CalcS6

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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}}<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).

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