# What is the convergence radius for the series sum_(r=0)^oo((x+1)/3)^r ?

Nov 20, 2016

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

#### Explanation:

You can write this series as

${\sum}_{r = 0}^{\infty} {\left(\frac{x + 1}{3}\right)}^{r}$ Calling now $y = \frac{x + 1}{3}$ we have equivalently

${\lim}_{n \to \infty} {\sum}_{r = 0}^{n} {y}^{r} = {\lim}_{n \to \infty} \frac{{y}^{n} - 1}{y - 1}$.

We are using the fact

$1 + y + {y}^{2} + \cdots + {y}^{n} = \frac{{y}^{n + 1} - 1}{y - 1}$

Now, if $\left\mid y \right\mid < 1$ we have ${\lim}_{n \to \infty} {y}^{n + 1} = 0$ so assuming $\left\mid y \right\mid < 1$

${\lim}_{n \to \infty} {\sum}_{r = 0}^{n} {y}^{r} = \frac{1}{1 - y} = \frac{1}{1 - \frac{x + 1}{3}} = \frac{3}{2 - x}$ this of course for $\left\mid x + 1 \right\mid < 3$