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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 $$\require{bbox} \bbox[2pt,border:1px solid black]{{R = 0}}$$ .