# 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 $\sum _{{n=0}}^{\infty }{\frac {n(3x-4)^{n}}{{\sqrt[ {3}]{n^{4}}}(2x)^{{n-1}}}}=2x\sum _{{n=0}}^{\infty }\left({\frac {3x-4}{2x}}\right)^{n}n^{{-{\frac {1}{3}}}}$. 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 $x\in \left[{\frac {4}{5}},4\right)$.