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Section 10.14 : Power Series

2. For the following power series determine the interval and radius of convergence.

\[\sum\limits_{n = 0}^\infty {\frac{{{n^{2n + 1}}}}{{{4^{3n}}}}{{\left( {2x + 17} \right)}^n}} \]

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Okay, let’s start off with the Root Test to get our hands on \(L\).

\[L = \mathop {\lim }\limits_{n \to \infty } {\left| {\frac{{{n^{2n + 1}}}}{{{4^{3n}}}}{{\left( {2x + 17} \right)}^n}} \right|^{\frac{1}{n}}} = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{n^{2 + \frac{1}{n}}}}}{{{4^3}}}\left( {2x + 17} \right)} \right| = \left| {2x + 17} \right|\mathop {\lim }\limits_{n \to \infty } \frac{{{n^{2 + \frac{1}{n}}}}}{{{4^3}}}\]

Okay, we can see that , in this case, \(L\) will be infinite provided \(x \ne - \frac{{17}}{2}\) and so the series will diverge for \(x \ne - \frac{{17}}{2}\). We also know that the power series will converge for \(x = -\frac{{17}}{2}\) (this is the value of \(a\) for this series!).

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Therefore, we know that the interval of convergence is \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = - \frac{{17}}{2}}}\) and the radius of convergence is \(\require{bbox} \bbox[2pt,border:1px solid black]{{R = 0}}\) .