Question: What is the center of the interval of convergence for the power series :

 $\displaystyle\sum\limits_{n=0}^\infty \frac{(ln5)^{n}}{(n+2)^{\frac{1}{7}}}(6x-2)^{n}$

Just to clarify things, when the question is asking for the center of the interval of convergence, that essentially implies to find what the radius of convergence is? Usually the radius of convergence is referred as the center of the interval so thats why they call it that?

Attempt at solution:

 \displaystyle\frac{(ln5)^{n}}{(n+2)^{\frac{1}{7}}}  (6x-2)^{n} = a_{n}(x - x_{0})^{n}

 \displaystyle\frac{(ln5)^{n}}{(n+2)^{\frac{1}{7}}}  (6(x-\frac{1}{3})^{n} = \frac{ln(5)^{n}\cdot 6^{n}}{(n+2)^{\frac{1}{7}}}(x-\frac{1}{3})^{n} = \frac{6^{n}ln(5)^{n}}{(n+2)^{\frac{1}{7}}}(x-\frac{1}{3})^{n}

Where  a_{n} = \displaystyle\frac{6^{n}(ln5)^{n}}{(n+2)^{\frac{1}  {7}}} and  t_{0} = \frac{1}{3}

 L = \displaystyle\lim_{n\to\infty}|\frac{6^{n+1}(ln5)^  {n+1}}{((n+1)+2)^{\frac{1}{7}}}\cdot \frac{(n+2)^{\frac{1}{7}}}{6^{n}(ln5)^{n}}|  = \lim_{n\to\infty}|\frac{6(ln5)(n+2)^{\frac{1}{7}}}  {(n+3)^{\frac{1}{7}}}}|

 L = \displaystyle 6(ln5)\lim_{n\to\infty}|\frac{(n+2)^{\frac{1}{7}}}  {(n+3)^{\frac{1}{7}}}}|

 L = 6(ln5)

 R = \displaystyle\frac{1}{6(ln5)} since  L > 0 and  R = \frac{1}{L}

So, all my work is here, i don't know whats wrong with it. Someone please help me, all help is appreciated!