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

Jan 26, 2018

The radius of convergence is $\left\mid x \right\mid < 1$.

#### Explanation:

We will use the ratio test for convergence. We have that:

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

The ratio test tells us that the sum convergence if:

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

In our case: ${a}_{n} = 2 {x}^{n}$

So:

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

$= {\lim}_{n \to \infty} \left\mid {x}^{n + 1 - 1} \right\mid = {\lim}_{n \to \infty} \left\mid x \right\mid = \left\mid x \right\mid$

So, by the ratio test, for convergence:

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