Section 10.14 : Power Series
For each of the following power series determine the interval and radius of convergence.
- \( \displaystyle \sum\limits_{n = 0}^\infty {\frac{{{6^{1 - n}}}}{{{{\left( { - 2} \right)}^{3 - 2n}}}}{{\left( {x + 4} \right)}^n}} \)
- \( \displaystyle \sum\limits_{n = 1}^\infty {\frac{{{{\left( {10x - 1} \right)}^n}}}{{{n^{3 + n}}}}} \)
- \( \displaystyle \sum\limits_{n = 0}^\infty {\frac{{\left( {3n} \right)!}}{{\left( {2n - 2} \right)!}}{{\left( {6x - 9} \right)}^n}} \)
- \( \displaystyle \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{n^2}}}{{4n + 1}}{{\left( {5x + 20} \right)}^n}} \)
- \( \displaystyle \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}\,{8^{1 + n}}}}{{n + 4}}{{\left( {x - 7} \right)}^n}} \)
- \( \displaystyle \sum\limits_{n = 1}^\infty {\frac{{{2^{1 + 2n}}}}{{{{\left( { - 3} \right)}^{1 + 2n}}\,{n^2}}}{{\left( {4x + 2} \right)}^n}} \)
- \( \displaystyle \sum\limits_{n = 0}^\infty {\frac{{{{\left( {x + 12} \right)}^{2 + n}}}}{{{{\left( { - 16} \right)}^{5 + n}}}}} \)