# How do you find the radius of convergence of the power series Sigma 2^n n^3 x^n from n=[0,oo)?

Feb 20, 2017

The series:

${\sum}_{n = 0}^{\infty} {2}^{n} {n}^{3} {x}^{n}$

has radius of convergence $R = \frac{1}{2}$

#### Explanation:

Given the series:

${\sum}_{n = 0}^{\infty} {2}^{n} {n}^{3} {x}^{n}$

apply the ratio test to determine the radius of convergence.
Evaluate:

${\lim}_{n \to \infty} \left\mid \frac{{2}^{n + 1} {\left(n + 1\right)}^{3} {x}^{n + 1}}{{2}^{n} {n}^{3} {x}^{n}} \right\mid = {\lim}_{n \to \infty} 2 {\left(\frac{n + 1}{n}\right)}^{3} \left\mid x \right\mid = 2 \left\mid x \right\mid$

So the series is absolutely convergent for:

$2 \left\mid x \right\mid < 1$

that is:

$\left\mid x \right\mid < \frac{1}{2}$