Power Series Question 5

The radius of convergence for the power power-series $\displaystyle \sum_{n=1}^\infty \frac{(x-3)^{2n}}{n}$ is equal to 1. What is the interval of convergence?

-4 \leq x \lt -2

-1 \lt x \lt 1

2 \lt x \lt 4

2 \leq x \lt 4

Approach

The Taylor power-series is centered around $x=3$ , so the interval of convergence is $(2,4)$ given a radius of 1.

Checking the endpoints:

x=2

\sum_{n=1}^\infty \frac{(2-3)^{2n}}{n}

= \sum_{n=1}^\infty \frac{(-1)^{2n}}{n}

= \sum_{n=1}^\infty \frac{1}{n}

Diverges ( $p$ -power-series where $p\leq1$ ).

x=4

\sum_{n=1}^\infty \frac{(4-3)^{2n}}{n}

= \sum_{n=1}^\infty \frac{(1)^{2n}}{n}

= \sum_{n=1}^\infty \frac{1}{n}

Diverges ( $p$ -power-series where $p\leq1$ ).