8 Σ n=1 (x-4)-1 3 n...

Given expression is

$\sum _{n=1}^{\infty }\frac{{(x-4)}^{n-1}}{3^{n}n}$

Let 

$u_n=\frac{{(x-4)}^{n-1}}{3^{n}n}$

then

$u_{n+1}=\frac{{(x-4)}^n}{3^{n+1}(n+1)}$

Then

$$\begin{gathered} \underset{n\to \infty }{\lim }\frac{u_{n+1}}{u_n} \\ =\underset{n\to \infty }{\lim }\frac{(x-4)}{3^{n+1}(n+1)}\times \frac{3^{n}n}{{(x-4)}^{n-1}} \\ =\underset{n\to \infty }{\lim }\frac{n}{(n+1)}\times \left|\frac{(x-4)}{3}\right| \\ =\left|\frac{(x-4)}{3}\right| \end{gathered}$$

For convergence

$$\begin{gathered} \left|\frac{x-4}{3}\right|<1 \\ =\left|x-4\right|\mathbf{<}\mathbf{3} \left\{\because \left|x\right|<R, R \mathrm{is} \mathrm{radius} \mathrm{of} \mathrm{convergence}.\right\} \\ =-3\le x-4<3 \\ =-3+4\le x-4+4<3+4 \\ =\mathbf{1}\mathbf{\le }\mathit{x}\mathbf{<}\mathbf{7} \end{gathered}$$

So interval of convergence is $\mathbf{[}\mathbf{1}\mathbf{,}\mathbf{7}\mathbf{)}$

and radius of convergence $\mathbf{3}$.