Question

The radius of convergence for the power series ${\left(x-3\right)}^{\raisebox{1ex}{$2n$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}$ from $n=1$ to infinity is equal to $1.$

What is the interval of convergence?

Answer 1

Compute the interval of convergence:

The radius of convergence of the power series ${\left(x-3\right)}^{\raisebox{1ex}{$2n$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}$ is given as $1$.

The interval of convergence of a function say $\left|x-a\right|$is defined as $-R<\left|x-a\right|<R$ where $a$ is an integer and $R$ is the radius of convergence.

The given power series ${\left(x-3\right)}^{\raisebox{1ex}{$2n$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}$ will converge for ${\left|x-3\right|}^{\raisebox{1ex}{$2n$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}<1$.

This implies that ${\left|x-3\right|}^2<1$.

Take square roots on both sides to get $\left|x-3\right|<1$.

The modulus property for $|x-a|<b$ is equivalent to both of $x-a<b\mathrm{or}-\left(x-a\right)<b$.

Apply the modulus property on $\left|x-3\right|<1$.

$$\begin{gathered} x-3<1\mathrm{or}-\left(x-3\right)<1 \\ \Rightarrow x-3<1\mathrm{or}x-3>-1 \\ \therefore x<4\mathrm{or}x>2\left(\mathrm{Add}3\mathrm{to}\mathrm{both}\mathrm{sides}\right) \end{gathered}$$

Therefore, the interval of convergence is $2<x<4.$