Power Series Question 7

What is the interval of convergence of the power series $\displaystyle \sum_{n=1}^\infty \frac{(x-3)^n}{n\cdot 2^n}$ ?

1\lt x\lt 5

1\leq x\lt 5

1\leq x\leq 5

2\leq x\leq 4

Approach

Perform the ratio test.

\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| \lt 1

\lim_{n\to\infty}\frac{(x-3)^{n+1}}{(n+1)\cdot 2^{n+1}}\cdot \frac{n\cdot 2^n}{(x-3)^n}

=\lim_{n\to\infty}\frac{n(x-3)}{2(n+1)}

\left| \frac{x-3}{2} \right|\lt 1

\left|x-3\right| \lt 2

1\lt x \lt 5

Testing the endpoints:

x=1

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

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

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

Which converges conditionally through the alternating series test.

x=5

\sum_{n=1}^\infty \frac{(5-3)^n}{n\cdot 2^n}

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

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

Which is a $p$ -series that diverges.