# What is the radius of convergence of sum_1^oo ((2x)^n ) / 8^n?

Mar 6, 2016

The radius of convergence $R$ of a power series of the form ${\sum}_{n = 0}^{\infty} {a}_{n} {x}^{n}$ is given by $R = \frac{1}{\lim {\text{sup}}_{n \to \infty} \sqrt[n]{\left\mid {a}_{n} \right\mid}} .$

In your case, $R = \frac{8}{2}$ and ${\sum}_{n = 1}^{\infty} \frac{{\left(2 x\right)}^{n}}{8} ^ n$ converges $\forall x \in \left(- \frac{8}{2} , \frac{8}{2}\right) .$

#### Explanation:

The series ${\sum}_{n = 1}^{\infty} \frac{{\left(2 x\right)}^{n}}{8} ^ n$ is already written in the wished form :

${\sum}_{n = 1}^{\infty} \frac{{\left(2 x\right)}^{n}}{8} ^ n = {\sum}_{n = 1}^{\infty} {\left(\frac{2}{8}\right)}^{n} {x}^{n} = {\sum}_{n = 1}^{\infty} {a}_{n} {x}^{n} ,$ ${a}_{n} = {\left(\frac{2}{8}\right)}^{n} .$

Let's compute $\lim {\text{sup}}_{n \to \infty} \sqrt[n]{\left\mid {a}_{n} \right\mid}$ :

$\sqrt[n]{\left\mid {a}_{n} \right\mid} = \sqrt[n]{\left\mid {\left(\frac{2}{8}\right)}^{n} \right\mid} = \sqrt[n]{{\left(\frac{2}{8}\right)}^{n}} = \frac{2}{8} \forall n \in \mathbb{N} .$

Therefore, $\lim {\text{sup}}_{n \to \infty} \sqrt[n]{\left\mid {a}_{n} \right\mid} = {\lim}_{n \to \infty} \sqrt[n]{\left\mid {a}_{n} \right\mid} = \frac{2}{8} \implies R = \frac{1}{\frac{2}{8}} = \frac{8}{2.}$

Thus, ${\sum}_{n = 1}^{\infty} \frac{{\left(2 x\right)}^{n}}{8} ^ n$ converges $\forall x \in \left(- \frac{8}{2} , \frac{8}{2}\right) .$

You should now check the endpoints of the interval.

• If $x = - \frac{8}{2}$ :

${\sum}_{n = 1}^{\infty} {\left(\frac{2}{8}\right)}^{n} {\left(- \frac{8}{2}\right)}^{n} = {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n}$ is not defined.

• If $x = \frac{8}{2}$ :

${\sum}_{n = 1}^{\infty} {\left(\frac{2}{8}\right)}^{n} {\left(\frac{8}{2}\right)}^{n} = {\sum}_{n = 1}^{\infty} 1$ diverges.

Thus, ${\sum}_{n = 1}^{\infty} \frac{{\left(2 x\right)}^{n}}{8} ^ n$ converges $\forall x \in \left(- \frac{8}{2} , \frac{8}{2}\right) .$