Question

determine the radius and interval of convergence of the given power series
∞
∑ nx^n/3^n
n
=
0

Answer 1

We should calculate the radius of convergence of the series

$\sum\limits_{n=0}^{\infty} \dfrac{nx^n}{3^n} = \sum\limits_{n=0}^{\infty} n \left(\dfrac{x}{3}\right)^n.$

We may use the D'Alembert's rule and calculate the radius of convergence as

$R = \lim\limits_{n\to\infty} \left| \dfrac{n \left(\dfrac{1}{3}\right)^n}{(n+1) \left(\dfrac{1}{3}\right)^{n+1}} \right| = 3.$

If $x=3,$ then $\sum\limits_{n=0}^{\infty} n\left(\dfrac{x}{3}\right)^n = \sum\limits_{n=0}^{\infty} n = +\infty.$

If $x=-3,$ then $\sum\limits_{n=0}^{\infty} n\left(\dfrac{x}{3}\right)^n = \sum\limits_{n=0}^{\infty} (-1)^nn,$ this series is not convergent, therefore the interval of convergence is (-3, 3).