Radius of convergence query

Question

Find the radius of convergence of the series of $\frac{2^n(4z-8)^n}{n}$

My answer: $(4z-8)^n=4^n(z-2)^n=2^{2n}(z-2)^n$. Let $c_{n}=\frac{2^{3n}}{n}$. Then $\frac{c_{n}}{c_{n+1}}=\frac{n+1}{2n}$ so radius of convergence is 1/2. This disagrees with 'Pauls online notes, which says it is 1/8.

Answer 1

Your ratio is incorrect. You should have:

$$\frac{c_n}{c_{n+1}}=\frac{2^{3n}(n+1)}{2^{3(n+1)}n}=\frac{2^{3n}(n+1)}{2^{3n+3}n}=\frac{n+1}{2^3n}$$

Answer 2

By the root test we have

$$\left|2^n\frac{(4z-8)^n}n\right|^{1/n}\xrightarrow{n\to\infty}2|4z-8|<1\iff|z-2|<\frac18\iff z\in B\left(2,\frac18\right)$$ so the radius is $\frac18$.