# CalcS6

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}.

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 $x\in\left[\frac{4}{5},4\right)$.

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