# What is the radius of convergence?

## What is the radius of convergence of ${\sum}_{n = 1}^{\in} \frac{\left({3}^{n + 2}\right) \left({x}^{n}\right)}{n + 1}$?

Apr 4, 2018

The radius of convergence is $R = \frac{1}{3}$

#### Explanation:

Apply the ratio test by evaluating:

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

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

${\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid = {\lim}_{n \to \infty} 3 \left\mid x \right\mid \frac{n + 1}{n + 2}$

${\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid = {\lim}_{n \to \infty} 3 \left\mid x \right\mid$

The series then is absolutely convergent for:

$3 \left\mid x \right\mid < 1$

$\left\mid x \right\mid < \frac{1}{3}$

therefore the radius of convergence is $R = \frac{1}{3}$.