# Find the radius of convergence, R, of the series. \Sigma_{n = 1}^\infty \frac{n^2x^n}{6 \cdot...

## Question:

Find the radius of convergence, R, of the series.

{eq}\Sigma_{n = 1}^\infty \frac{n^2x^n}{6 \cdot 12 \cdot 18 \cdot \cdot \cdot \cdot \cdot (6n)} {/eq}

R = _____

Find the interval, I, of convergence of the series. (Enter your answer using interval notation.)

I = _____

## Ratio Test:

Given {eq}\displaystyle{\sum_{n=1}^{\infty}}{u_n} {/eq} a series with nonzero terms and {eq}G=\displaystyle{\lim_{n \to \infty }} \left|\frac{u_{n + 1}}{u_n}\right| {/eq}:

1) If {eq}G<1 {/eq}, the series is absolutely convergent ({eq}\Rightarrow {/eq} convergent).

2) If {eq}G>1 {/eq}, the series is divergent.

3) If {eq}G=1 {/eq}, no result from this test.

## Answer and Explanation:

Given that: {eq}\displaystyle \sum\limits_{n = 1}^\infty {\frac{{{n^2}{x^n}}}{{6\cdot12\cdot18\cdot\cdot\cdot\cdot\cdot(6n)}}} {/eq}

{eq}\displaystyle\ \eqalign{ & {\text{Let,}} \cr & {a_n} = \frac{{{n^2}{x^n}}}{{6\cdot12\cdot18\cdot\cdot\cdot\cdot\cdot(6n)}} \cr & \frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{{{\left( {n + 1} \right)}^2}{x^{n + 1}}}}{{6\cdot12\cdot18\cdot\cdot\cdot\cdot\cdot(6n)(6n + 6)}}\frac{{6\cdot12\cdot18\cdot\cdot\cdot\cdot\cdot(6n)}}{{{n^2}{x^n}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\left( {n + 1} \right)}^2}x}}{{(6n + 6)}}\frac{1}{{{n^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Cancel the common term}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{x}{{(6n + 6)}}\frac{{{n^2}{{\left( {1 + \frac{1}{n}} \right)}^2}}}{{{n^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Taking Common highest term}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{x}{{(6n + 6)}}{\left( {1 + \frac{1}{n}} \right)^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Cancel the common term}}} \right) \cr & \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{x}{{(6n + 6)}}{{\left( {1 + \frac{1}{n}} \right)}^2}} \right| \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left| x \right|\left| {\mathop {\lim }\limits_{n \to \infty } \frac{1}{{(6n + 6)}}{{\left( {1 + \frac{1}{n}} \right)}^2}} \right| \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left| x \right|\left| {(0){{\left( {1 + 0} \right)}^2}} \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} = 0} \right) \cr & \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = 0 \cr & \cr & {\text{The interval of convergence of the series is between }} - \infty {\text{ to }}\infty {\text{.}} \cr & \cr & {\text{The radius of convergence of the series is }}\infty {\text{.}} \cr & \cr & {\text{R = }}\infty \cr & \cr & I = \left( { - \infty ,\infty } \right) \cr} {/eq}

#### Learn more about this topic:

from AP Calculus BC: Exam Prep

Chapter 21 / Lesson 4