Rightarrow-1<\frac{3}{2}-\frac{2}{x}<1$	$\Rightarrow-\frac{5}{2}<-\frac{2}{x}<-\frac{1}{2}$
	$\Rightarrow\frac{1}{4}<\frac{1}{x}<\frac{5}{4}$
	$\Rightarrow\frac{4}{5}<x<4$ .

We need to check both endpoints for convergence as well. If $x=4\,$ , then the power series becomes $\sum_{n=0}^\infty\frac{8}{n^\frac{1}{3}}$ . Note this is a $p$ -series with $p < 1$ , implying divergence. If $x=\frac{4}{5}$ , the power series becomes $\sum_{n=0}^\infty\frac{8}{5}(-1)^n\frac{1}{n^\frac{1}{3}}$ . This is an alternating $p$ -series, implying conditional convergence. Thus, the interval of convergence is $x\in\left[\frac{4}{5},4\right)$ .

Back to Calculus

Back to Main Page

CalcS6

From Exampleproblems

Views

Personal tools

Example problems .com

Search

Toolbox